Question

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

• A. $\sqrt{3}x+y=0$
• B. $x+\frac{\sqrt{3}}{2}y=0$
• C. $\frac{\sqrt{3}}{2}x+y=0$
• D. $x+\sqrt{3}y=0$

Answer 1

The correct option is A $\sqrt{3}x+y=0$

Equation of the line joining the points $P(-1,0)$ and $Q(0,0)$ is

$y=0$ ............. (1)

Equation of the line joining the points $R(3,3\sqrt{3})$ and $Q(0,0)$ will be of the form $y=mx$, where $m$ is the slope of the line.

Point $R(3,3\sqrt{3})$ will satisfy the equation $y=mx$,

So,

$3\sqrt{3}=3m$

$\Rightarrow m=\sqrt{3}$

Equation of the line joining the points $R(3,3\sqrt{3})$ and $Q(0,0)$ is

$y=\sqrt{3}x$

$y-\sqrt{3}x=0$ .................. (2)

Any point $(x,y)$ on the angle bisectors of angles between lines (1) and (2) will be at Equal perpendicular distance from the two lines.

So, equations of the acute and obtuse angle bisectors are

$\frac{y}{\sqrt{0^2+1^2}}=\pm \frac{-\sqrt{3}x+y}{\sqrt{(-\sqrt{3}{)}^2+1^2}}$

$2y=\pm (-\sqrt{3}x+y)$

$\sqrt{3}x+y=0$ and $-\sqrt{3}x+3y=0$

Bisector of the angle $PQR$ will have negative slope, so the answer is

$\sqrt{3}x+y=0$