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(B) Given vertices of $	riangle ABC$ are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$.Let the angle bisector of $\angle B$ meet side $AC$ at point $D$.By the

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$x - 7y + 2 = 0$

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(B) Given vertices of $	riangle ABC$ are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$.Let the angle bisector of $\angle B$ meet side $AC$ at point $D$.By the Angle Bisector Theorem,$\frac{AD}{DC} = \frac{BA}{BC}$.Calculate lengths $BA$ and $BC$:$BA = \sqrt{(5 - (-1))^2 + (1 - (-7))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.$BC = \sqrt{(5 - 1)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$.Thus,$\frac{AD}{DC} = \frac{10}{5} = 2$.Point $D$ divides $AC$ in the ratio $2:1$ internally.Using the section formula,$D = \left(\frac{2(1) + 1(-1)}{2+1}, \frac{2(4) + 1(-7)}{2+1}ight) = \left(\frac{1}{3}, \frac{1}{3}ight)$.The angle bisector is the line passing through $B(5, 1)$ and $D(\frac{1}{3}, \frac{1}{3})$.Slope $m = \frac{1 - 1/3}{5 - 1/3} = \frac{2/3}{14/3} = \frac{2}{14} = \frac{1}{7}$.The equation of the line $BD$ is $y - 1 = \frac{1}{7}(x - 5)$.$7y - 7 = x - 5$.$x - 7y + 2 = 0$.

If the vertices of a triangle are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$,then the equation of the angle bisector of $\angle ABC$ is:

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