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

It is clear that the lines lie on opposite side of the origin $O$. Let the equation of any line through $O$ be  $\frac{x}{\cos \theta }=\frac{y}{\sin \theta }.\text{ If }OP=r_1$ If $OP=r_1\text{ and }OQ=r_2$ then the coordinates  of $p$ are $\left(r_1\cos \theta ,r_1\sin \theta \right)$ and that of $Q$ are 

$\left(-r_2\cos \theta ,-r_2\sin \theta \right)$

Since $P$ lies on $4x+2y=9,2r_1(2\cos \theta +\sin \theta )=9$

and $Q$ lies on $2x+y+6=0,-r_2(2\cos \theta +\sin \theta )=-6$

so that $\frac{r_1}{r_2}=\frac{9}{12}=\frac{3}{4}$

and the required ratio is thus 3 : 4

Alternately Let the equation of the line through $O$ be  $y = mx$, 

then coordinates of $P$ and $Q$ are

respectively $\left(\frac{9}{4+2m},\frac{9m}{4+2m}\right)\text{ and }\left(\frac{-6}{2+m},\frac{-6m}{2+m}\right)$ so that

$\frac{OP}{OQ}=\frac{9}{|4+2m|}\times \frac{|2+m|}{6}=\frac{3}{4}$