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

(D) Given lines are:
$5x - y + 4 = 0$ ..... $(1)$
$3x + 4y - 4 = 0$ ..... $(2)$
Let the required line intersect the lines $(1)$ and $(2)$ at points $P(\alpha_{1}, \beta_{1})$ and $Q(\alpha_{2}, \beta_{2})$ respectively.
Since $P$ lies on $(1)$,we have $\beta_{1} = 5\alpha_{1} + 4$.
Since $Q$ lies on $(2)$,we have $\beta_{2} = \frac{4 - 3\alpha_{2}}{4}$.
The midpoint of $PQ$ is $(1, 5)$,so $\frac{\alpha_{1} + \alpha_{2}}{2} = 1$ and $\frac{\beta_{1} + \beta_{2}}{2} = 5$.
From the first,$\alpha_{2} = 2 - \alpha_{1}$.
Substituting into the second: $\frac{(5\alpha_{1} + 4) + \frac{4 - 3(2 - \alpha_{1})}{4}}{2} = 5$.
$20\alpha_{1} + 16 + 4 - 6 + 3\alpha_{1} = 40 \implies 23\alpha_{1} = 26 \implies \alpha_{1} = \frac{26}{23}$.
Then $\beta_{1} = 5(\frac{26}{23}) + 4 = \frac{130 + 92}{23} = \frac{222}{23}$.
The line passes through $(1, 5)$ and $(\frac{26}{23}, \frac{222}{23})$.
The slope $m = \frac{\frac{222}{23} - 5}{\frac{26}{23} - 1} = \frac{222 - 115}{26 - 23} = \frac{107}{3}$.
The equation is $y - 5 = \frac{107}{3}(x - 1) \implies 3y - 15 = 107x - 107$.
Thus,the equation is $107x - 3y - 92 = 0$.