Let $p=(-1,0)$, $q=(0,0)$ and $r=(3,\sqrt 3)$ be three points. Find the equation of angle bisector of $PQR$

Question

Slope of $PQ$ is $0$

Then equation of $PQ$ is $$y-0=0$$ $$y=0$$ Slope of $QR$ is $\sqrt 3$ $$y-0=\sqrt 3(x-0)$$ $$\sqrt 3x-y=0$$ Equation of angle bisector $$\frac y1 =\frac{\sqrt 3 x-y}{2}$$ $$2y=\sqrt 3x-y$$ $$\sqrt 3x-3y =0$$ But the answer is $\sqrt 3x+y=0$ what’s going wrong?

Answer 1

The slope of $QR$ is $\frac{\sqrt 3}{3}$, that's your mistake.

Answer 2

It is very simple making a sketch, and reasoning with polar angles:

Since the slope of the line $(QR)$ is $\frac{\sqrt 3}3$, its polar angle is $\frac\pi 6$, so the polar angle of the bissectrix of angle $\widehat{xQR}$ is $\frac\pi{12}$, and the polar angle of the bissectrix of angle $\widehat{PQR}$ is $\frac\pi{12}+\frac\pi 2$.

Therefore, the slope of this bissectrix is $$\tan\bigl(\tfrac\pi{12}+\tfrac\pi 2\bigr)=-\frac 1{\tan\frac\pi{12}}=-(2+\sqrt 3),$$ so its equation is $$(2+\sqrt 3)x+y=0.$$