Question

Let $P = ( - 1,0),Q = (0,0),R = (3,3\sqrt 3 )$ be three points. Then the equation of the bisector of the $\angle PQR$ is
A. $y = \sqrt 3 x$
B. $\sqrt 3 y = x$
C. $y = - \sqrt 3 x$
D. $\sqrt 3 y = - x$

Answer 1

Hint: In this question, we are given three points, to find the bisector of the angle formed by these three points we have to first find the angle formed by these three points, as $Q$ lies at origin, so we have to find the angle between line $PQ$ and $QR$ . As the point $P$ lies on the negative x-axis, so we will find the angle formed by $QR$ with x-axis by finding the slope of the line $QR$ , that can be easily found as the coordinates of the endpoints are given. Then we will find the angle that the bisector forms with the x-axis.

Complete step by step answer:

Image: The angle bisector of PQR
$
  m = \dfrac{{3\sqrt 3 - 0}}{{3 - 0}} \\
   \Rightarrow \tan \theta = \sqrt 3 \\
   \Rightarrow \theta = 60^\circ \\
 $
So, $QR$ makes an angle of $60^\circ $ with the x-axis, so we get:
$
  \angle PQR = 180^\circ - 60^\circ \\
   \Rightarrow \angle PQR = 120^\circ \\
 $
Now, $QM$ is the bisector of $\angle PQR$ , so $\angle MQR = \dfrac{{120^\circ }}{2} = 60^\circ $
The angle that $QM$ makes with the x-axis is the sum of $\angle MQR$ and the angle that $QR$ makes with the x-axis, so it is $60^\circ + 60^\circ = 120^\circ $ .
Slope of line $QM = \tan 120^\circ = - \sqrt 3 $
Now, $QM$ passes through the origin, so using point slope from of line (that is $y - {y_1} = m(x - {x_1})$), we get:
$
  y - 0 = - \sqrt 3 (x - 0) \\
   \Rightarrow y = - \sqrt 3 x \\
 $
The correct option is option C.

Note:
Angle bisector of an angle is defined as a line that divides that angle into two halves, that’s why we take $\angle MQR = \dfrac{{\angle PQR}}{2}$ . Equation of a line can be obtained by using various methods, as here only the slope and one passing point of the bisector line are given, so we used the point-slope form of the line.