Question

A line which makes an acute angle $\theta$ with the positive direction of the $x$-axis is drawn through the point $P(3,4)$ to meet the line $x=6$ at $R$ and $y=8$ at $S$. Then,

• A. $P R=3 \sec \theta$
• B. $P S=4 \operatorname{cosec} \theta$
• C. $P R+P S=\frac{2(3 \sin \theta+4 \cos \theta)}{\sin 2 \theta}$
• D. $\frac{9}{(P R)^{2}}+\frac{16}{(P S)^{2}}=1$

Answer 1

Answer: (A), (B), (C), (D)

The equation of any line making an acute angle $\theta$ with the positive direction of the $x$-axis and passing through $P(3,4)$ is $\frac{x-3}{\cos \theta}=\frac{y-4}{\sin \theta}=r$
where $|r|$ is the distance of any point $(x, y)$ from $P$. Therefore, $A(r \cos \theta+3, r \sin \theta+4)$ is a general point on line (i). If $A$ is $R$, then
$r \cos \theta+3=6 \text { or } r=\frac{3}{\cos \theta}=3 \sec \theta$
Since $\theta$ is acute, $\cos \theta>0$. Therefore,
$P R=r=3 \sec \theta$
If $A \equiv S, r \sin \theta+4=8$. Therefore,
$r=4 \operatorname{cosec} \theta$
$\therefore P S=4 \operatorname{cosec} \theta$
Also, $P R+P S=\frac{3}{\cos \theta}+\frac{4}{\sin \theta}$
$=\frac{2(3 \sin \theta+4 \cos \theta)}{\sin 2 \theta}$
and $\frac{9}{(P R)^{2}}+\frac{16}{(P S)^{2}}=\cos ^{2} \theta+\sin ^{2} \theta=1$
Therefore, (1), (2), (3), and (4) all are correct.