Given that, $\quad P=(-1,0)$
$Q=(0,0)$
and $R=(3,3 \sqrt{3})$
In $\triangle A Q R$,
$\tan \phi=\frac{3 \sqrt{3}}{3}$
$\Rightarrow \tan \phi=\sqrt{3}$
$\Rightarrow \tan \phi=\tan 60^{\circ}$
$\Rightarrow \phi=\frac{\pi}{3}$
Slope of line $Q R$
$m_{1}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}$
$\Rightarrow m_{1}=\sqrt{3}$
Slope of line QP
$m_{2}=\frac{0-0}{-1-0}=0$
$\Rightarrow m_{2}=0$
Let the angle between $P Q R$ is $\theta$, then
$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} \cdot m_{2}}\right|$
$\Rightarrow \tan \theta =\left|\frac{\sqrt{3}-0}{1+0}\right|=\sqrt{3}$ or $-\sqrt{3}$
$\Rightarrow \tan \theta=\tan \frac{2 \pi}{3}$
$\Rightarrow \theta=\frac{2 \pi}{3}$
[$\because(\tan \theta \neq \sqrt{3})$ from figure]
Hence, $\angle R Q B=\frac{\theta}{2}=\frac{2 \pi}{6}=\frac{\pi}{3}$
So, the angle bisector making an angle with
$(+)$ ive $x$ -axis, is $\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$
The equation of angle bisector is
$\Rightarrow y =m x=\tan \frac{2 \pi}{3} \cdot x$
$\Rightarrow y=-\sqrt{3} \cdot x$
$\Rightarrow \sqrt{3} \cdot x +y=0$
