Let the ABC forms a triangle having $A(3, –1)$ as one vertex, $x – 4y + 10 = 0 \mathrm{and}$ $6x + 10y – 59= 0$ are the equations of an angle bisector and a median respectively drawn from different vertices then

Let the vertices of $∆$ABC be $A(3, –1), B(x_1, y_1) and C(x_2, y_2).$
Let the equation of the median through vertex B be $6x + 10y –59=0$

and the bisector of $\angle C be x –3y + 10 = 0$

Since, $C(x_2, y_2)$lies on $x –4y + 10 = 0$ and $E\left(\frac{x_2+3}{2},\frac{y_2-1}{2}\right)$  lies on$6x + 10y – 59 = 0$

$\therefore x_2-4y_2+10=0\text{ and }3x_2+9+5y_2-5-59=0$

$\Rightarrow x_2-4y_2+10=0\text{ and }3x_2+5y_2-55=0$

$\Rightarrow x_2=10,y_2=5$

So, the coordinates of C are $(10, 5).$
The equation of AC is $y+1=\frac{5+1}{10-3}(x-3)$ or, $6x – 7y = 25$

Let the slope of BC be m. Then, its equation is   $y– 5 = m(x–10) ... \left(i\right)$

Since BC and AC are equally inclined to $x – 4y + 10 = 0$ i.e., the angle bisector. Therefore,$\frac{\frac{1}{4}-m}{\frac{1}{4}+m}=\frac{\frac{6}{7}-\frac{1}{4}}{1+\frac{6}{7}\times \frac{1}{4}}\Rightarrow \frac{1-4m}{4+m}=\frac{17}{34}\Rightarrow \frac{-2}{9}$

Substituting the value of m in (i), we obtain that the equation of BC is $2x + 9y = 65$.
Since $B(x_1, y_1)$ lies on BC and the median through B i.e., $6x + 10y –59 = 0$. Therefore,
$2x_1 + 9y_1 = 65$ and $6 x_1+10 y_1–59=0$

$\Rightarrow x_1=\frac{-7}{2},y_1=8$

So, the coordinates of B are $\left(-\frac{7}{2},8\right)$ and so the equation of A is

$y+1=\frac{8+1}{\frac{-7}{2}-3}(x-3),\text{ or }18x+13y=41$