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(B) The given parallel lines are $4x + 2y - 9 = 0$ and $2x + y + 6 = 0$.We can rewrite the first equation as $2x + y = \frac{9}{2}$.The second equatio

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$3: 4$

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(B) The given parallel lines are $4x + 2y - 9 = 0$ and $2x + y + 6 = 0$.We can rewrite the first equation as $2x + y = \frac{9}{2}$.The second equation is $2x + y = -6$.Let the line through the origin be $y = mx$. Substituting this into the equations:For $P$: $2x + mx = \frac{9}{2} \Rightarrow x_P = \frac{9}{2(2+m)}$,$y_P = \frac{9m}{2(2+m)}$.For $Q$: $2x + mx = -6 \Rightarrow x_Q = \frac{-6}{2+m}$,$y_Q = \frac{-6m}{2+m}$.The ratio in which $O(0,0)$ divides $PQ$ is given by $\frac{OP}{OQ} = \frac{|x_P|}{|x_Q|} = \frac{9/2(2+m)}{6/(2+m)} = \frac{9}{12} = \frac{3}{4}$.Thus,the point $O$ divides $PQ$ in the ratio $3: 4$.

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

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