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(C) The given lines are $L_1: 4x + 3y - 12 = 0$ and $L_2: 3x + 4y - 12 = 0$.Any line passing through their intersection is given by $L_1 + \lambda L_2

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$6xy = 7(x + y)$

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(C) The given lines are $L_1: 4x + 3y - 12 = 0$ and $L_2: 3x + 4y - 12 = 0$.Any line passing through their intersection is given by $L_1 + \lambda L_2 = 0$,which is $(4x + 3y - 12) + \lambda(3x + 4y - 12) = 0$.Rearranging,we get $x(4 + 3\lambda) + y(3 + 4\lambda) - 12(1 + \lambda) = 0$.The line meets the axes at $A\left(\frac{12(1 + \lambda)}{4 + 3\lambda}, 0\right)$ and $B\left(0, \frac{12(1 + \lambda)}{3 + 4\lambda}\right)$.Let the midpoint of $AB$ be $(h, k)$. Then $h = \frac{6(1 + \lambda)}{4 + 3\lambda}$ and $k = \frac{6(1 + \lambda)}{3 + 4\lambda}$.From these,$\frac{1}{h} = \frac{4 + 3\lambda}{6(1 + \lambda)}$ and $\frac{1}{k} = \frac{3 + 4\lambda}{6(1 + \lambda)}$.Adding these,$\frac{1}{h} + \frac{1}{k} = \frac{7 + 7\lambda}{6(1 + \lambda)} = \frac{7}{6}$.Thus,$\frac{h + k}{hk} = \frac{7}{6}$,which implies $6(h + k) = 7hk$.The locus is $6(x + y) = 7xy$.

If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$ meets the coordinate axes at $A$ and $B$ $(A \neq B)$,then the locus of the midpoint of $AB$ is

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