Equation of a Line — Exercise 14(C)

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

1. x-axis

2. y-axis

3. x + y = 0

4. 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.

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

1. x + y = 2

2. y - x = 2

3. x - y = 2

4. 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.

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

1. (0, -4)

2. (4, 0)

3. (4, 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.

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

1. x - y = 7

2. y - x = 7

3. x + y + 7 = 0

4. 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.

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

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

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

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

4. 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.

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.

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.

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 = $\sqrt{3}x + 4 - 5\sqrt{3}$.

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.

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).

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).

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 = $\sqrt{3}x - 3\sqrt{3} + 4$.

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.

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.

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.

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.

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.

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.

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.

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.

Find the equation of the line, whose :

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

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

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.

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.

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.

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.

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,

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.

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.

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).

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.

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.

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.