Let \(P \equiv(-1,0), Q \equiv(0,0)\) and \(R=(3,3)\) be three points. The equation of the bisector of the angle PQR is

\(\begin{array}{ll}\text { Given that, } & \mathrm{P}=(-1,0) \\ \text { and } & \mathrm{Q}=(0,0) \\ & \mathrm{R}=(3,3 \sqrt{3})\end{array}\)



In \(\triangle \mathrm{AQR}\),
\(
\begin{aligned}
\tan \phi &=\frac{3 \sqrt{3}}{3} \\
\Rightarrow \quad \tan \phi &=\sqrt{3}
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow & & \tan \phi &=\tan 60^{\circ} \\
\Rightarrow & \phi &=\frac{\pi}{3}
\end{aligned}
\)
Slope of line QR
\(\mathrm{m}_{1}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}\) \(\Rightarrow \quad \mathrm{m}_{1}=\sqrt{3}\) Slope of line \(\mathrm{QP}\) \(\mathrm{m}_{2}=\frac{0-0}{-1-0}=0 \Rightarrow \mathrm{m}_{2}=0\)
Slope of line QP
\(
\mathrm{m}_{2}=\frac{0-0}{-1-0}=0 \Rightarrow \mathrm{m}_{2}=0
\)
Let the angle between PQR is \(\theta\), then
\(
\begin{aligned}
\tan \theta &=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \cdot \mathrm{m}_{2}}\right| \\
\Rightarrow \quad \tan \theta &=\left|\frac{\sqrt{3}-0}{1+0}\right|=\sqrt{3} \text { or }-\sqrt{3} \\
\Rightarrow \quad & \tan \theta=\tan \frac{2 \pi}{3} \Rightarrow \theta=\frac{2 \pi}{3} \\
& {[\because(\tan \theta \neq \sqrt{3}) \text { from figure }] }
\end{aligned}
\)
Hence, \(\quad \angle \mathrm{RQB}=\frac{\theta}{2}=\frac{2 \pi}{6}=\frac{\pi}{3}\)
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
\(
\begin{array}{ll}
\Rightarrow & \mathrm{y}=\mathrm{mx}=\tan \frac{2 \pi}{3} \cdot \mathrm{x} \\
\Rightarrow & \mathrm{y}=-\sqrt{3} \cdot \mathrm{x} \Rightarrow \sqrt{3} \cdot \mathrm{x}+\mathrm{y}=0
\end{array}
\)