Question -
Answer -
Given:
The equation isperpendicular to √3x – y + 5 = 0 equation and cuts off anintercept of 4 units with the negative direction of y-axis.
The line perpendicularto √3x – y + 5 = 0 is x + √3y + λ = 0
It is given that theline x + √3y + λ = 0 cuts off an intercept of 4 units with thenegative direction of the y-axis.
This means that theline passes through (0,-4).
So,
Let us substitute thevalues in the equation x + √3y + λ = 0, we get
0 – √3 (4) + λ = 0
λ = 4√3
Now, substitute thevalue of λ back, we get
x + √3y + 4√3 = 0
∴ The required equationof line is x + √3y + 4√3 = 0.