Question -
Answer -
Given:
The equations, 2x – 7y+ 11 = 0 and x + 3y – 8 = 0
The equation of thestraight 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 requiredline is parallel to the x-axis. So, the coefficient of x should be zero.
2 + λ = 0
λ = -2
Now, substitute thevalue of λ back in equation, we get
0 +(− 7 − 6)y + 11 + 16 = 0
13y − 27 = 0
∴ The equation of therequired line is 13y − 27 = 0
(ii) The requiredline is parallel to the y-axis. So, the coefficient of y should be zero.
-7 + 3λ = 0
λ = 7/3
Now, substitute thevalue of λ back in equation, we get
(2 + 7/3)x + 0 + 11 –8(7/3) = 0
13x – 23 = 0
∴ The equation of therequired line is 13x – 23 = 0