question_answer

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]

A)            \[\frac{\sqrt{3}}{2}x+y=0\] 

B)            \[x+\sqrt{3}y=0\]

C)            \[\sqrt{3}x+y=0\]                   

D)            \[x+\frac{\sqrt{3}}{2}y=0\]

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.,\[\sqrt{3}x+y=0\].