Q. Let $P = ( - 1,\ 0)\mathcal{,\ Q} = (0,\ 0)$ and $R = (3,\ 3\sqrt{3})$be three points. The equation of the bisector of the angle $P\mathcal{Q}R$ is
A) $\sqrt{3}x + y = 0$
B) $x + \frac{\sqrt{3}}{2}y = 0$
C) $\frac{\sqrt{3}}{2}x + y = 0$
D) $x + \sqrt{3}y = 0$
Explanation
(a)
$Now,\ slope\ of\ \mathcal{Q}R = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3} = \tan\theta$
$\Rightarrow \ \ \theta = \frac{\pi}{3}$
[image:10_image36.emf]
$\therefore\ The\ angle\ between\ P\mathcal{Q}R\ is\frac{2\pi}{3}\ ,\ so\ the\ line\ \mathcal{Q}M\ makes\ an\ angle\frac{2\pi}{3}\ from\ positive\$
direction of $x$-axis.
$Slope\ of\ the\ line\ \mathcal{Q}M = \tan{\frac{2\pi}{3} = - \sqrt{3}}$
Hence, equation of line $\mathcal{Q}M$ is $y = - \sqrt{3}x$
or $\sqrt{3}x + y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.