Question

A line parallel to $y=\sqrt{3}x$ passes through $Q(2,3)$ and intersects the line $2x+4y-27=0$ at $P$. If $PQ=r,$ then the value of $r^2+2r+9$ is equal to

Answer 1

Given line : $y=\sqrt{3}x\Rightarrow \tan \theta =\sqrt{3}\Rightarrow \theta ={60}^{\circ }$
$\therefore P\equiv (2+r\cos {60}^{\circ },3+r\sin {60}^{\circ })\equiv (2+\frac{r}{2},3+\frac{\sqrt{3}r}{2})$
$P$ lies on the line $2x+4y-27=0$
$\Rightarrow 2(2+\frac{r}{2})+4(3+\frac{\sqrt{3}r}{2})=27$
$r=\frac{11}{2\sqrt{3}+1}=2\sqrt{3}-1$
$\therefore r+1=2\sqrt{3}\Rightarrow r^2+2r+1=12$
$\Rightarrow r^2+2r+9=20$