$A\left(1,7\right), B\left(-5,-1\right), C\left(7,4\right)$

Then the bisector of angle $B$ meets side $AC$ at $D$ then

$\frac{AD}{DC}=\frac{BA}{BC}$

By using distance formula,

 ${\left(BA\right)}^2=36+64=100\Rightarrow BA=10$

$B{C}^2={\left(7-\left(-5\right)\right)}^2+{\left(4-\left(-1\right)\right)}^2=144+25=169$

$\Rightarrow BC=13$

So $\frac{AD}{DC}=\frac{10}{13}$

Hence $D\left(\frac{83}{23},\frac{131}{23}\right)$

Then the equation of a bisector of angle $\angle ABC$ is

$y+1=\left(\frac{-1-{\displaystyle \frac{131}{23}}}{-5-{\displaystyle \frac{83}{23}}}\right)\left(x+5\right)$

$77x-99y+286=0$

$7x-9y+26=0$