Equation of a Line — Exercise 14(D)

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

1. 2x - y = 5

2. x + y = 5

3. x - y = 5

4. 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.

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

1. equal

2. having same sign

3. equal in magnitude with opposite signs

4. 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,

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.

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

1. 2x - y = 3

2. 2x - y + 3 = 0

3. 2x + y = 3

4. 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.

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

1. x + y = 3

2. x - y = 3

3. y - x = 3

4. 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.

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

1. x - y = 12

2. x - y = 0

3. x + y = 6

4. 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.

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 = $-\dfrac{5}{4}$.

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°.

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.

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.

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.

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.

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

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 $\dfrac{3}{2}$.

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 $x - \dfrac{y}{2} + 3 = 0 \text{ is } -\dfrac{1}{2}.$

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.

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.

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.

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

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°.

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.

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.

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.

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 (-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.

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.

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:

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.

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).

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

⇒ 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.

(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:

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.

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.

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).

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).

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

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.