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(B) Let the equation of the line passing through the origin be $y = mx$. The given parallel lines are $L_1: 4x + 2y - 9 = 0$ and $L_2: 2x + y + 6 = 0$

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$3 : 4$

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(B) Let the equation of the line passing through the origin be $y = mx$. The given parallel lines are $L_1: 4x + 2y - 9 = 0$ and $L_2: 2x + y + 6 = 0$. Note that $L_1$ can be written as $2(2x + y) = 9$,or $2x + y = 4.5$. Let the line $y = mx$ intersect $L_1$ at $P(x_1, y_1)$ and $L_2$ at $Q(x_2, y_2)$. For point $P$: $2x_1 + y_1 = 4.5$ and $y_1 = mx_1$. Substituting $y_1$,we get $x_1(2 + m) = 4.5$,so $x_1 = \frac{4.5}{2 + m}$. For point $Q$: $2x_2 + y_2 = -6$ and $y_2 = mx_2$. Substituting $y_2$,we get $x_2(2 + m) = -6$,so $x_2 = \frac{-6}{2 + m}$. Since $O(0,0)$ divides $PQ$ in ratio $k:1$,we have $0 = \frac{k x_2 + 1 x_1}{k + 1}$. This implies $k x_2 + x_1 = 0$,or $k = -\frac{x_1}{x_2}$. Substituting the values: $k = -\frac{4.5 / (2 + m)}{-6 / (2 + m)} = \frac{4.5}{6} = \frac{4.5}{6} = \frac{9}{12} = \frac{3}{4}$. Thus,the ratio is $3:4$.

$(a)$ Bacillus thuringiensis	$(i)$ Cloning vector
$(b)$ Thermus aquaticus	$(ii)$ Construction of first $rDNA$ molecule
$(c)$ Agrobacterium tumefaciens	$(iii)$ $DNA$ polymerase
$(d)$ Salmonella typhimurium	$(iv)$ Cry proteins

$A$ straight line passing through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at points $P$ and $Q$ respectively. In what ratio does the point $O$ divide the line segment $PQ$?

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