Chapter 14 - Equation of a Line Exercise Ex. 14(C)

Question 1

Find the equation of a line whose:

y-intercept = 2 and slope = 3.

Solution 1

Given, y-intercept = c = 2 and slope = m = 3.

Substituting the values of c and m in the equation y = mx + c, we get,

y = 3x + 2, which is the required equation.

Question 2

Find the equation of a line whose:

y-intercept = -1 and inclination = 45^o.

Solution 2

Given, y-intercept = c = -1 and inclination = 45^o.

Slope = m = tan 45^o = 1

Substituting the values of c and m in the equation y = mx + c, we get,

y = x - 1, which is the required equation.

Question 3

Find the equation of the line whose slope is  and which passes through (-3, 4).

Solution 3

Given, slope = 

The equation passes through (-3, 4) = (x₁, y₁)

Substituting the values in y - y₁ = m(x - x₁), we get,

y - 4 = (x + 3)

3y - 12 = -4x - 12

4x + 3y = 0, which is the required equation.

Question 4

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

Solution 4

Slope of the line = tan 60^o = 

The line passes through the point (5, 4) = (x₁, y₁)

Substituting the values in y - y₁ = m(x - x₁), we get,

y - 4 = (x - 5)

y - 4 = x - 5

y =x + 4 - 5, which is the required equation.

Question 5

Find the equation of the line passing through:

(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)

Solution 5

(i) Let (0, 1) = (x₁, y₁) and (1, 2) = (x₂, y₂)

The required equation of the line is given by:

y - y₁ = m(x - x₁)

y - 1 = 1(x - 0)

y - 1 = x

y = x + 1

(ii) Let (-1, -4) = (x₁, y₁) and (3, 0) = (x₂, y₂)

The required equation of the line is given by:

y - y₁ = m(x - x₁)

y + 4 = 1(x + 1)

y + 4 = x + 1

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.

Solution 6

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

(i) Gradient of PQ = 

(ii) The equation of the line PQ is given by:

y - y₁ = m(x - x₁)

y - 6 = (x - 2)

5y - 30 = x - 2

5y = x + 28

(iii) Let the line PQ intersects the x-axis at point A (x, 0).

Putting y = 0 in the equation of the line PQ, we get,

0 = x + 28

x = -28

Thus, the co-ordinates of the point where PQ intersects the x-axis are A (-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.

Solution 7

(i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).

Slope =

The equation of the line AB is given by:

y - y₁ = m(x - x₁)

y + 1 = -1(x - 2)

y + 1 = -x + 2

x + y = 1

(ii) Let the line AB intersects the y-axis at point (0, y).

Putting x = 0 in the equation of the line, we get,

0 + y = 1

y = 1

Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).

Question 8

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

Solution 8

Slope of line AB = tan 45^o = 1

The line AB passes through P (3, 4). So, the equation of the line AB is given by:

y - y₁ = m(x - x₁)

y - 4 = 1(x - 3)

y - 4 = x - 3

y = x + 1

Slope of line CD = tan 60^o = 

The line CD passes through P (3, 4). So, the equation of the line CD is given by:

y - y₁ = m(x - x₁)

y - 4 =  (x - 3)

y - 4 = x - 3 

y = x + 4 - 3 

Question 9

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

Solution 9

Question 10

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

Solution 10

Since, ABCD is a parallelogram,

B = 180^o - 60^o = 120^o

Slope of BC = tan 120^o = tan (90^o + 30^o) = cot30^o = 

Equation of the line BC is given by:

y - y₁ = m(x - x₁)

y - 5 = (x - 7)

y - 5 = x - 7

y = x + 5 - 7

Since, CD || AB and AB || x-axis, slope of CD = Slope of AB = 0

Equation of the line CD is given by:

y - y₁ = m(x - x₁)

y - 5 = 0(x - 7)

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.

Solution 11

The given equations are:

x + 2y = 7 ....(1)

x - y = 4 ....(2)

 

Subtracting (2) from (1), we get,

3y = 3

y = 1

 

From (2), x = 4 + y = 4 + 1 = 5

 

The required line passes through (0, 0) and (5, 1).

 

Required equation of the line is given by:

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.

Solution 12

Given, the co-ordinates of vertices A, B and C of a triangle ABC are (4, 7), (-2, 3) and (0, 1) respectively.

Let AD be the median through vertex A.

Co-ordinates of the point D are

Slope of AD = 

The equation of the median AD is given by:

y - y₁ = m(x - x₁)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

The slope of the line which is parallel to line AC will be equal to the slope of AC.

Slope of AC = 

The equation of the line which is parallel to AC and passes through B is given by:

y - 3 = (x + 2)

2y - 6 = 3x + 6

2y = 3x + 12

Question 13

A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.

Solution 13

Slope of BC = 

Slope of line perpendicular to BC = 

The equation of the line through A and perpendicular to BC is given by:

y - y₁ = m(x - x₁)

y - 3 = 1(x - 0)

y - 3 = x

y = x + 3

Question 14

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

Solution 14

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

Slope of AB = 

Slope of equation perpendicular to AB = 

The equation of the perpendicular drawn through C onto AB is given by:

y - y₁ = m(x - x₁)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

Question 15

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

Solution 15

(i) When x-intercept = 5, corresponding point on x-axis is (5, 0)

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

Let (x₁, y₁) = (5, 0) and (x₂, y₂) = (0, 3)

Slope = 

 

The required equation is:

y - y₁ = m(x - x₁)

y - 0 = (x - 5)

5y = -3x + 15

3x + 5y = 15

 

(ii) When x-intercept = -4, corresponding point on x-axis is (-4, 0)

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

Let (x₁, y₁) = (-4, 0) and (x₂, y₂) = (0, 6)

Slope = 

 

The required equation is:

y - y₁ = m(x - x₁)

y - 0 = (x + 4)

2y = 3x + 12

 

(iii) When x-intercept = -8, corresponding point on x-axis is (-8, 0)

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

Let (x₁, y₁) = (-8, 0) and (x₂, y₂) = (0, -4)

Slope = 

 

The required equation is:

y - y₁ = m(x - x₁)

y - 0 =  (x + 8)

2y = -x - 8

x + 2y + 8 = 0

Question 16

Find the equation of the line whose slope is  and x-intercept is 6.

Solution 16

Since, x-intercept is 6, so the corresponding point on x-axis is (6, 0).

Slope = m = 

Required equation of the line is given by:

y - y₁ = m(x - x₁)

y - 0 =  (x - 6)

6y = -5x + 30

5x + 6y = 30

Question 17

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

Solution 17

Since, x-intercept is 5, so the corresponding point on x-axis is (5, 0).

The line also passes through (-3, 2).

Slope of the line = 

Required equation of the line is given by:

y - y₁ = m(x - x₁)

y - 0 =  (x - 5)

4y = -x + 5

x + 4y = 5

Question 18

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

Solution 18

Since, y-intercept = 5, so the corresponding point on y-axis is (0, 5).

The line passes through (1, 3).

Slope of the line = 

Required equation of the line is given by:

y - y₁ = m(x - x₁)

y - 5 = -2(x - 0)

y - 5 = -2x

2x + y = 5

Question 19

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

Solution 19

Let AB and CD be two equally inclined lines.

For line AB:

Slope = m = tan 45^o = 1

(x₁, y₁) = (-2, 0)

Equation of the line AB is:

y - y₁ = m(x - x₁)

y - 0 = 1(x + 2)

y = x + 2

For line CD:

Slope = m = tan (-45^o) = -1

(x₁, y₁) = (-2, 0)

Equation of the line CD is:

y - y₁ = m(x - x₁)

y - 0 = -1(x + 2)

y = -x - 2

x + y + 2 = 0

Question 20

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.

Solution 20

(i)

 

 

(ii)

 

 

(iii)

Question 21

Write down the equation of the line whose gradient is  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.

Solution 21

Given, P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.

Co-ordinates of point P are

Slope = m = (Given)

Thus, the required equation of the line is

y - y₁ = m(x - x₁)

y + 2 = (x - 10)

5y + 10 = -2x + 20

2x + 5y = 10

Question 22

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.

Solution 22

(i) Co-ordinates of the centroid of triangle ABC are

 

(ii) Slope of AB = 

Slope of the line parallel to AB = Slope of AB = -1

 

Thus, the required equation of the line is

y - y₁ = m(x - x₁)

Question 23

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.

Solution 23

Given, AP: CP = 2: 3

Co-ordinates of P are

Slope of BP = 

Required equation of the line passing through points B and P is

y - y₁ = m(x - x₁)

y - 1 = 0(x - 3)

y = 1