Question

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

Answer 1

The line perpendicular to $\sqrt{3}x-y+5=0$ is $x+\sqrt{3}y+\lambda =0$.

It is given that the line $x+\sqrt{3}y+\lambda =0$ cuts off an intercept of 4 units with the negative direction of the y-axis.
This means that the line passes through $\left(0,-4\right)$.

$$\begin{gathered} \therefore 0-\sqrt{3}\times 4+\lambda =0 \\ \Rightarrow \lambda =4\sqrt{3} \end{gathered}$$

Substituting the value of $\lambda$, we get $x+\sqrt{3}y+4\sqrt{3}=0$, which is the equation of the required line.