Question

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

• A. $\mathrm{y}=\surd 3\mathrm{x}$
• B. $\surd 3y=x$
• C. $\mathrm{y}=–\surd 3\mathrm{x}$
• D. $\surd 3\mathrm{y}=-\mathrm{x}$

Answer 1

The correct option is C

$\mathrm{y}=–\surd 3\mathrm{x}$

Determine the equation of the bisector of the $\angle \mathrm{PQR}$.

Given three vertices of a triangle are $\mathrm{P}(-1,0)$, $\mathrm{Q}(0,0)$, and $\mathrm{R}(3,3\sqrt{3})$

The figure for the given problem is as shown:

Slope of a line is given by:

$\mathrm{Slope},\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

Therefore the slope of a line $\mathrm{QR}$ is given by:

$$\begin{gathered} {\mathrm{m}}_{\mathrm{QR}}=\frac{3\sqrt{3}-0}{3-0} \\ =\sqrt{3} \end{gathered}$$

Now we also know that, $\mathrm{Slope},\mathrm{m}=\tan \left(\mathrm{\theta }\right)$

Therefore, the angle made by the line $\mathrm{QR}$ with the positive x-axis is:

$$\begin{gathered} \tan \left(\mathrm{\theta }\right)=\sqrt{3} \\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\sqrt{3}\right) \\ \Rightarrow \mathrm{\theta }={60}^0 \end{gathered}$$

Therefore $\angle \mathrm{PQR}=180-60$

$\Rightarrow \angle \mathrm{PQR}={120}^0$

Therefore the line $\mathrm{QM}$ which is a bisector of $\angle \mathrm{PQR}$ will make an angle of $60+60={120}^0$ with the positive x-axis.

Therefore the equation of a line $\mathrm{QM}$ is:

$$\begin{gathered} \mathrm{y}={\mathrm{m}}_{\mathrm{QM}}\mathrm{x}+\mathrm{c} \\ \Rightarrow \mathrm{y}=\tan {\left(120\right)}^0\mathrm{x}+0 \\ \Rightarrow \mathrm{y}=-\sqrt{3}\mathrm{x} \end{gathered}$$

Hence, option C is the correct answer.