Chapter 14: Equation of a Line

Exercise 14(A)

Question 1(a)

The point (m, -4) lies on the line x + y = 4, then the value of m is :

-6
6
-8
8

Answer

Given,

Point (m, -4) lies on the line x + y = 4.

Substituting value of point in equation, we get :

⇒ m + (-4) = 4

⇒ m - 4 = 4

⇒ m = 8.

Hence, Option 4 is the correct option.

Question 1(b)

The line kx - y = 9 passes through the point (6, 3), the value of k is :

2
-2
$\dfrac{1}{2}$
$-\dfrac{1}{2}$

Answer

Given,

Line kx - y = 9 passes through the point (6, 3).

Substituting value of point in equation, we get :

⇒ 6k - 3 = 9

⇒ 6k = 9 + 3

⇒ 6k = 12

⇒ k = $\dfrac{12}{6}$ = 2.

Hence, Option 1 is the correct option.

Question 1(c)

The line 3x - $\dfrac{y}{2}$ = 10 passes through the point (2, -4); is this statement true ?

no
yes
neither true nor false
none of the above

Answer

Substituting value of point in L.H.S. of the equation 3x - $\dfrac{y}{2}$ = 10, we get :

⇒ 3x - $\dfrac{y}{2}$ = 3 × 2 - $\dfrac{-4}{2}$

= 6 - (-2)

= 6 + 2

= 8.

Since,

8 ≠ 10.

∴ The statement, line 3x - $\dfrac{y}{2}$ = 10 passes through the point (2, -4) is false.

Hence, Option 1 is the correct option.

Question 1(d)

The point of intersection of the lines x + y = 8 and x - y = 0 lies on the line mx - 2y = 0; the value of m is :

3
4
2
-2

Answer

Given,

1st Equation :

⇒ x + y = 8

⇒ x = 8 - y ......(1)

2nd equation :

⇒ x - y = 0

⇒ x = y .........(2)

Substituting value of x from equation (2) in (1), we get :

⇒ y = 8 - y

⇒ y + y = 8

⇒ 2y = 8

⇒ y = $\dfrac{8}{2}$ = 4.

From equation (2), we get :

⇒ x = y = 4.

Point of intersection of lines x + y = 8 and x - y = 0 is (4, 4).

Given, point of intersection lies on the line mx - 2y = 0.

Substituting values we get :

⇒ 4m - 2 × 4 = 0

⇒ 4m - 8 = 0

⇒ 4m = 8

⇒ m = $\dfrac{8}{4}$ = 2.

Hence, Option 3 is the correct option.

Question 1(e)

The line x + y = 4 bisects the line segment joining the points (0, k) and (4, 0), the value of k is :

2
4
-4
-2

Answer

Given,

Points : (0, k) and (4, 0)

By formula,

Mid-point : $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get :

$\text{Mid-point } = \Big(\dfrac{0 + 4}{2}, \dfrac{k + 0}{2}\Big) \\[1em] = \Big(\dfrac{4}{2}, \dfrac{k}{2}\Big) \\[1em] = \Big(2, \dfrac{k}{2}\Big).$

Given,

Line x + y = 4 bisects the line segment joining the points (0, k) and (4, 0).

∴ Line x + y = 4 passes through the point $\Big(2, \dfrac{k}{2}\Big)$ .

Substituting value we get :

$\Rightarrow 2 + \dfrac{k}{2} = 4 \\[1em] \Rightarrow \dfrac{k}{2} = 4 - 2 \\[1em] \Rightarrow \dfrac{k}{2} = 2 \\[1em] \Rightarrow k = 4.$

Hence, Option 2 is the correct option.

Question 2

Find which of the following points lie on the line x - 2y + 5 = 0 :

(i) (1, 3)

(ii) (0, 5)

(iii) (-5, 0)

(iv) (5, 5)

Answer

(i) Substituting x = 1 and y = 3 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 1 - 2 × 3 + 5

= 1 - 6 + 5

= -5 + 5

= 0.

Since, L.H.S. = R.H.S.

∴ Point (1, 3) satisfies the equation.

Hence, (1, 3) lies on the line represented by the equation x - 2y + 5 = 0.

(ii) Substituting x = 0 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 0 - 2 × 5 + 5

= 0 - 10 + 5

= -5

Since, L.H.S. ≠ R.H.S.

∴ Point (0, 5) does not satisfies the equation.

Hence, (0, 5) does not lies on the line represented by the equation x - 2y + 5 = 0.

(iii) Substituting x = -5 and y = 0 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = -5 - 2 × 0 + 5

= -5 - 0 + 5

= -5 + 5

= 0.

Since, L.H.S. = R.H.S.

∴ Point (-5, 0) satisfies the equation.

Hence, (-5, 0) lies on the line represented by the equation x - 2y + 5 = 0.

(iv) Substituting x = 5 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 5 - 2 × 5 + 5

= 5 - 10 + 5

= 10 - 10

= 0.

Since, L.H.S. = R.H.S.

∴ Point (5, 5) satisfies the equation.

Hence, (5, 5) lies on the line represented by the equation x - 2y + 5 = 0.

Question 3(i)

State true or false :

the line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ passes through the point (2, 3).

Answer

Substituting x = 2 and y = 3 in the L.H.S. of the equation $\dfrac{x}{2} + \dfrac{y}{3} = 0$ , we get :

L.H.S. = $\dfrac{2}{2} + \dfrac{3}{3}$

= 1 + 1

= 2

Since, L.H.S. ≠ R.H.S.

∴ The line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ does not passes through the point (2, 3).

Hence, the statement is false.

Question 3(ii)

State true or false :

if the point (2, a) lies on the line 2x - y = 3, then a = 5.

Answer

Given,

(2, a) lies on the line 2x - y = 3.

∴ Substituting (2, a) in 2x - y = 3, satisfies the equation.

⇒ 2(2) - a = 3

⇒ 4 - a = 3

⇒ a = 4 - 3 = 1.

Hence, the statement is false.

Question 4

For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?

Answer

In order for the point (3, -k) to lie on the line 9x + 4y = 3, it must satisfy the equation.

On putting x = 3 and y = -k, we have

⇒ 9(3) + 4(-k) = 3

⇒ 27 – 4k = 3

⇒ 4k = 27 – 3 = 24

⇒ k = $\dfrac{24}{4}$

⇒ k = 6.

Hence, k = 6.

Question 5

The line $\dfrac{3x}{5} - \dfrac{2y}{3} + 1 = 0$ contains the point (m, 2m - 1); calculate the value of m.

Answer

Since, (m, 2m - 1) lies on the line $\dfrac{3x}{5} - \dfrac{2y}{3} + 1 = 0$ , it will satisfy the equation.

Substituting x = m, y = 2m - 1 in the equation $\dfrac{3x}{5} - \dfrac{2y}{3} + 1 = 0$ we get,

$\Rightarrow \dfrac{3m}{5} - \dfrac{2(2m - 1)}{3} + 1 = 0 \\[1em] \Rightarrow \dfrac{(3m \times 3) - [5 \times 2(2m - 1)] + 15}{15} = 0 \\[1em] \Rightarrow 9m - 10(2m - 1) + 15 = 0 \\[1em] \Rightarrow 9m - 20m + 10 + 15 = 0 \\[1em] \Rightarrow -11m = -25 \\[1em] \Rightarrow 11m = 25 \\[1em] \Rightarrow m = \dfrac{25}{11} = 2\dfrac{3}{11}.$

Hence, m = $2\dfrac{3}{11}.$

Question 6

Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?

Answer

By mid-point formula,

Mid-point of (5, -2) and (-1, 2) = $\dfrac{5 + (-1)}{2}, \dfrac{-2 + 2}{2}$

= $\dfrac{4}{2}, \dfrac{0}{2}$

= (2, 0).

Substituting x = 2 and y = 0 in L.H.S. of the equation 3x - 5y = 6.

L.H.S. = 3 × 2 - 5 × 0

= 6.

Since, L.H.S. = R.H.S.,

∴ (2, 0) lies on the line 3x - 5y = 6.

Hence, the line 3x – 5y = 6 bisects the join of (5, -2) and (-1, 2).

Question 7(i)

The line y = 3x - 2 bisects the join of (a, 3) and (2, -5), find the value of a.

Answer

By mid-point formula,

Mid-point of (a, 3) and (2, -5) = $\dfrac{a + 2}{2}, \dfrac{3 + (-5)}{2}$

= $\dfrac{a + 2}{2}, \dfrac{-2}{2}$

= $\Big(\dfrac{a + 2}{2}, -1\Big)$ .

Given, line y = 3x - 2 bisects the join of (a, 3) and (2, -5).

∴ $\Big(\dfrac{a + 2}{2}, -1\Big)$ satisfies the equation y = 3x - 2.

$\therefore -1 = 3 \times \dfrac{a + 2}{2} - 2 \\[1em] \Rightarrow -1 = \dfrac{3a + 6}{2} - 2 \\[1em] \Rightarrow -1 = \dfrac{3a + 6 - 4}{2} \\[1em] \Rightarrow -2 = 3a + 2 \\[1em] \Rightarrow 3a = -4 \\[1em] \Rightarrow a = -\dfrac{4}{3}.$

Hence, a = $-\dfrac{4}{3}$ .

Question 7(ii)

The line x - 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.

Answer

By mid-point formula,

Mid-point of (8, -1) and (0, k) = $\dfrac{8 + 0}{2}, \dfrac{-1 + k}{2}$

= $\dfrac{8}{2}, \dfrac{k - 1}{2}$

= $\Big(4, \dfrac{k - 1}{2}\Big)$ .

Given, line x - 6y + 11 bisects the join of (8, -1) and (0, k).

∴ $\Big(4, \dfrac{k - 1}{2}\Big)$ satisfies the equation x - 6y + 11 = 0.

$\therefore 4 - 6 \times \dfrac{k - 1}{2} + 11 = 0 \\[1em] \Rightarrow 4 - 3(k - 1) + 11 = 0 \\[1em] \Rightarrow 4 - 3k + 3 + 11 = 0 \\[1em] \Rightarrow 18 - 3k = 0 \\[1em] \Rightarrow 3k = 18 \\[1em] \Rightarrow k = \dfrac{18}{3} \\[1em] \Rightarrow k = 6.$

Hence, k = 6.

Question 8(i)

The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.

Answer

Since, (-3, 2) lies on the line ax + 3y + 6 = 0, so it satisfies the equation.

Substituting x = -3 and y = 2 in the equation ax + 3y + 6 = 0 we have,

⇒ -3a + 3(2) + 6 = 0

⇒ -3a + 6 + 6 = 0

⇒ -3a + 12 = 0

⇒ 3a = 12

⇒ a = $\dfrac{12}{3}$

⇒ a = 4.

Hence, a = 4.

Question 8(ii)

The line y = mx + 8 contains the point (-4, 4), calculate the value of m.

Answer

Since, (-4, 4) lies on the line y = mx + 8, so it satisfies the equation.

Substituting x = -4 and y = 4 in the equation y = mx + 8 we have,

⇒ 4 = -4m + 8

⇒ 4m = 8 - 4

⇒ 4m = 4

⇒ m = $\dfrac{4}{4}$

⇒ m = 1.

Hence, m = 1.

Question 9

The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x - 5y + 15 = 0 ?

Answer

By section-formula co-ordinates of,

P = $\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$P = \Big(\dfrac{2 \times -3 + 3 \times 2}{2 + 3}, \dfrac{2 \times 6 + 3 \times 1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{-6 + 6}{5}, \dfrac{12 + 3}{5}\Big) \\[1em] = \Big(\dfrac{0}{5}, \dfrac{15}{5}\Big) \\[1em] = (0, 3).$

If P will lie on the line x - 5y + 15 = 0, it will satisfy the equation.

Substituting x = 0 and y = 3 in L.H.S. of the equation x - 5y + 15 = 0.

L.H.S. = 0 - 5 × 3 + 15

= 0 - 15 + 15

= 0.

Since, L.H.S. = R.H.S.

∴ P lies on the line x - 5y + 15 = 0.

Hence, P = (0, 3) and it lies on the line x - 5y + 15 = 0.

Question 10

The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1 : 2. Does the line x - 2y = 0 contain Q ?

Answer

By section-formula co-ordinates of,

Q = $\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$Q = \Big(\dfrac{1 \times 2 + 2 \times 5}{1 + 2}, \dfrac{1 \times 2 + 2 \times -4}{1 + 2}\Big) \\[1em] = \Big(\dfrac{2 + 10}{3}, \dfrac{2 + (-8)}{3}\Big) \\[1em] = \Big(\dfrac{12}{3}, \dfrac{-6}{3}\Big) \\[1em] = (4, -2).$

If Q will lie on the line x - 2y = 0, it will satisfy the equation.

Substituting x = 4 and y = -2 in L.H.S. of the equation x - 2y = 0.

L.H.S. = 4 - 2 × (-2)

= 4 + 4

= 8.

Since, L.H.S. ≠ R.H.S.

∴ Q does not lies on the line x - 2y = 0.

Hence, Q = (4, -2) and it does not lie on the line x - 2y = 0.

Question 11

Find the point of intersection of the lines 4x + 3y = 1 and 3x - y + 9 = 0. If this point lies on the line (2k - 1)x - 2y = 4; find the value of k.

Answer

Solving,

4x + 3y = 1 .........(1)

3x - y + 9 = 0 ..........(2)

⇒ y = 3x + 9

Substituting value of y in equation 1 we get,

⇒ 4x + 3(3x + 9) = 1

⇒ 4x + 9x + 27 = 1

⇒ 13x = -26

⇒ x = $\dfrac{-26}{13}$

⇒ x = -2.

Substituting x = -2, in y = 3x + 9 we get,

⇒ y = 3(-2) + 9 = -6 + 9 = 3.

Point of intersection = (-2, 3).

Given, (-2, 3) lies on the line (2k - 1)x - 2y = 4

∴ (2k - 1)(-2) - 2 × 3 = 4

⇒ -4k + 2 - 6 = 4

⇒ -4k - 4 = 4

⇒ 4k = -8

⇒ k = $\dfrac{-8}{4}$

⇒ k = -2.

Hence, point of intersection = (-2, 3) and k = -2.

Question 12

Show that the lines 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Answer

When lines are concurrent they intersect at same point.

Solving, 2x + 5y = 1 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 6 + 3y

Substituting x = 3y + 6 in 2x + 5y = 1 we get,

⇒ 2(3y + 6) + 5y = 1

⇒ 6y + 12 + 5y = 1

⇒ 11y = -11

⇒ y = -1.

Substituting y = -1 in x = 6 + 3y we get,

x = 6 + 3(-1) = 6 - 3 = 3.

∴ 2x + 5y = 1 and x - 3y = 6 intersect in the point (3, -1).

Solving, x + 5y + 2 = 0 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 3y + 6

Substituting x = 3y + 6 in x + 5y + 2 = 0 we get,

⇒ 3y + 6 + 5y + 2 = 0

⇒ 8y + 8 = 0

⇒ 8y = -8

⇒ y = -1.

Substituting y = -1 in x = 3y + 6 we get,

x = 3(-1) + 6 = -3 + 6 = 3.

∴ x + 5y + 2 = 0 and x - 3y = 6 intersect in the point (3, -1).

Hence, proved that 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Exercise 14(B)

Question 1(a)

The slope of line is $\sqrt{3}$ , its inclination is :

30°
45°
60°
90°

Answer

Given,

⇒ Slope = $\sqrt{3}$

⇒ tan θ = $\sqrt{3}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, Option 3 is the correct option.

Question 1(b)

The slope of a line is 5, the slope of its perpendicular is :

5
-5
$\dfrac{1}{5}$
$-\dfrac{1}{5}$

Answer

We know that,

Product of slope of two perpendicular lines = -1.

Let slope of perpendicular line be m.

∴ 5 × m = -1

⇒ m = $-\dfrac{1}{5}$ .

Hence, Option 4 is the correct option.

Question 1(c)

The slope of the line passing through the origin and the point (-3, 4) is :

$\dfrac{4}{3}$
$-\dfrac{4}{3}$
$\dfrac{3}{4}$
$-\dfrac{3}{4}$

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of line passing through (0, 0) and (-3, 4)

= $\dfrac{4 - 0}{-3 - 0} = \dfrac{4}{-3} = -\dfrac{4}{3}$ .

Hence, Option 2 is the correct option.

Question 1(d)

The inclination of a line passing through the points (4, 3) and (5, 4) is :

1
-1
45°
60°

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of line passing through (4, 3) and (5, 4) = $\dfrac{4 - 3}{5 - 4} = \dfrac{1}{1}$ = 1.

By formula,

⇒ Slope = tan θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, Option 3 is the correct option.

Question 1(e)

The slope of the line which is perpendicular to the line segment joining the points (8, -5) and (-4, 7) is :

-1
1
45°
-45°

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of line passing through (8, -5) and (-4, 7)

= $\dfrac{7 - (-5)}{-4 - 8} = \dfrac{7 + 5}{-12} = \dfrac{12}{-12}$ = -1.

Let slope of line perpendicular to line segment joining points (8, -5) and (-4, 7) be m.

We know that,

Product of slopes of two perpendicular lines = -1.

∴ m × -1 = -1

⇒ m = $\dfrac{-1}{-1}$ = 1.

Hence, Option 2 is the correct option.

Question 2

Find the slope and the inclination of the line AB if :

(i) A = (-3, -2) and B = (1, 2)

(ii) A = (0, - $\sqrt{3}$ ) and B = (3, 0)

(iii) A = (-1, $2\sqrt{3}$ ) and B = (-2, $\sqrt{3}$ )

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

(i) A = (-3, -2) and B = (1, 2)

$\text{Slope of AB} = \dfrac{2 - (-2)}{1 - (-3)} \\[1em] = \dfrac{2 + 2}{1 + 3} \\[1em] = \dfrac{4}{4} = 1. \\[1em]$

Let inclination be θ,

∴ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, slope = 1 and inclination = 45°.

(ii) A = (0, - $\sqrt{3}$ ) and B = (3, 0)

$\text{Slope of AB} = \dfrac{0 - (-\sqrt{3})}{3 - 0} \\[1em] = \dfrac{\sqrt{3}}{3} \\[1em] = \dfrac{1}{\sqrt{3}}. \\[1em]$

Let inclination be θ,

∴ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, slope = $\dfrac{\sqrt{3}}{3}$ and inclination = 30°.

(iii) A = (-1, $2\sqrt{3}$ ) and B = (-2, $\sqrt{3}$ )

$\text{Slope of AB} = \dfrac{\sqrt{3} - 2\sqrt{3}}{-2 - (-1)} \\[1em] = \dfrac{-\sqrt{3}}{-2 + 1} \\[1em] = \dfrac{-\sqrt{3}}{-1} = \sqrt{3}. \\[1em]$

Let inclination be θ,

∴ tan θ = $\sqrt3{}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, slope = $\sqrt{3}$ and inclination = 60°.

Question 3

The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Let m₁ be the slope of line passing through (0, 2) and (-3, -1), and m₂ be the slope of line passing through (-1, 5) and (4, a).

Since, lines are parallel.

∴ m₁ = m₂

$\Rightarrow \dfrac{-1 - 2}{-3 - 0} = \dfrac{a - 5}{4 - (-1)} \\[1em] \Rightarrow \dfrac{-3}{-3} = \dfrac{a - 5}{5} \\[1em] \Rightarrow 1 = \dfrac{a - 5}{5} \\[1em] \Rightarrow a - 5 = 5 \\[1em] \Rightarrow a = 10.$

Hence, a = 10.

Question 4

The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Let m₁ be the slope of line passing through (-4, -2) and (2, -3), and m₂ be the slope of line passing through (a, 5) and (2, -1).

Since, lines are perpendicular.

∴ m₁.m₂ = -1

$\Rightarrow \dfrac{-3 - (-2)}{2 - (-4)} \times \dfrac{-1 - 5}{2 - a} = -1 \\[1em] \Rightarrow \dfrac{-3 + 2}{2 + 4} \times \dfrac{-6}{2 - a} = -1 \\[1em] \Rightarrow \dfrac{-1}{6} \times \dfrac{-6}{2 - a} = -1 \\[1em] \Rightarrow \dfrac{1}{2 - a} = -1 \\[1em] \Rightarrow -(2 - a) = 1 \\[1em] \Rightarrow a - 2 = 1 \\[1em] \Rightarrow a = 1 + 2 = 3.$

Hence, a = 3.

Question 5

Without using the distance formula, show that the points A(4, -2), B(-4, 4) and C(10, 6) are the vertices of a right-angled triangle.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Without using the distance formula, show that the points A(4, -2), B(-4, 4) and C(10, 6) are the vertices of a right-angled triangle. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

$\text{Slope of AB }(m_1) = \dfrac{4 - (-2)}{-4 - 4} \\[1em] = \dfrac{4 + 2}{-8} \\[1em] = -\dfrac{6}{8}. \\[1em] \text{Slope of AC }(m_2) = \dfrac{6 - (-2)}{10 - 4} \\[1em] = \dfrac{6 + 2}{6} \\[1em] = \dfrac{8}{6}. \\[1em] m_1 \times m_2 = -\dfrac{6}{8} \times \dfrac{8}{6} = -1.$

Since, m₁.m₂ = -1.

∴ AB ⊥ AC.

Hence, proved that ABC is a right-angled triangle at A.

Question 6

Without using the distance formula, show that the points A(4, 5), B(1, 2), C(4, 3) and D(7, 6) are the vertices of a parallelogram.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Without using the distance formula, show that the points A(4, 5), B(1, 2), C(4, 3) and D(7, 6) are the vertices of a parallelogram. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

$\Rightarrow \text{Slope of AB} = \dfrac{2 - 5}{1 - 4} \\[1em] = \dfrac{-3}{-3} \\[1em] = 1. \\[1em] \Rightarrow \text{Slope of CD} = \dfrac{6 - 3}{7 - 4} \\[1em] = \dfrac{3}{3} \\[1em] = 1. \\[1em] \Rightarrow \text{Slope of BC} = \dfrac{3 - 2}{4 - 1} \\[1em] = \dfrac{1}{3}. \\[1em] \Rightarrow \text{Slope of AD} = \dfrac{6 - 5}{7 - 4} \\[1em] = \dfrac{1}{3}.$

From above,

Slope of AB = Slope of CD and Slope of BC = Slope of AD.

∴ AB || CD and BC || AD.

Hence, proved ABCD is a parallelogram.

Question 7

(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Answer

Let the given points be A(-2, 4), B(4, 8), C(10, 7) and D(11, -5).

And, let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

By mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$ .

So,

$\Rightarrow P = \Big(\dfrac{-2 + 4}{2}, \dfrac{4 + 8}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{12}{2}\Big) = (1, 6). \\[1em] \Rightarrow Q = \Big(\dfrac{4 + 10}{2}, \dfrac{8 + 7}{2}\Big) \\[1em] = \Big(\dfrac{14}{2}, \dfrac{15}{2}\Big) = (7, 7.5). \\[1em] \Rightarrow R = \Big(\dfrac{10 + 11}{2}, \dfrac{7 + (-5)}{2}\Big) \\[1em] = \Big(\dfrac{21}{2}, \dfrac{2}{2}\Big) = (10.5, 1). \\[1em] \Rightarrow S = \Big(\dfrac{11 + (-2)}{2}, \dfrac{-5 + 4}{2}\Big) \\[1em] = (4.5, -0.5).$

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

$\Rightarrow \text{Slope of PQ} = \dfrac{7.5 - 6}{7 - 1} \\[1em] = \dfrac{1.5}{6} \\[1em] = \dfrac{1}{4}. \\[1em] \Rightarrow \text{Slope of QR} = \dfrac{1 - 7.5}{10.5 - 7} \\[1em] = \dfrac{-6.5}{3.5} \\[1em] = -\dfrac{65}{35} = -\dfrac{13}{7}. \\[1em] \Rightarrow \text{Slope of RS} = \dfrac{-0.5 - 1}{4.5 - 10.5} \\[1em] = \dfrac{-1.5}{-6} \\[1em] = \dfrac{1}{4}. \\[1em] \Rightarrow \text{Slope of PS} = \dfrac{-0.5 - 6}{4.5 - 1} \\[1em] = \dfrac{-6.5}{3.5} = -\dfrac{65}{35} \\[1em] = -\dfrac{13}{7}.$

From above calculation we get,

Slope of PQ = Slope of RS and Slope of QR = Slope of PS

∴ PQ || RS and QR || PS.

Hence, proved that the quadrilateral, obtained on joining the mid-points of sides of quadrilateral with vertices (-2, 4), (4, 8), (10, 7) and (11, -5), is a parallelogram.

Question 8

Show that the points P(a, b + c), Q(b, c + a) and R(c, a + b) are collinear.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

$\text{Slope of PQ} = \dfrac{c + a - (b + c)}{b - a} \\[1em] = \dfrac{c + a - b - c}{b - a} \\[1em] = \dfrac{a - b}{b - a} \\[1em] = \dfrac{-(b - a)}{b - a} = -1. \\[1em] \text{Slope of QR} = \dfrac{a + b - (c + a)}{c - b} \\[1em] = \dfrac{a + b - c - a}{c - b} \\[1em] = \dfrac{b - c}{c - b} \\[1em] = \dfrac{-(c - b)}{c - b} = -1.$

Since, Slope of PQ = QR.

Hence, proved that P, Q and R are collinear.

Question 9

Find x, if the slope of the line joining (x, 2) and (8, -11) is $-\dfrac{3}{4}$ .

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Given, slope of line joining (x, 2) and (8, -11) is $-\dfrac{3}{4}$ .

$\therefore -\dfrac{3}{4} = \dfrac{-11 - 2}{8 - x} \\[1em] \Rightarrow -3(8 - x) = 4 \times -13 \\[1em] \Rightarrow -24 + 3x = -52 \\[1em] \Rightarrow 3x = -52 + 24 \\[1em] \Rightarrow 3x = -28 \\[1em] \Rightarrow x = -\dfrac{28}{3}.$

Hence, x = $-\dfrac{28}{3}$ .

Question 10

A(5, 4), B(-3, -2) and C(1, -8) are the vertices of a triangle ABC. Find :

(i) the slope of the altitude of AB,

(ii) the slope of the median AD and

(iii) the slope of the line parallel to AC.

Answer

(i) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

$\text{Slope of AB }= \dfrac{-2 - 4}{-3 - 5} \\[1em] = \dfrac{-6}{-8} \\[1em] = \dfrac{3}{4}.$

A(5, 4), B(-3, -2) and C(1, -8) are the vertices of a triangle ABC. Find (i) the slope of the altitude of AB, (ii) the slope of the median AD and (iii) the slope of the line parallel to AC. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of altitude = -1

⇒ $\dfrac{3}{4}$ x Slope of altitude = -1

⇒ Slope of altitude = $-\dfrac{4}{3}$

Hence, slope of the altitude of AB = $-\dfrac{4}{3}$ .

(ii) Since, AD is median. So, D is the mid-point of BC.

D = $\Big(\dfrac{-3 + 1}{2}, \dfrac{-2 + (-8)}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{-10}{2}\Big) = (-1, -5).$

$\text{Slope of AD }= \dfrac{-5 - 4}{-1 - 5} \\[1em] = \dfrac{-9}{-6} \\[1em] = \dfrac{3}{2}.$

Hence, slope of the median AD = $\dfrac{3}{2}$ .

(iii) $\text{Slope of AC }= \dfrac{-8 - 4}{1 - 5} \\[1em] = \dfrac{-12}{-4} \\[1em] = 3.$

Since, slope of parallel lines are equal.

Hence, slope of line parallel to AC = 3.

Question 11

The slope of the side BC of a rectangle ABCD is $\dfrac{2}{3}$ . Find :

(i) the slope of the side AB,

(ii) the slope of the side AD.

Answer

Rectangle ABCD is shown in the figure below:

The slope of the side BC of a rectangle ABCD is 2/3. Find (i) the slope of the side AB, (ii) the slope of the side AD. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Since, product of slope of perpendicular line = -1.

Slope of AB x Slope of BC = -1

$\Rightarrow \text{Slope of AB} \times \dfrac{2}{3} = -1 \\[1em] \Rightarrow \text{Slope of AB} = -\dfrac{3}{2}.$

Hence, slope of AB = $-\dfrac{3}{2}$ .

(ii) AD is parallel to BC and slope of parallel lines are equal.

Hence, slope of AD = $\dfrac{2}{3}$ .

Question 12

The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.

Answer

Let points be A(K, 3), B(2, -4) and C(-K + 1, -2).

Since, points are collinear.

∴ Slope of AB = Slope of BC

$\Rightarrow \dfrac{-4 - 3}{2 - K} = \dfrac{-2 - (-4)}{-K + 1 - 2} \\[1em] \Rightarrow \dfrac{-7}{2 - K} = \dfrac{2}{-K - 1} \\[1em] \Rightarrow -7(-K - 1) = 2(2 - K) \\[1em] \Rightarrow 7K + 7 = 4 - 2K \\[1em] \Rightarrow 7K + 2K = 4 - 7 \\[1em] \Rightarrow 9K = -3 \\[1em] \Rightarrow K = -\dfrac{3}{9} = -\dfrac{1}{3}.$

Hence, K = $-\dfrac{1}{3}$ .

Question 13

Plot the points A(1, 1), B(4, 7) and C(4, 10) on a graph paper. Connect A and B, and also A and C.

Which segment appears to have steeper slope, AB or AC?

Justify your conclusion by calculating the slopes of AB and AC.

Answer

From graph,

Plot the points A(1, 1), B(4, 7) and C(4, 10) on a graph paper. Connect A and B, and also A and C. Which segment appears to have steeper slope, AB or AC? Justify your conclusion by calculating the slopes of AB and AC. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

AC appears to be steeper.

$\text{Slope of AB } = \dfrac{7 - 1}{4 - 1} \\[1em] = \dfrac{6}{3} = 2. \\[1em] \text{Slope of AC } = \dfrac{10 - 1}{4 - 1} \\[1em] = \dfrac{9}{3} = 3.$

Since, AC has greater slope.

Hence, proved AC has steeper slope.

Exercise 14(C)

Question 1(a)

The line passing through the points (-7, 4) and (5, 4) is parallel to :

x-axis
y-axis
x + y = 0
x - y = 0

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of line passing through (-7, 4) and (5, 4)

= $\dfrac{4 - 4}{5 - (-7)} = \dfrac{0}{5 + 7} = \dfrac{0}{12}$ = 0.

We know that,

Slope of parallel lines are equal.

We know that,

Slope of x-axis = 0.

∴ Slope of line passing through (-7, 4) and (5, 4) = Slope of x-axis.

Hence, Option 1 is the correct option.

Question 1(b)

The line intersecting the y-axis at point (0, 2) and making an angle of 45° with x-axis has equation :

x + y = 2
y - x = 2
x - y = 2
x + y = 0

Answer

Given,

The line intersects the y-axis at point (0, 2) and makes an angle of 45° with x-axis.

Slope (m) = tan θ = tan 45° = 1.

Equation of line passing through a point is given by :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = 1(x - 0)

⇒ y - 2 = x

⇒ y - x = 2.

Hence, Option 2 is the correct option.

Question 1(c)

The line 5x - 2y + 8 = 0, intersects the y-axis at point A, the co-ordinates of point A are :

(0, -4)
(4, 0)
(4, 4)
(0, 4)

Answer

We know that,

x co-ordinate at y-axis = 0.

Let co-ordinates of A are (0, b).

Since,

Line 5x - 2y + 8 = 0, intersects the y-axis at point A(0, b).

∴ Point A satisfies the equation 5x - 2y + 8 = 0.

Substituting values we get :

⇒ 5(0) - 2b + 8 = 0

⇒ 0 - 2b + 8 = 0

⇒ 2b = 8

⇒ b = $\dfrac{8}{2}$ = 4.

∴ A = (0, b) = (0, 4).

Hence, Option 4 is the correct option.

Question 1(d)

The equation of a line with x-intercept 7 and y-intercept -7 is :

x - y = 7
y - x = 7
x + y + 7 = 0
x + y = 7

Answer

By intercept form,

Equation of line : $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Given,

x-intercept (a) : 7

y-intercept (b) : -7

Substituting values in equation, we get :

$\Rightarrow \dfrac{x}{7} + \dfrac{y}{-7} = 1 \\[1em] \Rightarrow \dfrac{x}{7} - \dfrac{y}{7} = 1 \\[1em] \Rightarrow x - y = 7.$

Hence, Option 1 is the correct option.

Question 1(e)

The slope and the x-intercept of the line 5x - 5y = 12 are :

slope = 5 and x-intercept = $\dfrac{12}{5}$
slope = -1 and x-intercept = $\dfrac{12}{5}$
slope = 1 and x-intercept = $\dfrac{12}{5}$
slope = -5 and x-intercept = $\dfrac{12}{5}$

Answer

Given, equation :

⇒ 5x - 5y = 12

⇒ -5y = -5x + 12

⇒ 5y = 5x - 12

⇒ y = $\dfrac{5x}{5} - \dfrac{12}{5}$

⇒ y = x - $\dfrac{12}{5}$

Comparing above equation with y = mx + c (where m is slope and c is the y-intercept), we get :

⇒ m = 1 and c = $-\dfrac{12}{5}$

To find x-intercept of the line, we put y = 0 in above equation:

0 = x - $\dfrac{12}{5}$

⇒ x = $\dfrac{12}{5}$

∴ Slope = 1 and x-intercept = $\dfrac{12}{5}$

Hence, Option 3 is the correct option.

Question 2

Find the equation of a line whose :

y-intercept = -1 and inclination = 45°.

Answer

m = tan θ = tan 45° = 1.

Equation of a line : y = mx + c

Substituting values we get,

y = x - 1.

Hence, equation of line is y = x - 1.

Question 3

Find the equation of the line whose slope is $-\dfrac{4}{3}$ and which passes through (-3, 4).

Answer

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 4 = $-\dfrac{4}{3}$ [x - (-3)]

⇒ 3(y - 4) = -4(x + 3)

⇒ 3y - 12 = -4x - 12

⇒ 4x + 3y = -12 + 12

⇒ 4x + 3y = 0.

Hence, equation of line is 4x + 3y = 0.

Question 4

Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.

Answer

m = tan θ = tan 60° = $\sqrt{3}$

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 4 = $\sqrt{3}(x - 5)$

⇒ y - 4 = $\sqrt{3}x - 5\sqrt{3}$

⇒ y = $\sqrt{3}x + 4 - 5\sqrt{3}$ .

Hence, equation of line is y = $\sqrt{3}x + 4 - 5\sqrt{3}$ .

Question 5

Find the equation of a line passing through :

(i) (0, 1) and (1, 2)

(ii) (-1, -4) and (3, 0)

Answer

(i) (0, 1) and (1, 2)

Slope of the line (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$m = \dfrac{2 - 1}{1 - 0} = \dfrac{1}{1} = 1. \\[1em] \text{Equation } : y - y_1= m(x - x_1) \\[1em] \Rightarrow y - 1 = 1(x - 0) \\[1em] \Rightarrow y - 1 = x \\[1em] \Rightarrow y = x + 1.$

Hence, equation of line is y = x + 1.

(ii) (-1, -4) and (3, 0)

Slope of the line (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$m = \dfrac{0 - (-4)}{3 - (-1)} = \dfrac{0 + 4}{3 + 1} = \dfrac{4}{4} = 1. \\[1em] \text{Equation } : y - y_1= m(x - x_1) \\[1em] \Rightarrow y - (-4) = 1(x - (-1)) \\[1em] \Rightarrow y + 4 = x + 1 \\[1em] \Rightarrow y = x + 1 - 4 \\[1em] \Rightarrow y = x - 3.$

Hence, equation of line is y = x - 3.

Question 6

The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :

(i) the gradient of PQ;

(ii) the equation of PQ;

(iii) the co-ordinates of the point where PQ intersects the x-axis.

Answer

(i) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of PQ } = \dfrac{5 - 6}{-3 - 2} \\[1em] = \dfrac{-1}{-5} \\[1em] = \dfrac{1}{5}.$

Hence, gradient of slope PQ = $\dfrac{1}{5}$ .

(ii) By point slope form,

Equation : y - y₁ = m(x - x₁)

⇒ y - 6 = $\dfrac{1}{5}$ (x - 2)

⇒ 5(y - 6) = x - 2

⇒ 5y - 30 = x - 2

⇒ 5y = x - 2 + 30

⇒ 5y = x + 28

Hence, equation of PQ is 5y = x + 28.

(iii) The point where PQ intersects x-axis, there y co-ordinate = 0.

Substituting y = 0 in equation of PQ we get,

⇒ 5 × 0 = x + 28

⇒ 0 = x + 28

⇒ x = -28.

Point = (x, y) = (-28, 0).

Hence, co-ordinates of the point where PQ intersects the x-axis = (-28, 0).

Question 7

The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :

(i) the equation of AB;

(ii) the co-ordinates of the point where the line AB intersects the y-axis.

Answer

(i) $\text{Slope of AB } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-1 - 4}{2 - (-3)} \\[1em] = \dfrac{-5}{5} \\[1em] = -1.$

Equation : y - y₁ = m(x - x₁)

⇒ y - 4 = (-1)[x - (-3)]

⇒ y - 4 = -1(x + 3)

⇒ y - 4 = -x - 3

⇒ y + x = -3 + 4

⇒ x + y = 1.

Hence, equation of AB is x + y = 1.

(ii) The point where AB intersects y-axis, there x co-ordinate = 0.

Substituting x = 0 in equation of AB we get,

⇒ 0 + y = 1

⇒ y = 1.

Point = (x, y) = (0, 1).

Hence, co-ordinates of the point where AB intersects the y-axis = (0, 1).

Question 8

The figure given alongside shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.

Answer

From figure,

Slope of AB = tan 45° = 1.

Slope of CD = tan 60° = $\sqrt{3}$ .

By point-slope form,

$y - y_1 = m(x - x_1)$

Since, line AB passes through point P(3, 4) and slope = 1. Substituting values in point-slope form,

⇒ y - 4 = 1(x - 3)

⇒ y - 4 = x - 3

⇒ y - x = -3 + 4

⇒ y - x = 1

⇒ y = x + 1.

Since, line CD passes through point P(3, 4) and slope = $\sqrt{3}$ . Substituting values in point-slope form,

⇒ y - 4 = $\sqrt{3}$ (x - 3)

⇒ y - 4 = $\sqrt{3}x - 3\sqrt{3}$

⇒ y = $\sqrt{3}x - 3\sqrt{3} + 4$ .

Hence, equation of AB is y = x + 1 and equation of CD is y = $\sqrt{3}x - 3\sqrt{3} + 4$ .

Question 9

In △ABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.

Answer

Let AD be the median.

Since, AD is the median so D will be the mid-point of BC.

In △ABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Co-ordinates of D = $\Big(\dfrac{7 + 1}{2}, \dfrac{8 + (-10)}{2}\Big) = \Big(\dfrac{8}{2}, \dfrac{-2}{2}\Big)$ = (4, -1).

$\text{Slope of AD } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-1 - 5}{4 - 3} \\[1em] = \dfrac{-6}{1} \\[1em] = -6.$

By point-slope form,

$y - y_1 = m(x - x_1)$

Substituting values we get,

⇒ y - 5 = -6(x - 3)

⇒ y - 5 = -6x + 18

⇒ y + 6x = 18 + 5

⇒ 6x + y = 23.

Hence, equation of median through A is 6x + y = 23.

Question 10

The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex C = (7, 5). Find the equations of BC and CD.

The figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex C = (7, 5). Find the equations of BC and CD. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

Given, ∠A = 60° and vertex C = (7, 5)

As, ABCD is a parallelogram, we have

∠A + ∠B = 180° [Sum of adjacent angles in a || gm = 180°]

∠B = 180° – 60° = 120°

So, the anticlockwise angle of BC from x-axis is (180° - 120°) = 60°.

Slope of BC = tan 60° = $\sqrt{3}$

By point-slope form,

Equation of line BC is :

⇒ y – y₁ = m(x – x₁)

⇒ y – 5 = $\sqrt{3}$ (x – 7)

⇒ y – 5 = $\sqrt{3}x - 7\sqrt{3}$

⇒ y = $\sqrt{3}x - 7\sqrt{3} + 5$

As, CD || AB and AB || x-axis

Slope of CD = Slope of AB = 0 [As slope of x-axis is zero]

By point-slope form,

Equation of line CD is :

⇒ y – y₁ = m(x – x₁)

⇒ y – 5 = 0(x – 7)

⇒ y = 5.

Hence, equation of BC is y = $\sqrt{3}x - 7\sqrt{3} + 5$ and CD is y = 5.

Question 11

Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.

Answer

Solving x + 2y = 7 and x - y = 4 simultaneously,

⇒ x + 2y = 7

⇒ x = 7 - 2y .......(1)

Substituting above value of x in x - y = 4 we get,

⇒ 7 - 2y - y = 4

⇒ -3y = 4 - 7

⇒ -3y = -3

⇒ y = 1.

Substituting y = 1 in equation 1 we get,

⇒ x = 7 - 2(1) = 5.

Point of intersection = (5, 1).

Slope of line passing through (0, 0) and (5, 1) = $\dfrac{1 - 0}{5 - 0} = \dfrac{1}{5}.$

By point-slope form,

$y - y_1 = m(x - x_1)$

Substituting values we get,

⇒ y - 0 = $\dfrac{1}{5}(x - 0)$

⇒ 5(y - 0) = 1(x - 0)

⇒ 5y = x

⇒ x - 5y = 0.

Hence, equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4 is x - 5y = 0.

Question 12

In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.

Also, find the equation of the line through vertex B and parallel to AC.

Answer

Let AD be the median through A. So, D will be the mid-point of BC.

In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Co-ordinates of D = $\Big(\dfrac{-2 + 0}{2}, \dfrac{3 + 1}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{4}{2}\Big) = (-1, 2)$ .

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

$\text{Slope of AD} = \dfrac{2 - 7}{-1 - 4} \\[1em] = \dfrac{-5}{-5} = 1.$

By point-slope form,

Equation : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 7 = 1(x - 4)

⇒ y - 7 = x - 4

⇒ x - y - 4 + 7 = 0

⇒ x - y + 3 = 0.

$\text{Slope of AC} = \dfrac{1 - 7}{0 - 4} \\[1em] = \dfrac{-6}{-4} = \dfrac{3}{2}.$

Since, parallel lines have equal slope, equation of line passing through B and parallel to AC is

⇒ y - 3 = $\dfrac{3}{2}$ [x - (-2)]

⇒ 2(y - 3) = 3(x + 2)

⇒ 2y - 6 = 3x + 6

⇒ 3x - 2y + 12 = 0.

Hence, equation of median through A is x - y + 3 = 0 and equation of line passing through B and parallel to AC is 3x - 2y + 12 = 0.

Question 13

Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).

Answer

Let P = (-1, 2)

Let A and B be the points (1, 4) and (2, 3).

$\text{Slope of AB }= \dfrac{3 - 4}{2 - 1} \\[1em] = \dfrac{-1}{1} = -1.$

We know that,

Product of slope of perpendicular lines is -1.

Let slope of line through P be m.

∴ m × Slope of AB = -1

⇒ m × -1 = -1

⇒ -m = -1

⇒ m = 1.

By point-slope form,

Equation of line through P,

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = 1[x - (-1)]

⇒ y - 2 = 1(x + 1)

⇒ y - 2 = x + 1

⇒ y - x = 1 + 2

⇒ y - x = 3

⇒ y = x + 3

Hence, equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3) is y = x + 3.

Question 14

Find the equation of the line, whose :

(i) x-intercept = 5 and y-intercept = 3

(ii) x-intercept = -4 and y-intercept = 6

(iii) x-intercept = -8 and y-intercept = -4

Answer

(i) x-intercept = 5 and y-intercept = 3

When x-intercept = 5; corresponding point on the x-axis = (5, 0)

When y-intercept = 3; corresponding point on the y-axis = (0, 3).

$\text{Slope }= \dfrac{3 - 0}{0 - 5} = -\dfrac{3}{5}.$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{5}$ (x - 5)

⇒ 5y = -3(x - 5)

⇒ 5y = -3x + 15

⇒ 3x + 5y = 15.

Hence, equation of line is 3x + 5y = 15.

(ii) x-intercept = -4 and y-intercept = 6

When x-intercept = -4; corresponding point on the x-axis = (-4, 0)

When y-intercept = 6; corresponding point on the y-axis = (0, 6).

$\text{Slope }= \dfrac{6 - 0}{0 - (-4)} = \dfrac{6}{4}.$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{6}{4}$ [x - (-4)]

⇒ y = $\dfrac{3}{2}$ (x + 4)

⇒ 2y = 3x + 12

⇒ 2y - 3x = 12

⇒ 2y = 3x + 12

Hence, equation of line is 2y = 3x + 12.

(iii) x-intercept = -8 and y-intercept = -4

When x-intercept = -8; corresponding point on the x-axis = (-8, 0)

When y-intercept = -4; corresponding point on the y-axis = (0, -4).

$\text{Slope }= \dfrac{-4 - 0}{0 - (-8)} = \dfrac{-4}{8} = -\dfrac{1}{2}.$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{1}{2}$ [x - (-8)]

⇒ y = $-\dfrac{1}{2}$ (x + 8)

⇒ 2y = -x - 8

⇒ 2y + x + 8 = 0.

Hence, equation of line is x + 2y + 8 = 0.

Question 15

Find the equation of the line whose slope is $-\dfrac{5}{6}$ and x-intercept is 6.

Answer

When x-intercept = 6; corresponding point on the x-axis = (6, 0).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{5}{6}(x - 6)$

⇒ 6y = -5(x - 6)

⇒ 6y = -5x + 30

⇒ 5x + 6y = 30.

Hence, equation of line is 5x + 6y = 30.

Question 16

Find the equation of the line with x-intercept 5 and a point on it (-3, 2).

Answer

When x-intercept = 5; corresponding point on the x-axis = (5, 0).

Slope of the line through (5, 0) and (-3, 2) = $\dfrac{2 - 0}{-3 - 5} = \dfrac{2}{-8} = -\dfrac{1}{4}$ .

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{1}{4}(x - 5)$

⇒ 4y = -1(x - 5)

⇒ 4y = -x + 5

⇒ x + 4y = 5.

Hence, equation of line is x + 4y = 5.

Question 17

Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.

Answer

When y-intercept = 5; corresponding point on the y-axis = (0, 5).

Slope of the line through (0, 5) and (1, 3) = $\dfrac{3 - 5}{1 - 0} = \dfrac{-2}{1} = -2$ .

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = -2(x - 0)

⇒ y - 5 = -2x

⇒ 2x + y = 5.

Hence, equation of line is 2x + y = 5.

Question 18

Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.

Answer

Let there be two lines AB and CD equally inclined to co-ordinate axes and passing through E(-2, 0).

From figure,

Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

AB is inclined at an angle of 45°.

Slope = tan 45° = 1.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1[x - (-2)]

⇒ y = x + 2

⇒ x - y + 2 = 0.

From figure,

CD is inclined at an angle of -45° (As measured clockwise).

Slope = tan (-45°) = -1.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = -1[x - (-2)]

⇒ y = -[x + 2]

⇒ y = -x - 2

⇒ x + y + 2 = 0.

Hence, equation of lines are x - y + 2 = 0 and x + y + 2 = 0.

Question 19

The line through P(5, 3) intersects y-axis at Q.

(i) Write the slope of the line.

(ii) Write the equation of the line.

(iii) Find the co-ordinates of Q.

The line through P(5, 3) intersects y-axis at Q. Write the slope of the line. Write the equation of the line. Find the co-ordinates of Q. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

(i) From figure,

Inclination of the line = 45°

Slope = tan 45° = 1.

Hence, slope of the line = 1.

(ii) By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 5)

⇒ y - 3 = x - 5

⇒ x - y - 5 + 3 = 0

⇒ x - y - 2 = 0

⇒ x - y = 2

⇒ y = x - 2.

Hence, equation of the line is y = x - 2.

(iii) Let co-ordinates of point Q be (0, a).

Substituting values in equation we get,

a = 0 - 2

⇒ a = -2.

Hence, co-ordinates of point Q = (0, -2).

Question 20

Write down the equation of the line whose gradient is $-\dfrac{2}{5}$ and which passes through point P, where P divides the line segment joining A(4, -8) and B(12, 0) in the ratio 3 : 1.

Answer

By section formula,

Co-ordinates of P = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$P = \Big(\dfrac{3 \times 12 + 1 \times 4}{3 + 1}, \dfrac{3 \times 0 + 1 \times -8}{3 + 1}\Big) \\[1em] = \Big(\dfrac{36 + 4}{4}, \dfrac{0 - 8}{4}\Big) \\[1em] = \Big(\dfrac{40}{4}, \dfrac{-8}{4}\Big) \\[1em] = (10, -2).$

By point-slope form,

Equation of line having slope = $-\dfrac{2}{5}$ and passing through P,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $-\dfrac{2}{5}$ [x - 10]

⇒ 5(y + 2) = -2(x - 10)

⇒ 5y + 10 = -2x + 20

⇒ 5y + 2x = 20 - 10

⇒ 2x + 5y = 10.

Hence, 2x + 5y = 10.

Question 21

A(1, 4), B(3, 2) and C(7, 5) are vertices of a triangle ABC. Find :

(i) the co-ordinates of the centroid of triangle ABC.

(ii) the equation of a line, through the centroid and parallel to AB.

Answer

(i) Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get,

$\text{Centroid } = \Big(\dfrac{1 + 3 + 7}{3}, \dfrac{4 + 2 + 5}{3}\Big) \\[1em] = \Big(\dfrac{11}{3}, \dfrac{11}{3}\Big).$

Hence, centroid of triangle = $\Big(\dfrac{11}{3}, \dfrac{11}{3}\Big).$

(ii) $\text{Slope of AB} = \dfrac{2 - 4}{3 - 1} \\[1em] = \dfrac{-2}{2} = -1.$

Slope of line parallel to AB will also be equal to -1.

By point-slope form,

Equation of a line, through the centroid and parallel to AB,

⇒ y - y₁ = m(x - x₁)

$\Rightarrow y - \dfrac{11}{3} = -1\Big(x - \dfrac{11}{3}\Big) \\[1em] \Rightarrow \dfrac{3y - 11}{3} = -1 \times \dfrac{3x - 11}{3} \\[1em] \Rightarrow 3y - 11 = -1(3x - 11) \\[1em] \Rightarrow 3y - 11 = -3x + 11 \\[1em] \Rightarrow 3y + 3x = 11 + 11 \\[1em] \Rightarrow 3x + 3y = 22.$

Hence, the equation of a line, through the centroid and parallel to AB is 3x + 3y = 22.

Question 22

A(7, -1), B(4, 1) and C(-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP : CP = 2 : 3.

Answer

By section formula,

Co-ordinates of P = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$P = \Big(\dfrac{2 \times -3 + 3 \times 7}{2 + 3}, \dfrac{2 \times 4 + 3 \times -1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{-6 + 21}{5}, \dfrac{8 - 3}{5}\Big) \\[1em] = \Big(\dfrac{15}{5}, \dfrac{5}{5}\Big) \\[1em] = (3, 1).$

Slope of BP = $\dfrac{1 - 1}{3 - 4} = 0.$

By point-slope form,

Equation of BP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = 0(x - 3)

⇒ y - 1 = 0

⇒ y = 1.

Hence, equation of BP is y = 1.

Exercise 14(D)

Question 1(a)

The equation of a line with x-intercept 5 and y-intercept also 5 is :

2x - y = 5
x + y = 5
x - y = 5
x + 2y = 5

Answer

By intercept form,

Equation of line : $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Given,

x-intercept (a) : 5

y-intercept (b) : 5

Substituting values in equation, we get :

$\Rightarrow \dfrac{x}{5} + \dfrac{y}{5} = 1 \\[1em] \Rightarrow x + y = 5.$

Hence, Option 2 is the correct option.

Question 1(b)

The inclination of a line, which is not passing through origin, is 1. Its intercept with both the axes are :

equal
having same sign
equal in magnitude with opposite signs
not equal

Answer

Let AB be the line not passing through the origin and intersecting x-axis at D and y-axis at E.

We know that,

y-coordinate = 0 at x axis and x-coordinate = 0 at y-axis.

Let D = (a, 0) and E = (0, b).

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

From figure,

The inclination of a line, which is not passing through origin, is 1. Its intercept with both the axes are : Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Slope of DE = Slope of AB = 1

Substituting values we get :

$\Rightarrow \text{Slope of DE} = \dfrac{b - 0}{0 - a} \\[1em] \Rightarrow 1 = \dfrac{b}{-a} \\[1em] \Rightarrow b = -a.$

From figure,

a and b are intercepts on x-axis and y-axis respectively.

Hence, Option 3 is the correct option.

Question 1(c)

The line passing through point (1, 1) and making y-intercept equal to 3 has equation :

2x - y = 3
2x - y + 3 = 0
2x + y = 3
2y + x = 3

Answer

By slope-intercept form :

Equation of line : y = mx + c ......(1)

Given,

Line passes through point (1, 1) and makes a y-intercept of 3.

Substituting value in equation (1), we get :

⇒ 1 = m(1) + 3

⇒ 1 = m + 3

⇒ m = 1 - 3 = -2.

∴ Slope of line (m) = -2 and y-intercept (c) = 3

Substituting value of m and c in equation (1), we get :

⇒ y = -2x + 3

⇒ 2x + y = 3.

Hence, Option 3 is the correct option.

Question 1(d)

Equation of a line passing through the intersection of the lines x - y = 3 and x + y = 0 with inclination 45° is :

x + y = 3
x - y = 3
y - x = 3
y = 3x + 1

Answer

Given,

Equations :

⇒ x - y = 3 ........(1)

⇒ x + y = 0 ........(2)

Adding equation (1) and (2), we get :

⇒ (x - y) + (x + y) = 3 + 0

⇒ x + x - y + y = 3

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$

Substituting value of x in equation (2), we get :

⇒ $\dfrac{3}{2}$ + y = 0

⇒ y = $-\dfrac{3}{2}$ .

∴ Point of intersection of lines x - y = 3 and x + y = 0 is $\Big(\dfrac{3}{2}, -\dfrac{3}{2}\Big)$ .

Given,

Inclination (θ) = 45°

Slope = tan 45° = 1.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Equation of line passing through $\Big(\dfrac{3}{2}, -\dfrac{3}{2}\Big)$ and slope = 1 is :

$\Rightarrow y - \Big(-\dfrac{3}{2}\Big) = 1\Big(x - \dfrac{3}{2}\Big) \\[1em] \Rightarrow y + \dfrac{3}{2} = x - \dfrac{3}{2} \\[1em] \Rightarrow x - y = \dfrac{3}{2} + \dfrac{3}{2} \\[1em] \Rightarrow x - y = \dfrac{6}{2} \\[1em] \Rightarrow x - y = 3.$

Hence, Option 2 is the correct option.

Question 1(e)

A line cuts equal intercepts with both the axes and passes through the point (6, 6). The equation of the line is :

x - y = 12
x - y = 0
x + y = 6
x + y = 12

Answer

By intercept form :

Equation of line : $\dfrac{x}{a} + \dfrac{y}{b} = 1$ .........(1)

Given,

Line cuts equal intercepts with both the axes.

∴ x-intercept (a) = y-intercept (b) = p (let)

Given,

Line passes through point (6, 6). So it will satisfy the equation (1).

Substituting values in equation we get :

$\Rightarrow \dfrac{6}{p} + \dfrac{6}{p} = 1 \\[1em] \Rightarrow \dfrac{12}{p} = 1 \\[1em] \Rightarrow p = 12.$

∴ a = b = 12.

Substituting value of a and b in equation (1), we get :

$\Rightarrow \dfrac{x}{12} + \dfrac{y}{12} = 1 \\[1em] \Rightarrow \dfrac{x + y}{12} = 1 \\[1em] \Rightarrow x + y = 12.$

Hence, Option 4 is the correct option.

Question 2

Find the slope and y-intercept of the line :

(i) y = 4

(ii) ax - by = 0

(iii) 3x - 4y = 5

Answer

(i) Given,

y = 4

Comparing it with y = mx + c, we get,

m = 0 and c = 4.

Hence, slope = 0 and y-intercept = 4.

(ii) Given,

⇒ ax - by = 0

⇒ by = ax

⇒ y = $\dfrac{a}{b}$ x

Comparing it with y = mx + c, we get,

m = $\dfrac{a}{b}$ and c = 0.

Hence, slope = $\dfrac{a}{b}$ and y-intercept = 0.

(iii) Given,

⇒ 3x - 4y = 5

⇒ 4y = 3x - 5

⇒ y = $\dfrac{3}{4}x - \dfrac{5}{4}$

Comparing it with y = mx + c, we get,

m = $\dfrac{3}{4}$ and c = $-\dfrac{5}{4}$ .

Hence, slope = $\dfrac{3}{4}$ and y-intercept = $-\dfrac{5}{4}$ .

Question 3

The equation of a line is x - y = 4. Find its slope and y-intercept. Also, find its inclination.

Answer

Given,

⇒ x - y = 4

⇒ -y = -x + 4

⇒ y = x - 4

Comparing it with y = mx + c, we get,

m = 1 and c = -4.

⇒ m = tan θ

⇒ 1 = tan θ

⇒ tan 45° = tan θ

⇒ θ = 45°.

Hence, slope = 1, y-intercept = -4 and inclination = 45°.

Question 4(i)

Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x - 21y + 50 = 0?

Answer

Given lines,

⇒ 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0

⇒ 4y = -3x - 7 and 21y = 28x + 50

⇒ y = $-\dfrac{3}{4}x - \dfrac{7}{4}$ and y = $\dfrac{28}{21}x + \dfrac{50}{21}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $-\dfrac{3}{4}$

Slope of 2nd line = $\dfrac{28}{21} = \dfrac{4}{3}$

Since,

Slope of 1st line × Slope of 2nd line = $-\dfrac{3}{4} \times \dfrac{4}{3} = -1.$

Hence, the lines 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0 are perpendicular to each other.

Question 4(ii)

Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?

Answer

Given lines,

⇒ x - 3y = 4 and 3x - y = 7

⇒ 3y = x - 4 and y = 3x - 7

⇒ y = $\dfrac{1}{3}x - \dfrac{4}{3}$ and y = 3x - 7

Comparing above equations with y = mx + c we get,

Slope of 1st line = $\dfrac{1}{3}$

Slope of 2nd line = 3

Since,

Slope of 1st line × Slope of 2nd line = $\dfrac{1}{3} \times 3 = 1$ which is not equal to -1.

Hence, the lines x - 3y = 4 and 3x - y = 7 are not perpendicular to each other.

Question 4(iii)

Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?

Answer

Given lines,

⇒ 3x + 2y = 5 and x + 2y = 1

⇒ 2y = -3x + 5 and 2y = -x + 1

⇒ y = $-\dfrac{3}{2}x + \dfrac{5}{2}$ and y = $-\dfrac{1}{2}x + \dfrac{1}{2}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $-\dfrac{3}{2}$

Slope of 2nd line = $-\dfrac{1}{2}$

Since,

Slope of 1st line ≠ Slope of 2nd line

Hence, the lines 3x + 2y = 5 and x + 2y = 1 are not parallel to each other.

Question 4(iv)

Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.

Answer

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow \text{Slope } = \dfrac{2 - 4}{x - 1} \\[1em] \Rightarrow 2 = \dfrac{-2}{x - 1} \\[1em] \Rightarrow 2(x - 1) = -2 \\[1em] \Rightarrow x - 1 = -1 \\[1em] \Rightarrow x = -1 + 1 = 0. \\[1em]$

Hence, x = 0.

Question 5(i)

Find the slope of the line which is parallel to x + 2y + 3 = 0.

Answer

Given,

⇒ x + 2y + 3 = 0

⇒ 2y = -x - 3

⇒ y = $-\dfrac{1}{2}x - \dfrac{3}{2}$

Comparing above equation with y = mx + c we get,

Slope = $-\dfrac{1}{2}$

Since, slope of parallel lines are equal.

Hence, slope of line parallel to x + 2y + 3 = 0 is $-\dfrac{1}{2}$ .

Question 5(ii)

Find the slope of the line which is parallel to $\dfrac{x}{2} - \dfrac{y}{3} - 1 = 0$

Answer

Given,

$\Rightarrow \dfrac{x}{2} - \dfrac{y}{3} - 1 = 0 \\[1em] \Rightarrow \dfrac{y}{3} = \dfrac{x}{2} - 1 \\[1em] \Rightarrow y = \dfrac{3}{2}x - 3.$

Comparing above equation with y = mx + c we get,

Slope = $\dfrac{3}{2}$

Since, slope of parallel lines are equal.

Hence, slope of line parallel to $\dfrac{x}{2} - \dfrac{y}{3} - 1 = 0$ is $\dfrac{3}{2}$ .

Question 6(i)

Find the slope of the line which is perpendicular to $x - \dfrac{y}{2} + 3 = 0$ .

Answer

Given,

$\Rightarrow x - \dfrac{y}{2} + 3 = 0 \\[1em] \Rightarrow \dfrac{y}{2} = x + 3 \\[1em] \Rightarrow y = 2x + 6.$

Comparing above equation with y = mx + c we get,

Slope = 2.

Let slope of perpendicular line be m.

Since, product of slopes of perpendicular lines = -1.

∴ m × 2 = -1

⇒ m = $-\dfrac{1}{2}$ .

Hence, slope of the line perpendicular to the line $x - \dfrac{y}{2} + 3 = 0 \text{ is } -\dfrac{1}{2}.$

Question 6(ii)

Find the slope of the line which is perpendicular to $\dfrac{x}{3} - 2y = 4$ .

Answer

Given,

$\Rightarrow \dfrac{x}{3} - 2y = 4 \\[1em] \Rightarrow 2y = \dfrac{x}{3} - 4 \\[1em] \Rightarrow y = \dfrac{1}{6}x - 2.$

Comparing above equation with y = mx + c we get,

Slope = $\dfrac{1}{6}$ .

Let slope of perpendicular line be m.

Since, product of slopes of perpendicular lines = -1.

∴ m × $\dfrac{1}{6}$ = -1

⇒ m = -6.

Hence, slope of the line perpendicular to the line $\dfrac{x}{3} - 2y = 4$ is -6.

Question 7(i)

Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.

Answer

Given lines,

⇒ 2x – by + 5 = 0 and ax + 3y = 2

⇒ by = 2x + 5 and 3y = -ax + 2

⇒ y = $\dfrac{2}{b}x + \dfrac{5}{2}$ and $y = -\dfrac{a}{3}x + \dfrac{2}{3}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $\dfrac{2}{b}$

Slope of 2nd line = $-\dfrac{a}{3}$

Since,

Slopes of parallel lines are equal.

$\therefore \dfrac{2}{b} = -\dfrac{a}{3} \\[1em] \Rightarrow 6 = -ab \\[1em] \Rightarrow ab = -6.$

Hence, relation connecting a and b is ab = -6.

Question 7(ii)

Lines mx + 3y = -7 and 5x - ny = 3 are perpendicular to each other. Find the relation connecting m and n.

Answer

Given lines,

⇒ mx + 3y = -7 and 5x - ny = 3

⇒ 3y = mx + 7 and ny = 5x - 3

⇒ y = $\dfrac{m}{3}x + \dfrac{7}{3}$ and $y = \dfrac{5}{n}x - \dfrac{3}{n}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $\dfrac{m}{3}$

Slope of 2nd line = $\dfrac{5}{n}$

Since, product of slopes of perpendicular lines = -1.

$\therefore \dfrac{m}{3} \times \dfrac{5}{n} = -1 \\[1em] \Rightarrow \dfrac{5m}{3n} = -1 \\[1em] \Rightarrow 5m = -3n \\[1em] \Rightarrow 5m + 3n = 0.$

Hence, relation connecting m and n is 5m + 3n = 0.

Question 8

Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

Answer

Given lines,

⇒ 2x - y + 5 = 0 and px + 3y = 4

⇒ y = 2x + 5 and 3y = -px + 4

⇒ y = 2x + 5 and y = $-\dfrac{p}{3}x + \dfrac{4}{3}$ .

Comparing above equations with y = mx + c we get,

Slope of 1st line = 2

Slope of 2nd line = $-\dfrac{p}{3}$

Since, product of slopes of perpendicular lines = -1.

$\therefore 2 \times -\dfrac{p}{3} = -1 \\[1em] \Rightarrow -p = \dfrac{-1 \times 3}{2} \\[1em] \Rightarrow -p = -\dfrac{3}{2} \\[1em] \Rightarrow p = \dfrac{3}{2}.$

Hence, p = $\dfrac{3}{2}$ .

Question 9

The equation of a line AB is 2x - 2y + 3 = 0.

(i) Find the slope of the line AB.

(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.

Answer

(i) Given,

⇒ 2x - 2y + 3 = 0

⇒ 2y = 2x + 3

⇒ y = $x + \dfrac{3}{2}$ .

Comparing above equation with y = mx + c we get,

Slope = 1.

Hence, slope of line AB = 1.

(ii) Let angle be θ.

By formula,

⇒ Slope = tan θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, angle that line makes with positive direction of x-axis = 45°.

Question 10

The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.

Answer

Given lines,

⇒ 4x + 3y = 9 and px - 6y + 3 = 0

⇒ 3y = -4x + 9 and 6y = px + 3

⇒ y = $-\dfrac{4}{3}x + 3$ and $y = \dfrac{p}{6}x + \dfrac{1}{2}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $-\dfrac{4}{3}$

Slope of 2nd line = $\dfrac{p}{6}$

Since,

Slopes of parallel lines are equal.

$\therefore -\dfrac{4}{3} = \dfrac{p}{6} \\[1em] \Rightarrow p = \dfrac{-4 \times 6}{3} \\[1em] \Rightarrow p = -8.$

Hence, p = -8.

Question 11

The line through A(-2, 3) and B(4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.

Answer

Given,

⇒ 2x - 4y = 5

⇒ 4y = 2x - 5

⇒ y = $\dfrac{2}{4}x - \dfrac{5}{4}$

⇒ y = $\dfrac{1}{2}x - \dfrac{5}{4}$

Comparing above equation with y = mx + c we get,

Slope (m₁) = $\dfrac{1}{2}$

$\text{Slope of AB }(m_2) = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{b - 3}{4 - (-2)} \\[1em] = \dfrac{b - 3}{6}.$

Since,

Product of slopes of perpendicular lines = -1.

$\therefore m_1 \times m_2 = -1 \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{b - 3}{6} = -1 \\[1em] \Rightarrow \dfrac{b - 3}{12} = -1 \\[1em] \Rightarrow b - 3 = -12 \\[1em] \Rightarrow b = -12 + 3 = -9.$

Hence, b = -9.

Question 12

Find the equation of the line passing through (-5, 7) and parallel to :

(i) x-axis

(ii) y-axis

Answer

(i) We know,

Slope of line parallel to x-axis = 0.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = 0[x - (-5)]

⇒ y - 7 = 0

⇒ y = 7.

Hence, equation of the line parallel to x-axis is y = 7.

(ii) We know that,

Eqiuation of line parallel to y-axis is given by,

⇒ x - x₁ = 0

⇒ x - (-5) = 0

⇒ x + 5 = 0.

Hence, equation of the line parallel to y-axis is x + 5 = 0.

Question 13

Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

Answer

Given,

⇒ x - 3y = 4

⇒ 3y = x - 4

⇒ y = $\dfrac{1}{3}x - \dfrac{4}{3}$

Comparing above equations with y = mx + c we get,

Slope = $\dfrac{1}{3}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to x - 3y = 4 is $\dfrac{1}{3}$ .

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $\dfrac{1}{3}(x - 5)$

⇒ 3(y + 3) = x - 5

⇒ 3y + 9 = x - 5

⇒ x - 3y - 5 - 9 = 0

⇒ x - 3y - 14 = 0.

Hence, equation of the line passing through (5, -3) and parallel to x - 3y = 4 is x - 3y - 14 = 0.

Find the equation of the line passing through (-5, 7) and parallel to :

(i) x-axis

(ii) y-axis

Answer

(i) We know,

Slope of line parallel to x-axis = 0.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = 0[x - (-5)]

⇒ y - 7 = 0

⇒ y = 7.

Hence, equation of the line parallel to x-axis is y = 7.

(ii) We know that,

Equation of line parallel to y-axis is given by,

⇒ x - x₁ = 0

⇒ x - (-5) = 0

⇒ x + 5 = 0.

Hence, equation of the line parallel to y-axis is x + 5 = 0.

Question 14

Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Answer

Given,

⇒ 4x + 5y = 6

⇒ 5y = -4x + 6

⇒ y = $-\dfrac{4}{5}x + \dfrac{6}{5}$

Comparing above equations with y = mx + c we get,

Slope (m₁) = $-\dfrac{4}{5}$

Let slope of line perpendicular to 4x + 5y = 6 be m₂.

Since, product of slopes of perpendicular lines be -1,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{4}{5} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -1 \times -\dfrac{5}{4} = \dfrac{5}{4}. \\[1em]$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $\dfrac{5}{4}$ [x - (-2)]

⇒ 4(y - 1) = 5(x + 2)

⇒ 4y - 4 = 5x + 10

⇒ 5x - 4y + 10 + 4 = 0

⇒ 5x - 4y + 14 = 0.

Hence, equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6 is 5x - 4y + 14 = 0.

Question 15

Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).

Answer

Let points be A(6, -3) and B(0, 3) and D be the mid-point of AB.

Co-ordinates of D = $\Big(\dfrac{6 + 0}{2}, \dfrac{-3 + 3}{2}\Big) = \Big(\dfrac{6}{2}, \dfrac{0}{2}\Big)$ = (3, 0).

Slope of AB = $\dfrac{3 - (-3)}{0 - 6} = \dfrac{3 + 3}{-6} = \dfrac{6}{-6}$ = -1.

Let slope of perpendicular bisector be m.

Then

⇒ m × Slope of AB = -1

⇒ m × -1 = -1

⇒ m = 1.

Perpendicular bisector of AB will pass through mid-point of AB i.e. D.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1(x - 3)

⇒ y = x - 3.

Hence, equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3) is y = x - 3.

Question 16

B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equations of diagonals BD and AC.

Answer

The rhombus ABCD is shown in the figure below:

B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equations of diagonals BD and AC. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of BD } = \dfrac{4 - 6}{1 - (-5)} \\[1em] = \dfrac{-2}{6} \\[1em] = -\dfrac{1}{3}.$

Equation of BD by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\dfrac{1}{3}$ [x - (-5)]

⇒ 3(y - 6) = -1(x + 5)

⇒ 3y - 18 = -x - 5

⇒ 3y + x - 18 + 5 = 0

⇒ x + 3y - 13 = 0

⇒ x + 3y = 13.

Since, diagonals of rhombus are perpendicular to each other.

So, product of their slopes will be -1.

Slope of AC × Slope of BD = -1

$\text{Slope of AC} \times -\dfrac{1}{3}$ = -1

Slope of AC = -1 × -3 = 3.

We know that,

Diagonals of rhombus bisect each other. Let diagonals meet at point O.

∴ Mid-point of AC = Mid-point of BD.

Co-ordinates of mid-point of BD (i.e. O)

= $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) = \Big(\dfrac{-5 + 1}{2}, \dfrac{6 + 4}{2}\Big) = \Big(\dfrac{-4}{2}, \dfrac{10}{2}\Big)$ = (-2, 5).

∴ Co-ordinates of mid-point of AC = (-2, 5).

Equation of AC by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = 3[x - (-2)]

⇒ y - 5 = 3[x + 2]

⇒ y - 5 = 3x + 6

⇒ 3x - y + 6 + 5 = 0

⇒ 3x - y + 11 = 0

⇒ y = 3x + 11.

Hence, equation of BD is x + 3y = 13 and equation of AC is y = 3x + 11.

Question 17

A(1, -5), B(2, 2) and C(-2, 4) are the vertices of triangle ABC. Find the equation of :

(i) the median of the triangle through A.

(ii) the altitude of the triangle through B.

(iii) the line through C and parallel to AB.

Answer

(i) Let AD be the median through A.

So, D will be the mid-point of BC.

Co-ordinates of D = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) = \Big(\dfrac{2 + (-2)}{2}, \dfrac{2 + 4}{2}\Big) = \Big(\dfrac{0}{2}, \dfrac{6}{2}\Big)$ = (0, 3).

A(1, -5), B(2, 2) and C(-2, 4) are the vertices of triangle ABC. Find the equation of (i) the median of the triangle through A. (ii) the altitude of the triangle through B. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

$\text{Slope of AD } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{3 - (-5)}{0 - 1} \\[1em] = -\dfrac{8}{1} \\[1em] = -8.$

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = -8(x - 1)

⇒ y + 5 = -8x + 8

⇒ 8x + y = 3.

Hence, equation of median through A is 8x + y = 3.

(ii) Let BF be the altitude.

$\text{Slope of AC } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{4 - (-5)}{-2 - 1} \\[1em] = \dfrac{9}{-3} \\[1em] = -3.$

Since, altitude is at 90°.

So, altitude through B (i.e. BF) will be perpendicular to AC.

Let slope of altitude through B be m.

∴ m × -3 = -1

⇒ m = $\dfrac{1}{3}$ .

Equation of altitude through B by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\dfrac{1}{3}$ (x - 2)

⇒ 3(y - 2) = x - 2

⇒ 3y - 6 = x - 2

⇒ x - 3y - 2 + 6 = 0

⇒ x - 3y + 4 = 0.

Hence, equation of altitude through B is x - 3y + 4 = 0.

(iii) $\text{Slope of AB } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{2 - (-5)}{2 - 1} \\[1em] = \dfrac{7}{1} \\[1em] = 7.$

Since, parallel lines have equal slopes.

So slope of line through C and parallel to AB = 7.

Equation of line through C by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = 7[x - (-2)]

⇒ y - 4 = 7(x + 2)

⇒ y - 4 = 7x + 14

⇒ 7x - y + 14 + 4 = 0

⇒ 7x - y + 18 = 0.

Hence, equation of line through C and parallel to AB is 7x - y + 18 = 0.

Question 18

(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.

(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.

Answer

(i) Below figure shows the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5:

Given,

⇒ 2y = 3x + 5

⇒ y = $\dfrac{3}{2}x + \dfrac{5}{2}$

Comparing above equation with y = mx + c we get,

m = $\dfrac{3}{2}$

Let slope of line AB be m₁.

Since, AB and line 2y = 3x + 5 are perpendicular.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

$\dfrac{3}{2} \times m_1 = -1$

$m_1 = -\dfrac{2}{3}$ .

By point-slope form,

Equation of AB : y - y₁ = m(x - x₁)

⇒ y - 2 = $-\dfrac{2}{3}$ (x - 3)

⇒ 3(y - 2) = -2(x - 3)

⇒ 3y - 6 = -2x + 6

⇒ 2x + 3y = 6 + 6

⇒ 2x + 3y = 12.

Hence, equation of AB is 2x + 3y = 12.

(ii) Below figure shows AB with its intercepts on x axis and y axis:

AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

At A,

y co-ordinate = 0 as it lies on x-axis.

Substituting y = 0 in equation of AB,

⇒ 2x + 3(0) = 12

⇒ 2x = 12

⇒ x = 6.

A = (x, 0) = (6, 0).

At B,

x co-ordinate = 0 as it lies on y-axis.

Substituting x = 0 in equation of AB,

⇒ 2(0) + 3y = 12

⇒ 3y = 12

⇒ y = 4.

B = (0, y) = (0, 4).

Area of right angle triangle OAB = $\dfrac{1}{2} \times OA \times OB$

= $\dfrac{1}{2} \times 6 \times 4$

= 12 sq. units.

Hence, A = (6, 0), B = (0, 4) and area of triangle OAB = 12 sq. units.

Question 19

The line 4x - 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.

Determine the equation of line through A and perpendicular to 4x - 3y + 12 = 0.

Answer

At A,

y co-ordinate = 0 as it lies on x-axis.

A = (x, 0)

Substituting y = 0 in equation 4x - 3y + 12 = 0 we get,

⇒ 4x - 3(0) + 12 = 0

⇒ 4x + 12 = 0

⇒ 4x = -12

⇒ x = -3.

A = (x, 0) = (-3, 0).

Given,

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x$ + 4

Comparing above equation with y = mx + c we get,

m = $\dfrac{4}{3}$ .

Let slope of line perpendicular to 4x - 3y + 12 = 0 be m₁.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

⇒ $\dfrac{4}{3} \times m_1 = -1$

⇒ $m_1 = -\dfrac{3}{4}$ .

By point slope form,

Equation of line through A and slope = $-\dfrac{3}{4}$ ,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{4}$ [x - (-3)]

⇒ 4y = -3(x + 3)

⇒ 4y = -3x - 9

⇒ 3x + 4y + 9 = 0.

Hence, equation of line through A and perpendicular to 4x - 3y + 12 = 0 is 3x + 4y + 9 = 0.

Question 20

The point P is the foot of perpendicular from A(-5, 7) to the line 2x - 3y + 18 = 0. Determine :

(i) the equation of the line AP

(ii) the co-ordinates of P

Answer

(i) Given,

⇒ 2x - 3y + 18 = 0

⇒ 3y = 2x + 18

⇒ y = $\dfrac{2}{3}x + 6$

Comparing above equation with y = mx + c we get,

m = $\dfrac{2}{3}$ .

Since, AP is perpendicular to 2x - 3y + 18 = 0.

∴ Product of slope of AP and 2x - 3y + 18 = 0 will be -1.

Let slope of AP = m₁.

∴ m × m₁ = -1

⇒ $\dfrac{2}{3} \times m_1 = -1$

⇒ $m_1 = -\dfrac{3}{2}$ .

By point-slope form, equation of AP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = $-\dfrac{3}{2}$ [x - (-5)]

⇒ 2(y - 7) = -3(x + 5)

⇒ 2y - 14 = -3x - 15

⇒ 3x + 2y - 14 + 15 = 0

⇒ 3x + 2y + 1 = 0.

Hence, equation of AP is 3x + 2y + 1 = 0.

(ii) P is the point where AP and 2x - 3y + 18 = 0 meets,

Solving 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0 simultaneously,

⇒ 3x + 2y + 1 = 0

⇒ 2y = -3x - 1

⇒ y = $\dfrac{-3x - 1}{2}$ .........(1)

Substituting value of y in 2x - 3y + 18 = 0 we get,

$\Rightarrow 2x - 3\Big(\dfrac{-3x - 1}{2}\Big) + 18 = 0 \\[1em] \Rightarrow 2x - \dfrac{-9x - 3}{2} + 18 = 0 \\[1em] \Rightarrow 2x + \dfrac{9x + 3}{2} + 18 = 0 \\[1em] \Rightarrow \dfrac{4x + 9x + 3 + 36}{2} = 0 \\[1em] \Rightarrow 13x + 39 = 0 \\[1em] \Rightarrow 13x = -39 \\[1em] \Rightarrow x = -3.$

Substituting x = -3 in (1) we get,

$y = \dfrac{-3 \times -3 - 1}{2} \\[1em] = \dfrac{9 - 1}{2} \\[1em] = \dfrac{8}{2} \\[1em] = 4.$

P = (-3, 4).

Hence, co-ordinates of P = (-3, 4).

Question 21

Th points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.

If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the coordinates of P and Q.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of AB} = \dfrac{2 - 0}{2 - 4} \\[1em] = \dfrac{2}{-2} \\[1em] = -1.$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = -1(x - 4)

⇒ y = -x + 4

⇒ x + y = 4.

$\text{Slope of BC} = \dfrac{6 - 2}{0 - 2} \\[1em] = \dfrac{4}{-2} \\[1em] = -2.$

By point-slope form,

Equation of BC is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = -2(x - 2)

⇒ y - 2 = -2x + 4

⇒ 2x + y = 6.

Let point P be (0, a) and Q be (b, 0).

Substituting value of P in equation of AB we get,

⇒ 0 + a = 4

⇒ a = 4.

Substituting value of Q in equation of BC we get,

⇒ 2b + 0 = 6

⇒ 2b = 6

⇒ b = 3.

Hence, equation of AB is x + y = 4 and BC is 2x + y = 6 and P = (0, 4) and Q = (3, 0).

Question 22

Match the equations A, B, C and D with the lines L₁, L₂, L₃ and L₄, whose graphs are roughly drawn in the given diagram.

A ≡ y = 2x;

B ≡ y - 2x + 2 = 0;

C ≡ 3x + 2y = 6;

D ≡ y = 2

Match the equations A, B, C and D with the lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

Putting x and y = 0 in y = 2x we get,

Both sides = 0.

So, line y = 2x passes through origin.

A → L₃.

Putting x = 0 in y - 2x + 2 = 0,

⇒ y - 2(0) + 2 = 0

⇒ y = -2.

Putting y = 0 in y - 2x + 2 = 0

⇒ 0 - 2x + 2 = 0

⇒ 2x = 2

⇒ x = 1.

So, x-intercept is positive and y-intercept is negative.

B → L₄.

Putting x = 0 in 3x + 2y = 6,

⇒ 3(0) + 2y = 6

⇒ 2y = 6

⇒ y = 3.

Putting y = 0 in 3x + 2y = 6

⇒ 3x + 2(0) = 6

⇒ 3x = 6

⇒ x = 2.

So, both intercept are positive.

C → L₂.

Comparing y = 2 with y = mx + c we get,

Slope (m) = 0.

So, the line y = 2 is parallel to x-axis.

D → L₁.

Hence, A → L₃, B → L₄, C → L₂ and D → L₁.

Test Yourself

Question 1(a)

The line 5x - ky + 3 = 0 passes through the point (x = y = 3). The value of k is :

-3
-6
6
3

Answer

Since, the line 5x - ky + 3 = 0 passes through point (x = y = 3).

∴ Point (3, 3) satisfies the equation 5x - ky + 3 = 0.

∴ 5(3) - 3k + 3 = 0

⇒ 15 - 3k + 3 = 0

⇒ -3k + 18 = 0

⇒ 3k = 18

⇒ k = $\dfrac{18}{3}$ = 6.

Hence, Option 3 is the correct option.

Question 1(b)

The inclination of the line $\sqrt{3}x - 3y + 7$ = 0 is :

30°
45°
60°
75°

Answer

Given, equation :

$\Rightarrow \sqrt{3}x - 3y + 7 = 0 \\[1em] \Rightarrow 3y = \sqrt{3}x + 7 \\[1em] \Rightarrow y = \dfrac{\sqrt{3}}{3}x + \dfrac{7}{3} \\[1em] \Rightarrow y = \dfrac{1}{\sqrt{3}}x + \dfrac{7}{3}.$

Comparing above equation with y = mx + c, we get :

⇒ m = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, Option 1 is the correct option.

Question 1(c)

Assertion (A) : 2x + y = 8 and 6x + 3y = 20 are two straight lines. To find whether the lines are parallel to each other or not, we need to find the slope of each line.

Reason (R) : If slopes of both the lines are same, they are parallel to each other.

A is true, R is false
A is true, R is true
A is false, R is false
A is false, R is true

Answer

We know that,

It is true that if slopes of both the lines are same, they are parallel to each other and in order to find whether two lines are parallel or not we find and compare their slopes.

Hence, Option 2 is the correct option.

Question 1(d)

Assertion (A) : y = x + 4 and y = $\sqrt{3}x + 5$ are two intersecting lines.

Reason (R) : The inclinations of both the given lines are not equal.

Answer

A is true, R is false
A is true, R is true
A is false, R is true
A is false, R is false

Answer

Given,

1st equation :

y = x + 4

Comparing above equation with y = mx + c, we get :

m = 1

Slope of 1st line = 1.

2nd equation :

y = $\sqrt{3}x + 5$

Comparing above equation with y = mx + c, we get :

m = $\sqrt{3}$

Slope of 2nd line = $\sqrt{3}$ .

Equating equations (1) and (2), we get :

$\Rightarrow x + 4 = \sqrt{3}x + 5 \\[1em] \Rightarrow x - \sqrt{3}x = 5 - 4 \\[1em] \Rightarrow x(1 - \sqrt{3}) = 1 \\[1em] \Rightarrow x = \dfrac{1}{1 - \sqrt{3}}$

Rationalising,

$\Rightarrow x = \dfrac{1}{1 - \sqrt{3}} \times \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}} \\[1em] \Rightarrow x = \dfrac{1 + \sqrt{3}}{1^2 - (\sqrt{3})^2} \\[1em] \Rightarrow x = \dfrac{1 + \sqrt{3}}{1 - 3} \\[1em] \Rightarrow x = \dfrac{1 + \sqrt{3}}{-2} \\[1em] \Rightarrow x = \dfrac{-(1 + \sqrt{3})}{2}$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{-(1 + \sqrt{3})}{2} + 4 \\[1em] \Rightarrow y = \dfrac{-1 - \sqrt{3} + 8}{2} \\[1em] \Rightarrow y = \dfrac{7 - \sqrt{3}}{2}.$

∴ y = x + 4 and y = $\sqrt{3}x + 5$ are two intersecting lines and also the inclinations of both the given lines are not equal.

Hence, Option 2 is the correct option.

Question 1(e)

If the lines 4x + 3y = 84 and 3x + ky + 7 = 0 are perpendicular to each other. Then the value of k is :

4
-4
$\dfrac{1}{4}$
$-\dfrac{1}{4}$

Answer

Given,

1st equation :

⇒ 4x + 3y = 84

⇒ 3y = -4x + 84

⇒ y = $-\dfrac{4}{3}x + \dfrac{84}{3}$

⇒ y = $-\dfrac{4}{3}x$ + 28

Comparing above equation with y = mx + c, we get :

Slope of first equation = $-\dfrac{4}{3}$

2nd equation :

⇒ 3x + ky + 7 = 0

⇒ ky = -3x - 7

⇒ y = $-\dfrac{3}{k}x - \dfrac{7}{k}$

Slope of second equation = $-\dfrac{3}{k}$

We know that,

Product of slope of two perpendicular lines = -1.

$\Rightarrow -\dfrac{4}{3} \times -\dfrac{3}{k} = -1 \\[1em] \Rightarrow \dfrac{4}{k} = -1 \\[1em] \Rightarrow -k = 4 \\[1em] \Rightarrow k = -4.$

Hence, Option 2 is the correct option.

Question 1(f)

The line 5x + 3y = 15 meets y-axis at point P. The co-ordinates of point P are :

(0, 5)
(0, 3)
(5, 0)
(3, 0)

Answer

We know that,

x-coordinate at y-axis = 0.

Let point P be (0, a).

Since,

Line 5x + 3y = 15 meets y-axis at point P.

∴ Point P(0, a) satisfies the equation 5x + 3y = 15.

⇒ 5(0) + 3a = 15

⇒ 0 + 3a = 15

⇒ 3a = 15

⇒ a = $\dfrac{15}{3}$ = 5.

∴ P = (0, a) = (0, 5).

Hence, Option 1 is the correct option.

Question 1(g)

The vertices A and C of rhombus ABCD are A = (3, -1) and C = (-4, -8). The slope of diagonal BD is :

7
-7
1
-1

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AC = $\dfrac{-8 - (-1)}{-4 - 3} = \dfrac{-8 + 1}{-7} = \dfrac{-7}{-7}$ = 1.

We know that,

Diagonals of rhombus are perpendicular to each other.

Let slope of diagonal of BD be x.

We know that,

Product of slope of two perpendicular lines = -1.

∴ Slope of AC × Slope of BD = -1

⇒ 1 × Slope of BD = -1

⇒ Slope of BD = -1.

Hence, Option 4 is the correct option.

Question 1(h)

The equation of line AB is x + 8 = 0. The slope of the line OP which bisects angle O, is :

1
-1
0
2

Answer

Given,

Line OP bisects angle O.

Since, ∠O = 90°.

∴ OP makes 45° with x-axis.

From figure,

OP lies in the third quadrant.

Since, tan is positive in third quadrant.

Slope = tan 45° = 1.

Hence, Option 1 is the correct option.

Question 1(i)

The line 2x - 3y = 12, meets x-axis at point A and y-axis at point B, then :

A = (6, 0) and B = (0, -4)
A = (0, -4) and B = (6, 0)
A = (0, -4) and B = (-6, 0)
A = (-6, 0) and B = (4, 0)

Answer

We know that,

y-coordinate at x-axis = 0.

Let point A be (a, 0).

Since,

Line 2x - 3y = 12 meets x-axis at point A.

∴ Point A(a, 0) satisfies the equation 2x - 3y = 12.

⇒ 2a - 3(0) = 12

⇒ 2a - 0 = 12

⇒ 2a = 12

⇒ a = $\dfrac{12}{2}$ = 6.

∴ A = (a, 0) = (6, 0).

We know that,

x-coordinate at y-axis = 0.

Let point B be (0, b).

Since,

Line 2x - 3y = 12 meets y-axis at point B.

∴ Point B(0, b) satisfies the equation 2x - 3y = 12.

⇒ 2(0) - 3b = 12

⇒ 0 - 3b = 12

⇒ -3b = 12

⇒ b = $\dfrac{12}{-3}$ = -4.

∴ B = (0, b) = (0, -4).

Hence, Option 1 is the correct option.

Question 1(j)

Two straight lines 3x - 2y = 15 and 2x + ky + 8 = 0.

Assertion (A) : The given two lines are perpendicular to each other and k = 3.

Reason (R) : If the inclination of two lines are α and β; then tan α = -cot β.

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Given, two lines: 3x - 2y = 15 and 2x + ky + 8 = 0

Converting the equation of first line in slope-intercept form (y = mx + c), we get :

⇒ 3x - 2y = 15

⇒ -2y = -3x + 15

⇒ y = $\dfrac{3}{2}x - \dfrac{15}{2}$

∴ Slope of the line (m₁) = $\dfrac{3}{2}$

Converting the equation of second line in slope-intercept form (y = mx + c), we get :

⇒ 2x + ky + 8 = 0

⇒ ky = -2x - 8

⇒ y = $-\dfrac{2}{k}x - \dfrac{8}{k}$

∴ Slope of the line (m₂) = $-\dfrac{2}{k}$

We know that,

The two lines are perpendicular if product of their slopes is -1.

$\Rightarrow m_1 \times m_2 = -1\\[1em] \Rightarrow \dfrac{3}{2} \times \Big(-\dfrac{2}{k}\Big) = -1\\[1em] \Rightarrow -\dfrac{3}{k} = -1\\[1em] \Rightarrow -3 = -k\\[1em] \Rightarrow k = 3\\[1em]$

So, assertion (A) is true.

If the inclination of two lines are α and β, then Slope of first line (m₁) = tan α and Slope of second line (m₂) = tan β.

⇒ m₁ x m₂ = -1

⇒ tan α x tan β = -1

⇒ tan α = $-\dfrac{1}{\text{tan } β}$

⇒ tan α = -cot β

So, reason (R) is true, but it is not the correct reason for assertion (A).

Hence, option 4 is the correct option.

Question 1(k)

A line 2x + 8y = 15.

Assertion (A) : The equation of line passing through origin and parallel to the given line 2x + 8y = 15 is x + 4y = 0.

Reason (R) : Equation of the line passing through the origin and parallel to ax + by + c = 0 is ax + by = 0.

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Given, 2x + 8y = 15

Converting in slope-intercept form (y = mx + c), we get :

⇒ 8y = -2x + 15

⇒ y = $-\dfrac{2}{8}x + \dfrac{15}{8}$

⇒ y = $-\dfrac{1}{4}x + \dfrac{15}{8}$

∴ Slope of the line (m) = $-\dfrac{1}{4}$

By point-slope formula,

⇒ (y - y₁) = m(x - x₁)

Equation of line parallel to the line 2x + 8y = 15 and passing through origin (0, 0) :

⇒ (y - 0) = $-\dfrac{1}{4}$ (x - 0)

⇒ y = $-\dfrac{1}{4}$ x

⇒ 4y = -x

⇒ x + 4y = 0

So, assertion (A) is true.

Given, equation of line : ax + by + c = 0

Converting in slope-intercept form (y = mx + c), we get :

⇒ by = -ax - c

⇒ y = $-\dfrac{a}{b}x - \dfrac{c}{b}$

∴ Slope of the line (m) = $-\dfrac{a}{b}$

By point-slope formula,

⇒ (y - y₁) = m(x - x₁)

Equation of line parallel to ax + by + c = 0 and passing through origin (0, 0) :

⇒ (y - 0) = $-\dfrac{a}{b}$ (x - 0)

⇒ y = $-\dfrac{a}{b}$ x

⇒ by = -ax

⇒ ax + by = 0

So, reason (R) is true, and it is the correct reason for assertion (A).

Hence, option 3 is the correct option.

Question 1(l)

Points P(x, 2), A(-2, 3) and B(-5, 4) are collinear.

Statement 1: Slope of PA = Slope of PB = Slope of AB.

Statement 2: x = 1.

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, P(x, 2), A(-2, 3) and B(-5, 4).

If three points are collinear, it means they lie on same line.

∴ Slope will be equal.

∴ Slope of PA = Slope of PB = Slope of AB

So, statement 1 is true.

Using formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = Slope of PA

$\Rightarrow \dfrac{4 - 3}{(-5) - (-2)} = \dfrac{3 - 2}{(-2) - x}\\[1em] \Rightarrow \dfrac{4 - 3}{-5 + 2} = \dfrac{3 - 2}{-2 - x}\\[1em] \Rightarrow \dfrac{1}{-3} = \dfrac{1}{-2 - x}\\[1em] \Rightarrow -2 - x = -3\\[1em] \Rightarrow x = 3 - 2\\[1em] \Rightarrow x = 1.$

So, statement 2 is true.

Hence, option 1 is the correct option.

Question 1(m)

Lines 2x - by + 7 = 0 and ax - 2y - 7 = 0 are perpendicular to each other.

Statement 1: $\dfrac{2}{a} = \dfrac{b}{2}$ .

Statement 2: Slope of line 2x - by + 7 = 0 is $\dfrac{2}{b}$ and slope of line ax - 2y - 7 = 0 is $\dfrac{a}{2}$ .

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given equation of line 2x - by + 7 = 0 and ax - 2y - 7 = 0

Converting the equation of first line in slope-intercept form (y = mx + c), we get :

⇒ 2x - by + 7 = 0

⇒ by = 2x + 7

⇒ y = $\dfrac{2}{b}x + \dfrac{7}{b}$

∴ Slope of the line (m₁) = $\dfrac{2}{b}$

Converting the equation of second line in slope-intercept form y = mx + c, we get :

⇒ ax - 2y - 7 = 0

⇒ 2y = ax - 7

⇒ y = $\dfrac{a}{2}x - \dfrac{7}{2}$

∴ Slope of the line (m₂) = $\dfrac{a}{2}$

So, statement 2 is true.

We know that,

The two lines are perpendicular if product of their slopes is -1.

$\Rightarrow m_1 \times m_2 = -1\\[1em] \Rightarrow \dfrac{2}{b} \times \dfrac{a}{2} = -1\\[1em] \Rightarrow \dfrac{2}{b} = -1 \times \dfrac{2}{a}\\[1em] \Rightarrow \dfrac{2}{b} = -\dfrac{2}{a}\\[1em]$

So, statement 1 is false.

Hence, option 4 is the correct option.

Question 2

Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3 : 5. Find its co-ordinates of point P.

Also, find the equation of the line through P and parallel to 3x + 5y = 7.

Answer

Given points, A (8, 0) and B (16, -8)

By section formula, the co-ordinates of the point P which divides AB in the ratio 3 : 5 is given by

$P = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{3 \times 16 + 5 \times 8}{3 + 5}, \dfrac{3 \times -8 + 5 \times 0}{3 + 5}\Big) \\[1em] = \Big(\dfrac{48 + 40}{8}, \dfrac{-24 + 0}{8}\Big) \\[1em] = \Big(\dfrac{88}{8}, \dfrac{-24}{8}\Big) \\[1em] = (11, -3).$

Given line equation is,

⇒ 3x + 5y = 7

⇒ 5y = -3x + 7

⇒ y = $-\dfrac{3}{5}x + \dfrac{7}{5}$

Comparing above equation with y = mx + c we get,

Slope = $-\dfrac{3}{5}$

The line parallel to the line 3x + 5y = 7 will have the same slope.

Hence, the slope of the required line = $-\dfrac{3}{5}$

By point-slope form, equation of the required line,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $-\dfrac{3}{5}$ (x - 11)

⇒ 5(y + 3) = -3(x - 11)

⇒ 5y + 15 = -3x + 33

⇒ 3x + 5y = 33 - 15

⇒ 3x + 5y = 18.

Hence, P = (11, -3) and equation of the required line is 3x + 5y = 18.

Question 3

The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.

Answer

Given points, A(3, -4) and B(-2, 1)

By section formula, the co-ordinates of the point P which divides AB in the ratio 1: 3 is given by

$P = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{1 \times -2 + 3 \times 3}{1 + 3}, \dfrac{1 \times 1 + 3 \times -4}{1 + 3}\Big) \\[1em] = \Big(\dfrac{-2 + 9}{4}, \dfrac{1 + -12}{4}\Big) \\[1em] = \Big(\dfrac{7}{4}, -\dfrac{11}{4}\Big)$

Given line equation is,

5x – 3y = 4

3y = 5x - 4

y = $\dfrac{5}{3}x - \dfrac{4}{3}$

So, the slope of this line (m) = $\dfrac{5}{3}$

Let slope of perpendicular line be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \dfrac{5}{3} = -1$

⇒ m₁ = $-\dfrac{3}{5}$ .

Slope of the required line = $-\dfrac{3}{5}$ .

By point-slope form,

Equation of line through P and slope = $-\dfrac{3}{5}$ is,

⇒ y – y₁ = m(x – x₁)

$\Rightarrow y - \Big(-\dfrac{11}{4}\Big) = -\dfrac{3}{5}\Big(x - \dfrac{7}{4}\Big) \\[1em] \Rightarrow \dfrac{4y + 11}{4} = -\dfrac{3}{5} \times \dfrac{4x - 7}{4} \\[1em] \Rightarrow 4y + 11 = -\dfrac{3}{5} \times (4x - 7) \\[1em] \Rightarrow 5(4y + 11) = -3(4x - 7) \\[1em] \Rightarrow 20y + 55 = -12x + 21 \\[1em] \Rightarrow 12x + 20y + 55 - 21 = 0 \\[1em] \Rightarrow 12x + 20y + 34 = 0 \\[1em] \Rightarrow 2(6x + 10y + 17) = 0 \\[1em] \Rightarrow 6x + 10y + 17 = 0.$

Hence, P = $\Big(\dfrac{7}{4}, -\dfrac{11}{4}\Big)$ the equation of required line is 6x + 10y + 17 = 0.

Question 4

A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x - 3y + 4 = 0.

Answer

As the point P lies on y-axis, the x co-ordinate of point P will be = 0.

Putting x = 0 in the equation 5x + 3y + 15 = 0, we get

⇒ 5(0) + 3y + 15 = 0

⇒ 3y + 15 = 0

⇒ 3y = -15

⇒ y = -5.

Hence, the co-ordinates of the point P are (0, -5).

Given line equation,

x - 3y + 4 = 0

3y = x + 4

y = $\dfrac{1}{3}x + \dfrac{4}{3}$

Comparing above equation with y = mx + c we get,

Slope of this line (m) = $\dfrac{1}{3}$

Product of slope of perpendicular lines = -1.

Let slope of line perpendicular to x - 3y + 4 = 0 be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \dfrac{1}{3} = -1$

⇒ m₁ = -3.

Slope of the required line = -3.

By point-slope form,

Equation of line through P and slope = -3 is,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = -3(x - 0)

⇒ y + 5 = -3x

⇒ 3x + y + 5 = 0.

Hence, P = (0, -5) and equation of required line is 3x + y + 5 = 0.

Question 5

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:

(i) The equation of the line.

(ii) The co-ordinates of A and B.

(iii) The co-ordinates of M.

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find (i) The equation of the line. (ii) The co-ordinates of A and B. (iii) The co-ordinates of M. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

(i) Given points, P (-1, 4) and Q (5, -2)

$\text{Slope of PQ } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-2 - 4}{5 - (-1)} \\[1em] = \dfrac{-6}{6} = -1.$

By point-slope form,

Equation of the line PQ,

⇒ y – y₁ = m(x – x₁)

⇒ y – 4 = -1[x - (-1)]

⇒ y – 4 = -[x + 1]

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y = 3.

Hence, equation of line is x + y = 3.

(ii) For point A (on x-axis), y = 0.

So, putting y = 0 in the equation of PQ, we have

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0).

For point B (on y-axis), x = 0.

So, putting x = 0 in the equation of PQ, we have

⇒ 0 + y = 3

⇒ y = 3

∴ B = (0, 3).

Hence, co-ordinates of A = (3, 0) and B = (0, 3).

(iii) M is the mid-point of AB.

∴ M = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

$= \Big(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\Big) \\[1em] = \Big(\dfrac{3}{2}, \dfrac{3}{2}\Big).$

Hence, M = $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$ .

Question 6

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.

(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.

(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Answer

Given, A (3, 2), B (6, -2) and C (2, -5)

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square. Find the co-ordinates of its fourth vertex D, if ABCD is a square. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

By distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{AB} = \sqrt{(6 - 3)^2 + (-2 - 2)^2} \\[1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \\[1.5em] \text{BC} = \sqrt{(2 - 6)^2 + (-5 - (-2))^2} \\[1em] = \sqrt{(-4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5$

Thus, AC = BC

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{-2 - 2}{6 - 3} = \dfrac{-4}{3}$

Slope of BC = $\dfrac{-5 - (-2)}{2 - 6} = \dfrac{-3}{-4} = \dfrac{3}{4}$

Slope of AB x Slope of BC = $\dfrac{-4}{3} \times \dfrac{3}{4}$ = -1

Hence, AB ⊥ BC

Therefore, A, B, C can be the vertices of a square.

(i) Slope of CD = Slope of AB = $-\dfrac{4}{3}$ (As they are parallel)

So, the equation of CD is

⇒ y – y₁ = m(x – x₁)

⇒ y - (-5) = $-\dfrac{4}{3}$ (x – 2)

⇒ 3(y + 5) = -4(x - 2)

⇒ 3y + 15 = -4x + 8

⇒ 4x + 3y + 7 = 0 .........(1)

Slope of AD = Slope of BC = $\dfrac{3}{4}$ (As these lines are parallel)

So, the equation of the line AD is

⇒ y – y₁ = m(x – x₁)

⇒ y - 2 = $\dfrac{3}{4}$ (x - 3)

⇒ 4(y – 2) = 3(x – 3)

⇒ 4y - 8 = 3x - 9

⇒ 3x - 4y = -8 + 9

⇒ 3x - 4y = 1 ......... (2)

Now, D is the point of intersection of CD and AD.

Solving (1) and (2),

Multiplying equation (1) by 4 and (2) by 3 and adding them we get,

⇒ 16x + 12y + 28 + 9x – 12y = 0 + 3

⇒ 25x = 3 - 28

⇒ 25x = -25

⇒ x = -1

Putting value of x in (1), we get

⇒ 4(-1) + 3y + 7 = 0

⇒ 3y - 4 + 7 = 0

⇒ 3y + 3 = 0

⇒ 3y = -3

⇒ y = -1

Hence, the co-ordinates of point D are (-1, -1).

(ii) From the equation (2)

The equation of the line AD is,

⇒ 3x – 4y = 1

⇒ 4y = 3x - 1.

Slope of AC = $\dfrac{-5 - 2}{2 - 3} = \dfrac{-7}{-1} = 7.$

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square. Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Since, diagonals of a square are perpendicular to each other. So, product of their slopes = -1.

∴ Slope of AC × Slope of BD = -1

⇒ $7 \times$ Slope of BD = -1

⇒ Slope of BD = - $\dfrac{1}{7}$ .

The equation of the diagonal BD is

⇒ y – y₁ = m(x – x₁)

⇒ y - (-2) = $-\dfrac{1}{7}$ (x - 6)

⇒ 7(y + 2) = –1(x – 6)

⇒ 7y + 14 = -x + 6

⇒ x + 7y + 14 - 6 = 0

⇒ x + 7y + 8 = 0.

Hence, equation of line BD is x + 7y + 8 = 0 and AD = 4y = 3x - 1.

Question 7

A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.

Answer

The given line equation is

⇒ x = 3y + 2 ..........(1)

⇒ 3y = x - 2

⇒ y = $\dfrac{1}{3}x - \dfrac{2}{3}$

Comparing above equation with y = mx + c we get,

Slope (m) = $\dfrac{1}{3}$

Let slope of perpendicular line be m₁.

⇒ m₁ × m = -1 (As product of slope of perpendicular lines = -1).

⇒ $m_1 \times \dfrac{1}{3}$ = -1

⇒ m₁ = -3.

Equation of the line passing through origin and with slope = -3, by point-slope form is:

⇒ y – y₁ = m(x – x₁)

⇒ y – 0 = -3(x – 0)

⇒ y = -3x

⇒ 3x + y = 0 ...........(2)

Next,

Point X is the intersection of the lines (1) and (2).

Substituting value of x from (1) in (2),

⇒ 3(3y + 2) + y = 0

⇒ 9y + 6 + y = 0

⇒ 10y = -6

⇒ y = $-\dfrac{6}{10} = -\dfrac{3}{5}$ .

⇒ x = 3y + 2 = $3 \times -\dfrac{3}{5} + 2 = \dfrac{-9 + 10}{5} = \dfrac{1}{5}$ .

Hence, the co-ordinates of the point X are $\Big(\dfrac{1}{5}, -\dfrac{3}{5}\Big)$ .

Question 8

A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer

Let P = (3, 2).

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).

A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Since, P is the mid-point of AB, we have:

$\Rightarrow P = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 3 \text{ and } \dfrac{y}{2} = 2\\[1em] \Rightarrow x = 6 \text{ and } y = 4.$

Thus, A = (6, 0) and B = (0, 4)

Slope of line AB = $\dfrac{4 - 0}{0 - 6} = \dfrac{4}{-6} = -\dfrac{2}{3}$ .

So, the required equation of the line AB is given by

⇒ y – y₁ = m(x – x₁)

⇒ y – 0 = $-\dfrac{2}{3}$ (x – 6)

⇒ 3y = -2x + 12

⇒ 2x + 3y = 12

Hence, equation of the required line is 2x + 3y = 12.

Question 9

Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.

Answer

Given line equations are,

7x + 6y = 71 .......(1)

and

5x - 8y = -23 .........(2)

Multiplying (1) by 4 and (2) by 3 we get,

⇒ 28x + 24y = 284 ......... (3)

⇒ 15x – 24y = -69 .........(4)

On adding (3) and (4), we get

⇒ 28x + 15x + 24y - 24y = 284 + (-69)

⇒ 43x = 215

⇒ x = $\dfrac{215}{43}$

⇒ x = 5.

From (2), we get

⇒ 8y = 5x + 23

⇒ 8y = 5(5) + 23

⇒ 8y = 25 + 23

⇒ 8y = 48

⇒ y = $\dfrac{48}{8}$

⇒ y = 6.

Hence, the required line passes through the point (5, 6).

Given, 4x – 2y = 1

⇒ 2y = 4x – 1

⇒ y = 2x – $\dfrac{1}{2}$ .

Comparing above equation with y = mx + c we get,

Slope (m) = 2

Let slope of required line be m₁.

As, the lines are perpendicular to each other so product of their slopes = -1.

⇒ m × m₁ = -1

⇒ 2 × m₁ = -1

⇒ m₁ = $-\dfrac{1}{2}$ .

Thus, equation of the line is,

⇒ y – y₁ = m(x – x₁)

⇒ y – 6 = $-\dfrac{1}{2}$ (x – 5)

⇒ 2(y – 6) = -1(x - 5)

⇒ 2y - 12 = -x + 5

⇒ 2y + x = 5 + 12

⇒ x + 2y = 17.

Hence, equation of required line is x + 2y = 17.

Question 10

Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.

Does the line 3x = y + 1 bisect the line segment joining the two given points?

Answer

Let A = (-2, 3) and B = (4, 1)

$\text{Slope of AB }(\text{m}_1) = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{1 - 3}{4 - (-2)} \\[1em] = \dfrac{1 - 3}{4 + 2} \\[1em] = \dfrac{-2}{6} \\[1em] = -\dfrac{1}{3}$

By point-slope form, the equation of line AB is

⇒ y – y₁ = m(x – x₁)

⇒ y – 3 = $-\dfrac{1}{3}$ [x - (-2)]

⇒ 3(y – 3) = -1(x + 2)

⇒ 3y - 9 = -x - 2

⇒ 3y + x = -2 + 9

⇒ x + 3y = 7 .........(1)

Given,

⇒ 3x = y + 1

⇒ y = 3x - 1

Comparing above equation with y = mx + c we get,

⇒ Slope (m₂) = 3

Since, m₁ × m₂ = $-\dfrac{1}{3} \times 3$ = -1.

Hence, the line through points A and B is perpendicular to the 3x = y + 1.

Given line is 3x = y + 1 ........(2)

Let P be the mid-point of AB,

The co-ordinates of the mid-point of AB (i.e. P) are

$\Big(\dfrac{-2 + 4}{2}, \dfrac{3 + 1}{2}\Big) = \Big(\dfrac{2}{2}, \dfrac{4}{2}\Big)$ = (1, 2).

Now, Let’s check if point P satisfies the line equation (2)

⇒ 3(1) = 2 + 1

⇒ 3 = 3

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.

Question 11

Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).

Answer

Given line equation,

⇒ x cos 30° + y sin 30° = 2

⇒ $\dfrac{\sqrt{3}}{2}x + \dfrac{y}{2} = 2$

⇒ y + $\sqrt{3}x$ = 4

⇒ y = - $\sqrt{3}x + 4$ .

Comparing above equation with y = mx + c we get,

Slope (m₁) = - $\sqrt{3}$ .

Since, slope of parallel lines are equal,

Slope of a line which is parallel to this given line = - $\sqrt{3}$ .

By slope-point form, the equation of the required line is,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = - $\sqrt{3}$ (x - 4)

⇒ y - 3 = $-\sqrt{3}x + 4\sqrt{3}$

⇒ $\sqrt{3}x + y = 4\sqrt{3} + 3$ .

Hence, the equation of the required line is $\sqrt{3}x + y = 4\sqrt{3} + 3$ .

Question 12

Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:

(i) perpendicular to the line 2x – y + 7 = 0

(ii) parallel to it.

Answer

Given line equation,

⇒ 2x - y + 7 = 0 ..........(1)

⇒ y = 2x + 7

Comparing above equation with y = mx + c we get,

Slope (m₁) = 2

Given line equation,

⇒ (k - 2)x + (k + 3)y - 5 = 0 ..........(2)

⇒ (k + 3)y = -(k - 2)x + 5

⇒ y = $-\dfrac{k - 2}{k + 3}x + \dfrac{5}{k + 3}$

Comparing above equation with y = mx + c we get,

Slope (m₂) = $-\dfrac{k - 2}{k + 3}$

(i) If lines (1) and (2) are perpendicular then product of their slopes = -1.

⇒ m₁ x m₂ = -1

$\Rightarrow 2 \times -\dfrac{k - 2}{k + 3} = -1 \\[1em] \Rightarrow -2(k - 2) = -1(k + 3) \\[1em] \Rightarrow -2k + 4 = -k - 3 \\[1em] \Rightarrow -k + 2k = 4 + 3 \\[1em] \Rightarrow k = 7.$

Hence, k = 7.

(ii) If lines 1 and 2 are parallel then their slopes will be equal.

⇒ $-\dfrac{k - 2}{k + 3}$ = 2

⇒ -(k - 2) = 2(k + 3)

⇒ -k + 2 = 2k + 6

⇒ 2k + k = 2 - 6

⇒ 3k = -4

⇒ k = $-\dfrac{4}{3}$ .

Hence, k = $-\dfrac{4}{3}$ .

Question 13

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :

(i) the equation of line through A and perpendicular to BC.

(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Answer

$\text{Slope of BC } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{7 - (-2)}{11 - (-1)} \\[1em] = \dfrac{7 + 2}{11 + 1} \\[1em] = \dfrac{9}{12} \\[1em] = \dfrac{3}{4}$

Then the equation of the line BC is

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $\dfrac{3}{4}$ [x - (-1)]

⇒ 4(y + 2) = 3(x + 1)

⇒ 4y + 8 = 3x + 3

⇒ 3x - 4y = 8 - 3

⇒ 3x - 4y = 5 ......... (1)

(i) Let slope of line perpendicular to BC = m₁

As product of slope of perpendicular lines is -1

∴ m x m₁ = -1

⇒ $\dfrac{3}{4}$ x m₁ = -1

⇒ m₁ = $-\dfrac{4}{3}$

So, the required equation of the line through A (0, 5) and perpendicular to BC is given by

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = $-\dfrac{4}{3}$ (x - 0)

⇒ 3y - 15 = -4x

⇒ 4x + 3y = 15 ......... (2)

Hence, equation of the required line is 4x + 3y = 15.

(ii) The required point P will be the point of intersection of lines (1) and (2).

3x - 4y = 5 ......... (1)

4x + 3y = 15 ......... (2)

Multiplying equation (1) by 3 and (2) by 4 and adding we get,

⇒ 9x - 12y + 16x + 12y = 15 + 60

⇒ 25x = 75

⇒ x = $\dfrac{75}{25}$

⇒ x = 3

Substituting the value of x in equation (1) we get,

3(3) - 4y = 5

⇒ 4y = 3(3) – 5

⇒ 4y = 9 - 5

⇒ 4y = 4

⇒ y = 1

Hence, the co-ordinates of the required point P is (3, 1).

Question 14

From the given figure, find:

(i) the co-ordinates of A, B and C.

(ii) the equation of the line through A and parallel to BC.

Answer

(i) From figure,

The co-ordinates of A = (2, 3), B = (-1, 2), C = (3, 0).

(ii) Slope of BC = $\dfrac{0 - 2}{3 - (-1)} = \dfrac{-2}{4} = -\dfrac{1}{2}$ .

Slope of the line which is parallel to BC = $-\dfrac{1}{2}$ (As parallel lines have equal slope).

Hence, the required equation of the line through A and parallel to BC is given by

⇒ y – y₁ = m(x – x₁)

⇒ y – 3 = $-\dfrac{1}{2}$ (x – 2)

⇒ 2(y – 3) = -1(x - 2)

⇒ 2y - 6 = -x + 2

⇒ x + 2y = 2 + 6

⇒ x + 2y = 8.

Hence, equation of line through A and parallel to BC is x + 2y = 8.

Question 15

A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.

Answer

ΔABC with vertices A (8, -6), B (-4, 2) and C (0, -10) is shown below:

A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

By Mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Co-ordinates of P = $\Big[\dfrac{8 + (-4)}{2}, \dfrac{-6 + 2}{2}\Big] = \Big(\dfrac{4}{2}, \dfrac{-4}{2}\Big) = (2, -2).$

Co-ordinates of Q = $\Big[\dfrac{8 + 0}{2}, \dfrac{-6 + (-10)}{2}\Big] = \Big(\dfrac{8}{2}, \dfrac{-16}{2}\Big)$ = (4, -8).

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of PQ = $\dfrac{-8 - (-2)}{4 - 2} = \dfrac{-6}{2}$ = -3.

Slope of BC = $\dfrac{-10 - 2}{0 - (-4)} = \dfrac{-12}{4}$ = -3.

Since, slope of PQ = slope of BC.

∴ PQ || BC.

From figure,

PBCQ is a trapezium.

Question 16

A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:

(i) the co-ordinates of A and B.

(ii) the equation of line through P and perpendicular to AB.

Answer

(i) Let’s assume the co-ordinates of point A, lying on x-axis be (x, 0) and the co-ordinates of point B (lying on y-axis) be (0, y).

Given,

P = (-4, -2) and AP : PB = 1 : 2

By section formula, we get

$\Rightarrow P = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{1 \times 0 + 2 \times x}{1 + 2}, \dfrac{1 \times y + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (-4, -2) = \Big(\dfrac{2x}{3}, \dfrac{y}{3}\Big) \\[1em] \Rightarrow -4 = \dfrac{2x}{3} \text{ and } -2 = \dfrac{y}{3} \\[1em] \Rightarrow x = \dfrac{-4 \times 3}{2} \text{ and } y = -6 \\[1em] \Rightarrow x = -6 \text{ and } y = -6.$

Hence, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{-6 - 0}{0 - (-6)} = \dfrac{-6}{6}$ = -1.

Let slope of perpendicular line be m.

⇒ m × -1 = -1

⇒ -m = -1

⇒ m = 1.

Therefore, the required equation of the line passing through P and perpendicular to AB is given by

⇒ y – y₁ = m(x – x₁)

⇒ y - (-2) = 1[x - (-4)]

⇒ y + 2 = x + 4

⇒ y = x + 2.

Hence, the equation of line through P and perpendicular to AB is y = x + 2.

Question 17

A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.

Answer

Let line intersect x-axis at point A.

So, A = (-2, 0)

Let line cuts off an intercept of 3 units from positive side of y-axis at point B.

So, co-ordinates of B are (0, 3).

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{3 - 0}{0 - (-2)} = \dfrac{3}{2}$ .

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{3}{2}$ [x - (-2)]

⇒ 2y = 3(x + 2)

⇒ 2y = 3x + 6

⇒ 3x - 2y + 6 = 0

⇒ 2y = 3x + 6.

Hence, equation of AB is 2y = 3x + 6.

Question 18

Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.

Answer

Let the line intersect x-axis at point A.

So, A = (4, 0).

Let B = (2, 3).

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{3 - 0}{2 - 4} = -\dfrac{3}{2}$ .

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{2}$ (x - 4)

⇒ 2y = -3(x - 4)

⇒ 2y = -3x + 12

⇒ 3x + 2y = 12.

Hence, the equation of required line is 3x + 2y = 12.

Question 19

The given figure (not drawn to scale) shows two straight lines AB and CD. If equation of the line AB is : y = x + 1 and equation of line CD is : y = $\sqrt{3}$ x - 1. Write down the inclination of lines AB and CD; also, find the angle θ between AB and CD.

Answer

Given,

Equation of AB :

⇒ y = x + 1

Comparing the above equation with y = mx + c we get,

Slope of AB = 1.

Let θ₁ be the inclination of AB then,

⇒ tan θ₁ = 1

⇒ tan θ₁ = tan 45°

⇒ θ₁ = 45°.

Equation of CD :

⇒ y = $\sqrt{3}$ x - 1

Comparing the above equation with y = mx + c we get,

Slope of CD = $\sqrt{3}$ .

Let θ₂ be the inclination of AB then,

⇒ tan θ₂ = $\sqrt{3}$

⇒ tan θ₂ = tan 60°

⇒ θ₂ = 60°.

From figure,

∠FGE = 180° - θ₂ = 180° - 60° = 120°.

In △EFG,

∠E + ∠F + ∠G = 180

⇒ θ₁ + θ + 120° = 180°

⇒ 45° + θ + 120° = 180°

⇒ θ + 165° = 180°

⇒ θ = 180° - 165° = 15°.

Hence, inclination of AB is 45°, inclination of CD is 60° and θ = 15°.

Question 20

Write down the equation of the line whose gradient is $\dfrac{3}{2}$ and which passes through P, where P divides the line segment joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.

Answer

By section-formula,

P = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$P = \Big(\dfrac{2 \times 3 + 3 \times -2}{2 + 3}, \dfrac{2 \times -4 + 3 \times 6}{2 + 3}\Big) \\[1em] = \Big(\dfrac{6 + (-6)}{5}, \dfrac{-8 + 18}{5}\Big) \\[1em] = \Big(\dfrac{0}{5}, \dfrac{10}{5}\Big) \\[1em] = (0, 2).$

By point-slope form,

Equation of line with slope = $\dfrac{3}{2}$ and passing through (0, 2) is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\dfrac{3}{2}$ (x - 0)

⇒ 2(y - 2) = 3x

⇒ 2y - 4 = 3x

⇒ 2y = 3x + 4.

Hence, equation of required line is 2y = 3x + 4.

Question 21

The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.

Answer

Let point A = (6, 4) and B = (7, -5).

Let the point P be (a, -23).

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{-5 - 4}{7 - 6} = \dfrac{-9}{1}$ = -9.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = -9(x - 6)

⇒ y - 4 = -9x + 54

⇒ 9x + y = 54 + 4

⇒ 9x + y = 58.

Since, P lies on AB so it will satisfy the equation.

Substituting values of P in equation we get,

⇒ 9(a) + (-23) = 58

⇒ 9a - 23 = 58

⇒ 9a = 58 + 23

⇒ 9a = 81

⇒ a = $\dfrac{81}{9}$

⇒ a = 9.

∴ P = (a, -23) = (9, -23).

Hence, co-ordinates of the required point are (9, -23).

Question 22

A and B are two points on the x-axis and y-axis respectively. P(2, -3) is the mid-point of AB. Find the

(i) the co-ordinates of A and B

(ii) slope of line AB

(iii) equation of line AB.

Answer

(i) Let point A be (x, 0) and B be (0, y). Since, P is the mid-point of AB.

$\therefore P = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (2, -3) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 2 \text{ and } \dfrac{y}{2} = -3 \\[1em] \Rightarrow x = 4 \text{ and } y = -6.$

∴ A = (x, 0) = (4, 0) and B = (0, y) = (0, -6).

Hence, co-ordinates of A = (4, 0) and B = (0, -6).

(ii) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of AB }= \dfrac{-6 - 0}{0 - 4} \\[1em] = \dfrac{-6}{-4} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}$

Hence, slope of AB = $1\dfrac{1}{2}$ .

(iii) By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{3}{2}$ (x - 4)

⇒ 2y = 3(x - 4)

⇒ 2y = 3x - 12

⇒ 3x - 2y = 12

⇒ 2y = 3x - 12

Hence, equation of AB is 2y = 3x - 12.

Question 23

Point A and B have co-ordinates (7, -3) and (1, 9) respectively. Find :

(i) the slope of AB.

(ii) the equation of perpendicular bisector of the line segment AB.

(iii) the value of 'p' if (-2, p) lies on it.

Answer

(i) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of AB } = \dfrac{9 - (-3)}{1 - 7} \\[1em] = \dfrac{12}{-6} \\[1em] = -2.$

Hence, slope of AB = -2.

(ii) Let O be the mid-point of AB.

By mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

O = $\Big(\dfrac{7 + 1}{2}, \dfrac{-3 + 9}{2}\Big) = \Big(\dfrac{8}{2}, \dfrac{6}{2}\Big)$ = (4, 3).

⇒ Slope of AB × Slope of perpendicular bisector = -1 (As product of slopes of perpendicular lines = -1)

⇒ -2 × Slope of perpendicular bisector = -1

⇒ Slope of perpendicular bisector = $\dfrac{1}{2}$ .

By point-slope form,

Equation of perpendicular bisector is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $\dfrac{1}{2}$ (x - 4)

⇒ 2(y - 3) = 1(x - 4)

⇒ 2y - 6 = x - 4

⇒ x - 2y - 4 + 6 = 0

⇒ x - 2y + 2 = 0.

Hence, the equation of perpendicular bisector of AB is x - 2y + 2 = 0.

(iii) Since, (-2, p) lies on perpendicular bisector. So, it will satisfy the equation.

⇒ -2 - 2p + 2 = 0

⇒ -2p = 0

⇒ p = 0.

Hence, p = 0.

Question 24

ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find :

(i) co-ordinates of A

(ii) equation of diagonal BD.

Answer

(i) Since, ABCD is a parallelogram.

So, AB || CD and AD || BC.

Since, AB || CD so slope of AB = slope of CD.

$\therefore \dfrac{8 - y}{5 - x} = \dfrac{7 - (-4)}{4 - 2} \\[1em] \Rightarrow \dfrac{8 - y}{5 - x} = \dfrac{11}{2} \\[1em] \Rightarrow 2(8 - y) = 11(5 - x) \\[1em] \Rightarrow 16 - 2y = 55 - 11x \\[1em] \Rightarrow 11x - 2y = 55 - 16 \\[1em] \Rightarrow 11x - 2y = 39 ..........(1)$

Since, AD || BC so slope of AD = slope of BC

$\therefore \dfrac{-4 - y}{2 - x} = \dfrac{7 - 8}{4 - 5} \\[1em] \Rightarrow \dfrac{-4 - y}{2 - x} = \dfrac{-1}{-1} \\[1em] \Rightarrow \dfrac{-4 - y}{2 - x} = 1 \\[1em] \Rightarrow -4 - y = 2 - x \\[1em] \Rightarrow x - y - 4 - 2 = 0 \\[1em] \Rightarrow x - y - 6 = 0 \\[1em] \Rightarrow x = y + 6 .........(2)$

Substituting value of x from (2) in (1) we get,

⇒ 11(y + 6) - 2y = 39

⇒ 11y + 66 - 2y = 39

⇒ 9y = 39 - 66

⇒ 9y = -27

⇒ y = $\dfrac{-27}{9}$

⇒ y = -3.

Substituting value of y in (2) we get,

⇒ x = -3 + 6 = 3.

∴ A = (3, -3).

Hence, co-ordinates of A = (3, -3).

(ii) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of BD }= \dfrac{-4 - 8}{2 - 5} \\[1em] = \dfrac{-12}{-3} = 4.$

By-point slope form,

Equation of BD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 8 = 4(x - 5)

⇒ y - 8 = 4x - 20

⇒ 4x - y = -8 + 20

⇒ 4x - y = 12.

Hence, equation of BD is 4x - y = 12.

Question 25

Given equation of line L₁ is y = 4.

(i) Write the slope of line L₂ if L₂ is the bisector of angle O.

(ii) Write the co-ordinates of point P.

(iii) Find the equation of L₂.

Answer

(i) The angle between x and y axis = 90°.

Given,

L₂ is the bisector of angle O. So, inclination of line L₂ is 45°.

Slope of L₂ = tan 45° = 1.

Hence, slope of L₂ = 1.

(ii) Equation of line L₂, passing through origin (0, 0) and slope = 1 is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1(x - 0)

⇒ y = x.

From figure,

P is the point of intersection of y = 4 and y = x.

Solving,

y = 4 ........(1)

y = x ........(2)

Substituting value of y from (1) in (2) we get,

x = 4.

So, point of intersection = (4, 4).

Hence, P = (4, 4).

(iii) From part (ii) we get equation of L₂ as,

y = x

Hence, equation of L₂ is y = x.

Question 26

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

Answer

Let point where line touches x-axis be A. So, A = (-3, 0)

Given equation of line,

⇒ 3x + 5y = 1

⇒ 5y = -3x + 1

⇒ y = $-\dfrac{3}{5}x + \dfrac{1}{5}$

Comparing above equation with y = mx + c we get,

Slope (m₁) = $-\dfrac{3}{5}$

Let slope of line perpendicular to 3x + 5y = 1 be m₂.

⇒ m₁ × m₂ = -1

⇒ $-\dfrac{3}{5} \times m_2 = -1$

⇒ $m_2 = \dfrac{5}{3}$ .

By point-slope form,

Equation of line with slope = $\dfrac{5}{3}$ and passing through (-3, 0) is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{5}{3}$ [x - (-3)]

⇒ 3y = 5[x + 3]

⇒ 3y = 5x + 15

⇒ 5x - 3y + 15 = 0.

Hence, equation of required line is 5x - 3y + 15 = 0.

Question 27

A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :

(i) the equation of the line.

(ii) the co-ordinates of points A and B.

(iii) the co-ordinates of point M.

Answer

(i) Slope of PQ = $\dfrac{-2 - 4}{5 - (-1)} = -\dfrac{6}{6}$ = -1.

By point-slope form,

Equation of PQ is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = -1[x - (-1)]

⇒ y - 4 = -1(x + 1)

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y = 3.

Hence, the equation of the line is x + y = 3.

(ii) Given,

PQ intersects x-axis at A and y-axis at B.

At x-axis, y co-ordinate = 0.

Substituting y = 0 in equation we get,

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0)

At y-axis, x co-ordinate = 0.

Substituting x = 0 in equation we get,

⇒ 0 + y = 3

⇒ y = 3.

∴ B = (0, 3)

Hence, co-ordinates of A = (3, 0) and B = (0, 3).

(iii) By formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$\text{M} = \Big(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\Big) \\[1em] = \Big(\dfrac{3}{2}, \dfrac{3}{2}\Big).$

Hence, co-ordinates of M = $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$ .

Question 28

In the given figure, line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point P. Find :

(i) equation of line AB.

(ii) equation of line CD.

(iii) co-ordinates of point E and D.

In the figure, line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point P. Find (i) equation of line AB. (ii) equation of line CD. (iii) co-ordinates of point E and D. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

(i) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\text{Slope of AB} = \dfrac{8 - 6}{-6 - 0} \\[1em] = \dfrac{2}{-6} \\[1em] = -\dfrac{1}{3}.$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\dfrac{1}{3}$ (x - 0)

⇒ 3(y - 6) = -1(x)

⇒ 3y - 18 = -x

⇒ x + 3y = 18.

Hence, equation of line AB is x + 3y = 18.

(ii) From figure,

CD is perpendicular to AB.

Slope of AB (m₁) = $-\dfrac{1}{3}$

Let slope of CD be m₂.

⇒ m₁ × m₂ = -1

⇒ $-\dfrac{1}{3} \times m_2 = -1$

⇒ m₂ = 3.

Equation of CD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 10 = 3(x - 2)

⇒ y - 10 = 3x - 6

⇒ 3x - y - 6 + 10 = 0

⇒ 3x - y + 4 = 0.

Hence, equation of CD is 3x - y + 4 = 0.

(iii) From figure,

E lies on x-axis. Let co-ordinates of E be (a, 0).

Since, E lies on line AB it will satisfy its equation.

Substituting value of E in AB we get,

⇒ a + 3(0) = 18

⇒ a = 18.

E = (18, 0)

D lies on x-axis. Let co-ordinates of D be (b, 0).

Since, D lies on CD it will satisfy its equation.

Substituting value of D in CD we get,

3b - 0 + 4 = 0

3b = -4

b = $-\dfrac{4}{3}$ .

D = $(-\dfrac{4}{3}, 0)$ .

Hence, co-ordinates of E = (18, 0) and D = $(-\dfrac{4}{3}, 0)$ .

Question 29

Find the equation of the line through the points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.

Answer

By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of AB = $\dfrac{2 - 3}{0 - (-1)} = \dfrac{-1}{1}$ = -1.

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -1[x - (-1)]

⇒ y - 3 = -1(x + 1)

⇒ y - 3 = -x - 1

⇒ x + y = -1 + 3

⇒ x + y = 2.

Slope of BC = $\dfrac{1 - 2}{1 - 0} = \dfrac{-1}{1}$ = -1.

Since, slope of AB = slope of BC.

Hence, proved that A, B and C are collinear and equation of AB is x + y = 2.

Question 30

In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.

(i) Write the co-ordinates of A.

(ii) Find the length of AB and AC.

(iii) Find the ratio in which Q divides AC.

(iv) Find the equation of the line AC.

In the figure, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively. (i) Write the co-ordinates of A. (ii) Find the length of AB and AC. Equation of a Line, Concise Mathematics Solutions ICSE Class 10.

Answer

(i) From graph,

The co-ordinates of A are (4, 0).

(ii) By distance formula,

Distance = $\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$

Substituting values we get,

$AB = \sqrt{(3 - 0)^2 + (-2 - 4)^2} \\[1em] = \sqrt{3^2 + (-6)^2} \\[1em] = \sqrt{9 + 36} \\[1em] = \sqrt{45} = 3\sqrt{5} \text{ units}. \\[1em] AC = \sqrt{(-4 - 0)^2 + (-2 - 4)^2} \\[1em] = \sqrt{(-4)^2 + (-6)^2} \\[1em] = \sqrt{16 + 36} \\[1em] = \sqrt{52} = 2\sqrt{13} \text{ units}.$

Hence, length of AB = $3\sqrt{5}$ units and AC = $2\sqrt{13}$ units.

(iii) From figure,

Q lies on y-axis.

∴ x co-ordinate of Q = 0.

Let co-ordinate of Q are (0, a).

Let ratio in which Q divides AC be k : 1.

By section-formula,

$Q = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \text{For x co-ordinate} \\[1em] \Rightarrow 0 = \dfrac{k \times -2 + 1 \times 4}{k + 1} \\[1em] \Rightarrow 0 = -2k + 4 \\[1em] \Rightarrow 2k = 4 \\[1em] \Rightarrow k = 2.$

k : 1 = 2 : 1.

Hence, Q divides AC in the ratio 2 : 1.

(iv) Slope of AC = $\dfrac{-4 - 0}{-2 - 4} = \dfrac{-4}{-6} = \dfrac{2}{3}$ .

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{2}{3}$ (x - 4)

⇒ 3y = 2x - 8

⇒ 2x - 3y = 8.

Hence, equation of AC is 2x - 3y = 8.

Question 31

A line AB meets X-axis at A and Y-axis at B. P(4, -1) divides AB in the ratio 1 : 2.

(i) Find the co-ordinates of A and B.

(ii) Find the equation of the line through P and perpendicular to AB.

Answer

(i) As A lies on x-axis let its co-ordinates be (a, 0) and B lies on y-axis so, co-ordinates = (0, b).

By section-formula,

$\Rightarrow P = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (4, -1) = \Big(\dfrac{1 \times 0 + 2 \times a}{1 + 2}, \dfrac{1 \times b + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (4, -1) = \Big(\dfrac{2a}{3}, \dfrac{b}{3}\Big) \\[1em] \Rightarrow \dfrac{2a}{3} = 4 \text{ and } \dfrac{b}{3} = -1 \\[1em] \Rightarrow a = \dfrac{12}{2} = 6 \text{ and } b = -3.$

Hence, A = (6, 0) and B = (0, -3).

(ii) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{-3 - 0}{0 - 6} = \dfrac{-3}{-6} = \dfrac{1}{2}$ .

Let slope of perpendicular line be m₁.

Since, slope of product of perpendicular lines = -1.

⇒ m₁ × Slope of AB = -1

⇒ m₁ $\times \dfrac{1}{2} = -1$

⇒ m₁ = -2.

By point-slope from,

Equation of line passing through P and slope = -2 is :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = -2(x - 4)

⇒ y + 1 = -2(x - 4)

⇒ y + 1 = -2x + 8

⇒ 2x + y = 7.

Hence, equation of required line is 2x + y = 7.

Chapter 14

Equation of a Line

Class - 10 Concise Mathematics Selina

Exercise 14(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3(i)

Question 3(ii)

Question 4

Question 5

Question 6

Question 7(i)

Question 7(ii)

Question 8(i)

Question 8(ii)

Question 9

Question 10

Question 11

Question 12

Exercise 14(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Exercise 14(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Exercise 14(D)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)