Question

$A$ is a point on either of two lines $y+\sqrt{3}|x|=2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of perpendicular from $A$ on the bisector of the angle between them are

Answer 1

Given : $A$ is a point on either of two lines $y+\sqrt{3}|x|=2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection.

To find : The coordinates of the foot of perpendicular from $A$ on the bisector of the angle between them.

$y+\sqrt{3}|x|=2$
When $x<0,y-\sqrt{3}x=2$
When $x\ge 0,y+\sqrt{3}x=2$


Point of intersection of two lines is $(0,2)$
Distance between the point $(0,2)$ and $(-\frac{2}{\sqrt{3}},0)$ is $\sqrt{(2-0{)}^2+{(-\frac{2}{\sqrt{3}})}^2}=\sqrt{4+\frac{4}{3}}=\frac{4}{\sqrt{3}}$
Distance between the point $(0,2)$ and $(\frac{2}{\sqrt{3}},0)$ is $\frac{4}{\sqrt{3}}$
So, the point $A$ will be either $(\frac{2}{\sqrt{3}},0)$ or $(-\frac{2}{\sqrt{3}},0)$
Angle bisector of line $y+\sqrt{3}|x|=2$ is $Y$ axis.
Foot of the perpendicular from the point $A$ to the angle bisector is $(0,0)$.