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(C) The given lines are $L_1: 4x + 3y - 10 = 0$ and $L_2: 8x + 6y + 5 = 0$.Note that $L_2$ can be written as $2(4x + 3y) + 5 = 0$,which means $4x + 3y

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$4:1$

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(C) The given lines are $L_1: 4x + 3y - 10 = 0$ and $L_2: 8x + 6y + 5 = 0$.Note that $L_2$ can be written as $2(4x + 3y) + 5 = 0$,which means $4x + 3y = -2.5$.Since the lines $4x + 3y = 10$ and $4x + 3y = -2.5$ are parallel,any line through the origin $O(0,0)$ will intersect them at points $A$ and $B$ such that the ratio of the distances $OA$ and $OB$ is equal to the ratio of the perpendicular distances from the origin to these lines.The perpendicular distance $d_1$ from $(0,0)$ to $4x + 3y - 10 = 0$ is $d_1 = \frac{|-10|}{\sqrt{4^2 + 3^2}} = \frac{10}{5} = 2$.The perpendicular distance $d_2$ from $(0,0)$ to $8x + 6y + 5 = 0$ is $d_2 = \frac{|5|}{\sqrt{8^2 + 6^2}} = \frac{5}{10} = 0.5 = \frac{1}{2}$.Since the lines are on opposite sides of the origin (as the constant terms have opposite signs relative to the origin),the origin $O$ divides the segment $AB$ internally in the ratio $OA:OB = d_1:d_2 = 2 : \frac{1}{2} = 4:1$.

$A$ straight line through the origin $O$ meets the lines $3y = 10 - 4x$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio:

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