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 = \(\frac{7}{3}\)
Hence, the equation of the required line is
\(\Big(2+\frac{7}{3}\Big)x+0+11-8\times\frac{7}{3}=0\)
⇒ 13x – 23 = 0
