Let $P \equiv(-3,0), Q \equiv(0,0)$ and $R \equiv(3,3 \sqrt{3})$ be three points. Then the equation of the…

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
  1. $\frac{\sqrt{3}}{2} x+y=0$
  2. $x+\sqrt{3} y=0$
  3. $\sqrt{3} x+y=0$
  4. $x+\frac{\sqrt{3}}{2} y=0$

Solution

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 $y=\tan 120^{\circ} x \Rightarrow y=-\sqrt{3} x \Rightarrow \sqrt{3} x+y=0$

Asked in: MHT CET 2023 (14 May Shift 2)

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