Equation of a Line — Exercise 14(A)

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

1. -6

2. 6

3. -8

4. 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.

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

1. 2

2. -2

3. $\dfrac{1}{2}$

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

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.

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

1. no

2. yes

3. neither true nor false

4. 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.

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 :

1. 3

2. 4

3. 2

4. -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.

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

1. 2

2. 4

3. -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.

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.

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.

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.

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.

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 = $2\dfrac{3}{11}.$

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).

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

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.

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.

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.

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.

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.

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.

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.