Question

Let $P = (-3, 0)$, $Q = (0,0)$ and $R = (3,3\sqrt{3})$ be three points. Then 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. $x + \frac{\sqrt{3{2}y = 0$

Hint:

For angle bisector problems involving the origin, visualize the angles formed by the lines with the x-axis. The angle of the bisector is the average of the angles of the two lines, or it can be found by adding/subtracting half the angle between the lines from one of the line's angles.

Answer 1

The Correct Option is A

Step 1: Slope of $QR$ Points: $Q(0,0)$, $R(3, 3\sqrt{3})$ \[ m_{QR} = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3} \]
Step 2: Angle made by $QR$ with the positive X-axis \[ m = \tan \theta \Rightarrow \tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ \]
Step 3: Find $\angle PQR$ Point $P(-3,0)$ lies on the negative X-axis $\Rightarrow 180^\circ$
Point $R(3,3\sqrt{3})$ makes an angle $60^\circ$ \[ \angle PQR = 180^\circ - 60^\circ = 120^\circ \]
Step 4: Angle bisector The bisector divides $120^\circ$ into two equal parts: \[ \frac{120^\circ}{2} = 60^\circ \] Thus, the bisector makes an angle: \[ 60^\circ + 60^\circ = 120^\circ \] with the positive X-axis.

Step 5: Equation of the angle bisector A line through origin making angle $\phi$ with X-axis: \[ y = (\tan \phi)x \] For $\phi = 120^\circ$: \[ y = \tan 120^\circ \cdot x = -\sqrt{3}x \] \[ \boxed{\sqrt{3}x + y = 0} \]