Question

The line perpendicular to $y=\sqrt{3}x+2$ and passing through (2,1) is rotated through an angle ${60}^{\circ }$ about (2,1). Area of triangle formed by these lines and $x$-axis is $A{\text{units}}^2$, then $4A^2$ is :

Answer 1

Line perpendicular to $y=\sqrt{3}x+2$ makes an angle ${150}^{\circ }$ with $+$ direction of $x$-axis
Now the required line which makes an angle ${60}^{\circ }$ with this,will have slope $\tan {90}^{\circ }$ or $\tan {30}^{\circ }$
$\therefore$ Required possible lines are $x=2$ or $y-1=\frac{1}{\sqrt{3}}(x-2)\Rightarrow x-\sqrt{3}y-2+\sqrt{3}=0$
$\therefore$ Area of triangle formed by
$x=2,x-\sqrt{3}y-2+\sqrt{3}=0$ and $x$-axis is
$A=\frac{1}{2}.b.h=\frac{\sqrt{3}}{2}{\text{units}}^2\therefore 4A^2=3$