Question

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

• A. $(-\frac{2}{\sqrt{3}},2)$
• B. $(0,0)$
• C. $(\frac{2}{\sqrt{3}},2)$
• D. $(0,4)$

Answer 1

The correct option is A $(-\frac{2}{\sqrt{3}},2)$

Point of intersection $I=(0,2)$

A can be either of two lines

(i) $A$ lies on $\sqrt{3}x+y=2$

$x^2+(y-2{)}^2=\frac{16}{3}$

$\Rightarrow x^2+(\sqrt{3}x{)}^2=\frac{16}{3}$

$\Rightarrow x=\frac{2}{\sqrt{3}},y=0$

(OR) $x=\frac{-2}{\sqrt{3}},y=4$

(ii) $A$ lies on $-\sqrt{3}x+y=2$

$x^2+(y-2{)}^2=\frac{16}{3}$

$\Rightarrow x^2+(\sqrt{3}x{)}^2=\frac{16}{3}$

$\Rightarrow x=\frac{2}{\sqrt{3}},y=4$

(OR) $x=\frac{-2}{\sqrt{3}},y=0$

Bisector of angle between lines are $x=0$ & $y=2$

$\therefore$ foot of perpendicular from $A$ are $(\frac{2}{\sqrt{3}},2),(\frac{-2}{\sqrt{3}},2),(0,0),(0,4)$