Question

If the straight line draw through the point $P(\sqrt{3},2)$ and inclined at an angle $\frac{\pi }{6}$ with the x axis meets the line $\sqrt{3}x-4y+8=0$ at $Q$. Find the length PQ ?

Answer 1

Equation of line through point $P(\sqrt{3},2)$

$\because m=tan\theta =tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$

since the equation of line is given by

$(y-y_1)=m(x-x_1)\Rightarrow y-2=\frac{1}{\sqrt{3}}(x-\sqrt{3})\Rightarrow \sqrt{3}y-2\sqrt{3}=x-\sqrt{3}\Rightarrow x-\sqrt{3}y=-\sqrt{3}⟶\left(1\right)$

also the given equation of line passing through point Q is

$\sqrt{3}x-4y+8=0\Rightarrow \sqrt{3}x-4y=-8⟶\left(2\right)solving\left(1\right)and\left(2\right)wegety=5andx=4\sqrt{3}hencecoordinateofpointQis(4\sqrt{3},5)\therefore lengthofPQ=\sqrt{{{(x}_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(4\sqrt{3}-\sqrt{3})}^2+{(5-2)}^2}=6units$