Question

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

• A. $x+\sqrt{3}y=0$
• B. $\sqrt{3}x+y=0$
• C. $x-\sqrt{3}y=0$
• D. $\sqrt{3}x-y=0$

Answer 1

Answer: (B)

Given that, $P=(-1,0)$
$Q=(0,0)$
and $R=(3,3\sqrt{3})$

In $△AQR$,
$\tan \varphi =\frac{3\sqrt{3}}{3}$
$\Rightarrow \tan \varphi =\sqrt{3}$
$\Rightarrow \tan \varphi =\tan 60^{\circ }$
$\Rightarrow \varphi =\frac{\pi }{3}$
Slope of line $QR$
$m_1=\frac{3\sqrt{3}-0}{3-0}=\sqrt{3}$
$\Rightarrow m_1=\sqrt{3}$
Slope of line QP
$m_2=\frac{0-0}{-1-0}=0$
$\Rightarrow m_2=0$
Let the angle between $PQR$ is $\theta$, then
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1\cdot m_2}\right|$
$\Rightarrow \tan \theta =\left|\frac{\sqrt{3}-0}{1+0}\right|=\sqrt{3}$ or $-\sqrt{3}$
$\Rightarrow \tan \theta =\tan \frac{2\pi }{3}$
$\Rightarrow \theta =\frac{2\pi }{3}$
[$\because (\tan \theta \ne \sqrt{3})$ from figure]
Hence, $\angle RQB=\frac{\theta }{2}=\frac{2\pi }{6}=\frac{\pi }{3}$
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
$\Rightarrow y=mx=\tan \frac{2\pi }{3}\cdot x$
$\Rightarrow y=-\sqrt{3}\cdot x$
$\Rightarrow \sqrt{3}\cdot x+y=0$