Let $P \equiv(-5,0), Q \equiv(0,0)$ and $R \equiv(2,2 \sqrt{3})$ be three points. Then the equation of the…

Let $P \equiv(-5,0), Q \equiv(0,0)$ and $R \equiv(2,2 \sqrt{3})$ be three points. Then the equation of the bisector of the angle $P Q R$ is
  1. $x-\frac{\sqrt{3}}{2} y=0$
  2. $\frac{\sqrt{3}}{2} x-y=0$
  3. $x+\sqrt{3} y=0$
  4. $\sqrt{3} x+y=0$

Solution


Slope of $\mathrm{QR}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}$ i.e., $\theta=60^{\circ}$ Clearly, $\angle \mathrm{PQR}=120^{\circ}$ OQ is the angle bisector of the angle PQR , so line OQ makes $120^{\circ}$ with the positive direction of 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 2024 (11 May Shift 1)

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