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(A) Let the line passing through the point $M(1, 5)$ be $\frac{x - 1}{\cos 	heta} = \frac{y - 5}{\sin 	heta} = r$. Since the segment is bisected at

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$83x - 35y + 92 = 0$

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(A) Let the line passing through the point $M(1, 5)$ be $\frac{x - 1}{\cos 	heta} = \frac{y - 5}{\sin 	heta} = r$. Since the segment is bisected at $M(1, 5)$,the endpoints $A$ and $B$ are at distances $r = d$ and $r = -d$ from $M$ respectively. Point $A$ is $(1 + d \cos 	heta, 5 + d \sin 	heta)$ and it lies on $5x - y - 4 = 0$. Substituting $A$ into the first line: $5(1 + d \cos 	heta) - (5 + d \sin 	heta) - 4 = 0 \implies 5d \cos 	heta - d \sin 	heta = 4$. Point $B$ is $(1 - d \cos 	heta, 5 - d \sin 	heta)$ and it lies on $3x + 4y - 4 = 0$. Substituting $B$ into the second line: $3(1 - d \cos 	heta) + 4(5 - d \sin 	heta) - 4 = 0 \implies 3 - 3d \cos 	heta + 20 - 4d \sin 	heta - 4 = 0 \implies -3d \cos 	heta - 4d \sin 	heta = -19 \implies 3d \cos 	heta + 4d \sin 	heta = 19$. We have the system: $5d \cos 	heta - d \sin 	heta = 4$ $3d \cos 	heta + 4d \sin 	heta = 19$ Solving for $d \cos 	heta$ and $d \sin 	heta$: Multiply the first by $4$: $20d \cos 	heta - 4d \sin 	heta = 16$. Adding to the second: $23d \cos 	heta = 35 \implies d \cos 	heta = \frac{35}{23}$. Substitute back: $5(\frac{35}{23}) - d \sin 	heta = 4 \implies d \sin 	heta = \frac{175}{23} - 4 = \frac{175 - 92}{23} = \frac{83}{23}$. The slope of the line is $m = \frac{\sin 	heta}{\cos 	heta} = \frac{83}{35}$. The equation of the line passing through $(1, 5)$ with slope $\frac{83}{35}$ is $y - 5 = \frac{83}{35}(x - 1)$. $35y - 175 = 83x - 83 \implies 83x - 35y + 92 = 0$.

$A$ line is such that its segment between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then its equation is

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