Straight Line Question 37

Question: Let $ P(-1,0), $ $ Q(0,0) $ and $ R(3,3\sqrt{3}) $ be three points. Then the equation of the bisector of the angle PQR is [IIT Screening 2002]

Options:

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 $

Show Answer

Answer:

Correct Answer: C

Solution:

  • Slope of QR = $ \frac{3\sqrt{3}-0}{3-0}=\sqrt{3} $ i.e., $ \theta =60^{o} $ Clearly, $ \angle PQR=120^{o} $ OQ is the angle bisector of the angle, so line OQ makes 120o with the positive direction of x-axis. Therefore equation of the bisector of $ \angle PQR $ is $ y=\tan 120^{o}x $ or $ y=-\sqrt{3}x $ i.e., $ \sqrt{3}x+y=0 $ .