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

Any line perpendicular to line

$\sqrt{3}x-y+5=0$

Will have the slope $\frac{-1}{m}$

Where,

$y=mx+c$

$y=\sqrt{3}x+5$

$m=\sqrt{3}$

Points is $\left(x_1{y}_1\right)=(3,3)$

$y-y_1=\frac{-1}{m}(x-x_1)$

$y-3=\frac{-1}{\sqrt{3}}(x-3)$

$x+\sqrt{3}y+6=0$

Point can be $(-3,-3)$

Then, equation is

$x+\sqrt{3}y-6$

$\therefore x+\sqrt{3}y\pm 6$