Question

Find the equation of a line perpendicular to the line $\sqrt{3}x-y+5=0$ and at a distance of 3 units from the origin.

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 $x+\sqrt{3}y+\lambda =0$ is at a distance of 3 units from the origin.

$$\begin{gathered} \therefore \left|\frac{\lambda }{\sqrt{1+3}}\right|=3 \\ \Rightarrow \lambda =\pm 6 \end{gathered}$$

Substituting the value of $\lambda$, we get $x+\sqrt{3}y\pm 6=0$, which is equation of the required line.