Question

A line making an angle $\theta$ with the $+$ive direction of $x-$axis passes through $P(5,6)$ to meet the line $x=6$ at $Q$ and $y=9$ at $R$ then $QR$ is :

• A. $\frac{2(\cos \theta +3\sin \theta )}{\sin 2\theta }$
• B. $\frac{2(3\cos \theta +\sin \theta )}{\sin 2\theta }$
• C. $\frac{2(3\cos \theta -\sin \theta )}{\sin 2\theta }$
• D. $\frac{2(\cos \theta -3\sin \theta )}{\sin 2\theta }$

Answer 1

The correct option is C $\frac{2(3\cos \theta -\sin \theta )}{\sin 2\theta }$

Ans. $\left(c\right)$
$\frac{x-5}{\cos \theta }=\frac{y-6}{\sin \theta }=r$
Any point on the line is $(r\cos \theta +5,r\sin \theta +6)$ and $Q$ lies on $x=6$.
$\therefore r\cos +5=6\therefore r=\frac{1}{\cos \theta }$
$\therefore y=r\sin \theta +6=\tan \theta +6$
$Q$ is $(6,6+\tan \theta )$
Similarly $R$ on $y=9$ is $(5+3\cot \theta ,9)$
$\therefore Q{R}^2=(1-3\cot \theta {)}^2+(\tan \theta -3{)}^2$
$=10-6(\tan \theta +\cot \theta )+9{\cot }^2\theta +{\tan }^2\theta$
$=9{\csc }^2\theta +{\sec }^2\theta -6.\frac{1}{\sin \theta \cos \theta }$
$=\frac{9{\csc }^2\theta +{\sec }^2\theta -6\sin \theta \cos \theta }{{\sin }^2\theta {\cos }^2\theta }$
$=\frac{4{(3\cos \theta -\sin \theta )}^2}{{\sin }^{2}2\theta }$
$\therefore QR=\frac{2(3\cos \theta -\sin \theta )}{\sin 2\theta }\Rightarrow \left(c\right)$