If (1, 5) is the midpoint of the segment of a | vedclass

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(B) Let the line pass through $A(1, 5)$ with slope $m = 	an 	heta$. The equation of the line is $\frac{y-5}{x-1} = m$, or $y - 5 = m(x - 1)$.Let the

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$83x - 35y + 92 = 0$

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(B) Let the line pass through $A(1, 5)$ with slope $m = 	an 	heta$. The equation of the line is $\frac{y-5}{x-1} = m$, or $y - 5 = m(x - 1)$.Let the line intersect $5x - y - 4 = 0$ at $P_1(x_1, y_1)$ and $3x + 4y - 4 = 0$ at $P_2(x_2, y_2)$.Since $(1, 5)$ is the midpoint, $P_1 = (1+r\cos	heta, 5+r\sin	heta)$ and $P_2 = (1-r\cos	heta, 5-r\sin	heta)$ for some distance $r$.Substituting $P_1$ into $5x - y - 4 = 0$: $5(1+r\cos	heta) - (5+r\sin	heta) - 4 = 0$ $\Rightarrow 5r\cos	heta - r\sin	heta = 4$ $\Rightarrow r = \frac{4}{5\cos	heta - \sin	heta}$.Substituting $P_2$ into $3x + 4y - 4 = 0$: $3(1-r\cos	heta) + 4(5-r\sin	heta) - 4 = 0$ $\Rightarrow 3 - 3r\cos	heta + 20 - 4r\sin	heta - 4 = 0$ $\Rightarrow 19 = r(3\cos	heta + 4\sin	heta)$ $\Rightarrow r = \frac{19}{3\cos	heta + 4\sin	heta}$.Equating the two expressions for $r$: $\frac{4}{5\cos	heta - \sin	heta} = \frac{19}{3\cos	heta + 4\sin	heta}$.$12\cos	heta + 16\sin	heta = 95\cos	heta - 19\sin	heta$.$35\sin	heta = 83\cos	heta \Rightarrow 	an	heta = \frac{83}{35}$.The equation of the line is $y - 5 = \frac{83}{35}(x - 1)$.$35y - 175 = 83x - 83 \Rightarrow 83x - 35y + 92 = 0$.

If $(1, 5)$ is the midpoint of the segment of a line between the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$, then the equation of the line is:

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