Q. 35.0( 4 Votes )

# Find the equation of the line passing through the point of intersection of 2x – 7y + 11 = 0 and x + 3y – 8 = 0 and is parallel to (i) x = axis (ii) y-axis.

Answer :

__Given:__

2x – 7y + 11 = 0 and x + 3y – 8 = 0

__To find:__

The equation of the line passing through the point of intersection of 2x – 7y + 11 = 0 and x + 3y – 8 = 0 and is parallel to (i) x = axis (ii) y-axis.

Explanation:

The equation of the straight line passing through the points of intersection of 2x − 7y + 11 = 0 and x + 3y − 8 = 0 is given below:

2x − 7y + 11 + λ(x + 3y − 8) = 0

⇒ (2 + λ)x + (− 7 + 3λ)y + 11 − 8λ = 0

(i) The required line is parallel to the x-axis. So, the coefficient of x should be zero.

∴ 2 + λ = 0

⇒ λ = -2

Hence, the equation of the required line is

0 + (− 7 − 6)y + 11 + 16 = 0

⇒ 13y − 27 = 0

(ii) The required line is parallel to the y-axis. So, the coefficient of y should be zero.

∴ -7 + 3λ = 0

⇒ λ

Hence, the equation of the required line is

⇒ 13x – 23 = 0

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