Question

A straight line through $Q(\sqrt{3},2)$ makes an angle $\frac{\pi }{6}$ with the positive direction of ther X-axis. If the straight line intersects the line $\sqrt{3}x-4y+8=0$ at P; find the distance of PQ.

Answer 1

Given the straight line makes an angle $\frac{\pi }{6}$ with x-axis and passes through $Q(\sqrt{3},2)$

The equation of straight line is $y=\tan \frac{\pi }{6}x+(2-\sqrt{3}\times \tan \frac{\pi }{6})⟹x-\sqrt{3}y+\sqrt{3}=0$

Point $P$ is the point of intersection of $\sqrt{3}x-4y+8=0,x-\sqrt{3}y+\sqrt{3}=0$ and it will be $(4\sqrt{3},5)$

$PQ=\sqrt{(4\sqrt{3}-\sqrt{3}{)}^2+(5-2{)}^2}=6$