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(C) The given lines are $L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - 9 = 0$. Note that $L_2$ can be written as $2(2x + y) - 9 = 0$,meaning $L_1$ and $L_2

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$4 : 3$

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(C) The given lines are $L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - 9 = 0$. Note that $L_2$ can be written as $2(2x + y) - 9 = 0$,meaning $L_1$ and $L_2$ are parallel.$A$ line perpendicular to these lines will have the form $x - 2y + k = 0$. Since it passes through the origin $(0, 0)$,we have $0 - 2(0) + k = 0$,so $k = 0$. The equation of the line is $x - 2y = 0$.Find the intersection point $P$ of $x - 2y = 0$ and $2x + y + 6 = 0$:Substitute $x = 2y$ into $2(2y) + y + 6 = 0$ $\Rightarrow 5y = -6$ $\Rightarrow y = -6/5, x = -12/5$. So $P = (-12/5, -6/5)$.Find the intersection point $Q$ of $x - 2y = 0$ and $4x + 2y - 9 = 0$:Substitute $x = 2y$ into $4(2y) + 2y - 9 = 0$ $\Rightarrow 10y = 9$ $\Rightarrow y = 9/10, x = 18/10 = 9/5$. So $Q = (9/5, 9/10)$.The origin $(0, 0)$ divides the segment $PQ$ in ratio $\lambda : 1$. Using the section formula for the $x$-coordinate:$0 = \frac{\lambda(9/5) + 1(-12/5)}{\lambda + 1}$ $\Rightarrow 9\lambda - 12 = 0$ $\Rightarrow \lambda = 12/9 = 4/3$.Thus,the ratio is $4 : 3$.

$A$ line passes through the origin and is perpendicular to two given lines $2x + y + 6 = 0$ and $4x + 2y - 9 = 0$. What is the ratio in which the origin divides the segment formed by the intersection points of this line with the two given lines?

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