Question

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

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

Answer 1

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

Given: The coordinates of points $P,Q,R$ are $(-1,0),(0,0),(3,3\sqrt{3})$ respectively.


$\text{Slope of QR}=\frac{y_2-y_1}{x_2-x_1}=\frac{3\sqrt{3}}{3}\Rightarrow tan\theta =\sqrt{3}\Rightarrow \theta =\frac{\pi }{3}\Rightarrow \angle RQX=\frac{\pi }{3}\therefore \angle RQP=\pi =\frac{\pi }{3}=\frac{2\pi }{3}$
Let $QM$ bisects the $\angle PQR.$
$\therefore$ Slope of the line $QM=tan\frac{2\pi }{3}=-\sqrt{3}$
$\therefore$ Equation of line $QM$ is $(y-0)=-\sqrt{3}(x-0)$
$\Rightarrow y=-\sqrt{3}x\Rightarrow \sqrt{3}x+y=0$