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(D) The equation of a family of lines passing through the intersection of $x-2y+3=0$ and $2x-y-1=0$ is given by $(x-2y+3) + K(2x-y-1) = 0$.Rearranging

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$14x+3xy-15y=0$

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(D) The equation of a family of lines passing through the intersection of $x-2y+3=0$ and $2x-y-1=0$ is given by $(x-2y+3) + K(2x-y-1) = 0$.Rearranging terms,we get $(1+2K)x - (2+K)y + (3-K) = 0$.This can be written as $\frac{(1+2K)}{K-3}x + \frac{-(2+K)}{K-3}y = 1$.The intercepts on the $X$ and $Y$ axes are $A\left(\frac{K-3}{1+2K}, 0\right)$ and $B\left(0, \frac{K-3}{-(2+K)}\right)$.Let the point dividing $AB$ in the ratio $-2:3$ be $(x, y)$. Using the section formula:$x = \frac{-2(0) + 3(\frac{K-3}{1+2K})}{3-2} = \frac{3(K-3)}{1+2K}$ and $y = \frac{-2(\frac{K-3}{-(2+K)}) + 3(0)}{3-2} = \frac{2(K-3)}{2+K}$.From $x = \frac{3K-9}{2K+1}$,we get $x(2K+1) = 3K-9 \Rightarrow K(2x-3) = -9-x \Rightarrow K = \frac{x+9}{3-2x}$.From $y = \frac{2K-6}{K+2}$,we get $y(K+2) = 2K-6 \Rightarrow K(y-2) = -6-2y \Rightarrow K = \frac{6+2y}{2-y}$.Equating the two expressions for $K$: $\frac{x+9}{3-2x} = \frac{6+2y}{2-y}$.$(x+9)(2-y) = (6+2y)(3-2x) \Rightarrow 2x - xy + 18 - 9y = 18 - 12x + 6y - 4xy$.Simplifying,we get $14x + 3xy - 15y = 0$.

If a variable straight line passing through the point of intersection of the lines $x-2y+3=0$ and $2x-y-1=0$ intersects the $X, Y$-axes at $A$ and $B$ respectively,then the equation of the locus of a point which divides the segment $AB$ in the ratio $-2:3$ is

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