Find the equation of the line passing through the point of intersectio

1.	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: The equations, 2x – 7y + 11 = 0 and x + 3y – 8 = 0 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 Now, substitute the value of λ back in equation, we get 0 + (− 7 − 6)y + 11 + 16 = 0 13y − 27 = 0 ∴ The equation of the required line is 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 = 7/3 Now, substitute the value of λ back in equation, we get (2 + 7/3)x + 0 + 11 – 8(7/3) = 0 13x – 23 = 0 ∴ The equation of the required line is 13x – 23 = 0

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:

The equations, 2x – 7y + 11 = 0 and x + 3y – 8 = 0

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

Now, substitute the value of λ back in equation, we get

0 + (− 7 − 6)y + 11 + 16 = 0

13y − 27 = 0

∴ The equation of the required line is 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 = 7/3
Now, substitute the value of λ back in equation, we get

(2 + 7/3)x + 0 + 11 – 8(7/3) = 0

13x – 23 = 0

∴ The equation of the required line is 13x – 23 = 0