# EQUATION OF LINE PASSING THROUGH INTERSECTION OF TWO LINES

Example 1 :

Find the equation of the straight line passing through point of intersection of the lines

2x + y – 3 = 0

5x + y – 6 = 0

and parallel to the line joining the points (1, 2) and (2, 1).

Solution :

First let us find the point of intersection of two lines

2x + y – 3 = 0 ----(1)

5x + y – 6 = 0 ----(2)

(1) - (2) :

2x + y – 3 - (5x + y – 6) = 0

2x - 5x + y - y - 3 + 6 = 0

-3x + 3 = 0

x = 1

By applying the value of x in (1), we get

2(1) + y – 3 = 0

2 + y – 3 = 0

-1 + y = 0

y = 1

The point of intersection of two lines (1, 1)

Slope of the line joining two points (1, 2) and (2, 1)

m = (y- y1)/(x- x1)

m = (1 - 2)/(2 - 1)

m = -1

Equation of required line :

(y - y1) = m(x - x1)

(y - 1) = -1(x – 1)

y - 1 = -x + 1

x + y – 1 – 1 = 0

x + y – 2 = 0

So, equation of the required line is x + y - 2 = 0.

Example 2 :

Find the equation of the straight line passing through the point of intersection of the

5x – 6y = 1

3x + 2y + 5 = 0

and is perpendicular to the straight line

3x – 5y + 11 = 0

Solution :

First let us find the point of intersection of two lines

5x - 6y = 1 ----(1)

3x + 2y = -5 ----(2)

(1) + 3(2) ==>

5x - 6y + 9x + 6y = 1 - 15

14x = -14

x = -1

By applying the value of x in (1), we get

5(-1) – 6y = 1

-5 – 6y = 1

-6y = 1 + 5

y = -1

The point of intersection of two lines (-1, -1).

Slope of the line 3x – 5y + 11 = 0

m = -3/(-5)

m = 3/5

Equation of required line :

y - y1 = m(x - x1)

y – (-1) = (3/5)(x – (-1))

y + 1 = (3/5)(x + 1)

5(y + 1) = 3(x + 1)

5y + 5 = 3x + 3

3x – 5y – 2 = 0

So, equation of the required line is 3x – 5y – 2 = 0.

Example 3 :

Find the equation of the straight line joining the point of intersection of the lines

3x – y + 9 = 0

x + 2y = 4

and the point of intersection of the lines

2x + y – 4 = 0

x – 2y + 3 = 0

Solution :

Point of intersection of the lines

3x – y = -9 ----(1)

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

2(1) + (2) ==>

6x – 2y + x + 2y = -18 + 4

7x = -14

x = -2

By applying the value of x in (1), we get

3(-2) – y = -9

-6 – y = -9

-y = -9 + 6

y = 3

So, point of intersection of first two lines is (-2, 3).

Point of intersection of the lines

2x + y = 4 ----(3)

x - 2y = -3 ----(4)

2(1) + (2) ==>

4x + 2y + (x - 2y) = 8 - 3

5x = 5

x = 1

Substitute x = 1 in (3).

2(1) + y = 4

2 + y = 4

y = 2

Point of intersection of first two lines is (2, 1).

Equation of the required line will pass through the points (-2, 3) and (2, 1)

(y - y1)/(y- y1) = (x - x1)/(x- x1)

(y - 3)/(1 - 3) = (x - (-2))/(2 - (-2))

(y - 3)/(-2) = (x + 2)/4

4(y - 3) = -2(x + 2)

4y – 12 = -2x – 4

2x + 4y – 12 + 4 = 0

2x + 4y – 8 = 0

x + 2y - 4 = 0

So, the equation of the required line is x + 2y – 4 = 0.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

## Recent Articles 1. ### Power Rule of Logarithms

Oct 04, 22 11:07 PM

Power Rule of Logarithms - Concept - Solved Problems

2. ### Product Rule of Logarithms

Oct 04, 22 11:07 PM

Product Rule of Logarithms - Concept - Solved Problems