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(C) The given lines are $L_1: 4x + 2y - 9 = 0$ and $L_2: 2x + y + 6 = 0$. We can rewrite $L_1$ as $2(2x + y) = 9$,which is $2x + y = 4.5$. Let the lin

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$3: 4$

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(C) The given lines are $L_1: 4x + 2y - 9 = 0$ and $L_2: 2x + y + 6 = 0$. We can rewrite $L_1$ as $2(2x + y) = 9$,which is $2x + y = 4.5$. Let the line passing through the origin be $y = mx$. Substituting $y = mx$ into $L_1$: $2x + mx = 4.5 \implies x_P = \frac{4.5}{2+m}$,$y_P = \frac{4.5m}{2+m}$. Substituting $y = mx$ into $L_2$: $2x + mx = -6 \implies x_Q = \frac{-6}{2+m}$,$y_Q = \frac{-6m}{2+m}$. Since $O$ is the origin $(0,0)$,the ratio in which $O$ divides $PQ$ is the ratio of the distances $OP$ and $OQ$. $OP = \sqrt{x_P^2 + y_P^2} = \frac{4.5}{|2+m|} \sqrt{1+m^2}$. $OQ = \sqrt{x_Q^2 + y_Q^2} = \frac{6}{|2+m|} \sqrt{1+m^2}$. The ratio $OP : OQ = 4.5 : 6 = 9 : 12 = 3 : 4$. Since $P$ and $Q$ lie on opposite sides of the origin,$O$ divides $PQ$ internally in the ratio $3:4$.

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

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