Question

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

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

Answer 1

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

The slope of QR is $\frac{(3\sqrt{3}-0)}{3-0}=\sqrt{3}$
i.e., $\theta ={60}^0$ Clearly, $\angle PQR={120}^0$ OQ is the angle bisector of the angle. So, line OQ makes ${120}^0$ with the positive direction of the x-axis.
Therefore, the equation of the bisector of $\angle PQRisy=xtan{120}^{0}ory=-\sqrt{3}x,i.e.,\sqrt{3}x+y=0.$