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(D) Let the line $L$ be $y - 5 = m(x - 1)$,which simplifies to $mx - y + (5 - m) = 0$.Let the points of intersection of $L$ with $L_1: 5x - y - 4 = 0$

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$83x - 35y + 92 = 0$

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(D) Let the line $L$ be $y - 5 = m(x - 1)$,which simplifies to $mx - y + (5 - m) = 0$.Let the points of intersection of $L$ with $L_1: 5x - y - 4 = 0$ and $L_2: 3x + 4y - 4 = 0$ be $A$ and $B$ respectively.Since $(1, 5)$ is the midpoint of $AB$,if $A = (x_1, y_1)$,then $B = (2 - x_1, 10 - y_1)$.Point $A$ lies on $5x - y - 4 = 0$,so $5x_1 - y_1 - 4 = 0 \implies y_1 = 5x_1 - 4$.Point $B$ lies on $3x + 4y - 4 = 0$,so $3(2 - x_1) + 4(10 - y_1) - 4 = 0$.$6 - 3x_1 + 40 - 4y_1 - 4 = 0 \implies 3x_1 + 4y_1 = 42$.Substitute $y_1 = 5x_1 - 4$ into the equation: $3x_1 + 4(5x_1 - 4) = 42$.$3x_1 + 20x_1 - 16 = 42 \implies 23x_1 = 58 \implies x_1 = \frac{58}{23}$.Then $y_1 = 5(\frac{58}{23}) - 4 = \frac{290 - 92}{23} = \frac{198}{23}$.The slope $m$ of the line passing through $(1, 5)$ and $(\frac{58}{23}, \frac{198}{23})$ is $m = \frac{\frac{198}{23} - 5}{\frac{58}{23} - 1} = \frac{198 - 115}{58 - 23} = \frac{83}{35}$.The equation of $L$ is $y - 5 = \frac{83}{35}(x - 1) \implies 35y - 175 = 83x - 83$.$83x - 35y + 92 = 0$.

If the intercept of a straight line $L$ made between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then the equation of $L$ is

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