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Let $P \equiv(-3,0), Q \equiv(0,0)$ and $R \equiv(3,3 \sqrt{3})$ be three points. Then the equation of the bisector of the angle $\mathrm{PQR}$ is
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The correct answer is:
$\sqrt{3} x+y=0$
Slope of $\mathrm{QR}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}$ i.e., $\theta=60^{\circ}$ Clearly, $\angle \mathrm{PQR}=120^{\circ}$ $\mathrm{OQ}$ is the angle bisector of the angle PQR, so line OQ makes $120^{\circ}$ with the positive direction of $\mathrm{X}$-axis.
Therefore, equation of the bisector of $\angle \mathrm{PQR}$ is $y=\tan 120^{\circ} x \Rightarrow y=-\sqrt{3} x \Rightarrow \sqrt{3} x+y=0$
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