Question

$P_1,P_2$ are points on either of the two lines $y-\sqrt{3}|x|=2$ at a distance of $5$ units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from $P_1,P_2$ on the bisector of the angle between the given lines.

Answer 1

Given lines : $y-\sqrt{3}|x|=2$

If $x\ge 0,$

$y-\sqrt{3}x=2$ ...$\left(1\right)$

If $x<0,$

$y+\sqrt{3}x=2$ ...$\left(2\right)$

$2y=4$

$\Rightarrow y=2$

$\Rightarrow x=0$

$\therefore$ The point of intersection of given lines is $(0,2)$


Now, slope of line

$y-\sqrt{3}x=2$ is

$m=\sqrt{3}$

$\Rightarrow \tan \theta =\sqrt{3}$

$\Rightarrow \theta ={60}^{\circ }$

$\Rightarrow \alpha ={30}^{\circ }$

So, in triangle $AB{P}_1,$ we get

$AB=A{P}_1\cos {30}^{\circ }$

$\Rightarrow AB=5\times \frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{2}$

So, the angle bisector of the given lines is $x=0$ or the $y-$axis.

So, the foot of perpendicular is $B$

Now, $A=(0,2),$

$AB=\frac{5\sqrt{3}}{2}$, then

Coordinate of $B$ is $(0,2+\frac{5\sqrt{3}}{2}).$

Hence, the coordinate of foot of perpendicular is $(0,2+\frac{5\sqrt{3}}{2})$