Question

A straight line through the origin $O$ meets the parallel lines $4x+2y=9$ and $2x+y+6=0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $PQ$ in the ratio.

• A. $1:2$
• B. $3:4$
• C. $2:1$
• D. $4:3$

Answer 1

The correct option is B $3:4$

Given:Line$4x+2y=9$

Let $(r_1\cos \theta ,r_1\sin \theta )$ is on the line.

$\therefore 4r_1\cos \theta +2r_1\sin \theta =9$

$\therefore r_1=\frac{9}{4\cos \theta +2\sin \theta }$

Let $(-r_2\cos \theta ,-r_2\sin \theta )$ lie on the line $2x+y+6=0$

$\therefore 2r_2\cos \theta +r_2\sin \theta =6$

$\therefore r_2=\frac{6}{2\cos \theta +\sin \theta }$

$\therefore \frac{r_1}{r_2}=\frac{\frac{9}{4\cos \theta +2\sin \theta }}{\frac{6}{2\cos \theta +\sin \theta }}$

$=\frac{9}{4\cos \theta +2\sin \theta }\times \frac{2\cos \theta +\sin \theta }{6}=\frac{3}{4}$

$\therefore \frac{OP}{OQ}=\frac{r_1}{r_2}=\frac{3}{4}$ or $3:4$