Find the equation of a line which is perpendicular to the line√3x �

1.	Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.
Answer» Given: The equation is perpendicular to √3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis. The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0 It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis. This means that the line passes through (0,-4). So, Let us substitute the values in the equation x + √3y + λ = 0, we get 0 – √3 (4) + λ = 0 λ = 4√3 Now, substitute the value of λ back, we get x + √3y + 4√3 = 0 ∴ The required equation of line is x + √3y + 4√3 = 0.

Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.

Answer»

Given:

The equation is perpendicular to √3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0

It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation x + √3y + λ = 0, we get

0 – √3 (4) + λ = 0
λ = 4√3

Now, substitute the value of λ back, we get

x + √3y + 4√3 = 0

∴ The required equation of line is x + √3y + 4√3 = 0.