Question

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

• A. $x+\sqrt{3}y=0$
• B. $\sqrt{3}x+y=0$
• C. $x+\frac{\sqrt{3}}{2}y=0$
• D. $\frac{\sqrt{3}}{2}x+y=0$

Answer 1

The correct option is B $\sqrt{3}x+y=0$

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

$\Rightarrow$ $QM$ is bisector of $\angle PQR$

$\Rightarrow$ $\tan \theta =\frac{3\sqrt{3}}{3}$

$\Rightarrow$ $\tan \theta =\sqrt{3}$

$\therefore$ $\theta ={\tan }^{-1}\left(\sqrt{3}\right)={60}^o$

$\Rightarrow \theta +\angle PQR={180}^o\Rightarrow {60}^0+\angle PQR={180}^0$

$\therefore$ $\angle PQR={120}^o$

Slop of the line $QM$ is $\tan {120}^o=-\sqrt{3}$

we know that equation of a line with slope $m$ and passing through $(x_1,y_1)$ is $(y-y_1)=m(x-x_1)$

Hence, equation of line bisector is $y=-\sqrt{3}x$

$\therefore$ $\sqrt{3}x+y=0$

$\therefore$ The equation of the bisector of the bisector of the $\angle PQR$ is $\sqrt{3}x+y=0$.