The angular points of a triangle are A(-1, -7 | vedclass

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(B) The side lengths are calculated as follows:$AB = \sqrt{(5 - (-1))^2 + (1 - (-7))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$$BC = \sqrt{(1 - 5)^2

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$7y = x + 2$

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(B) The side lengths are calculated as follows:$AB = \sqrt{(5 - (-1))^2 + (1 - (-7))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$$BC = \sqrt{(1 - 5)^2 + (4 - 1)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = 5$By the Angle Bisector Theorem, the bisector of $\angle ABC$ divides the opposite side $AC$ in the ratio of the adjacent sides $AB:BC = 10:5 = 2:1$.Let $P$ be the point on $AC$ that divides it in the ratio $2:1$. Using the section formula:$P = \left( \frac{2(1) + 1(-1)}{2 + 1}, \frac{2(4) + 1(-7)}{2 + 1} \right) = \left( \frac{2 - 1}{3}, \frac{8 - 7}{3} \right) = \left( \frac{1}{3}, \frac{1}{3} \right)$The equation of the line passing through $B(5, 1)$ and $P(\frac{1}{3}, \frac{1}{3})$ is given by:$y - 1 = \frac{\frac{1}{3} - 1}{\frac{1}{3} - 5}(x - 5)$$y - 1 = \frac{-2/3}{-14/3}(x - 5)$$y - 1 = \frac{1}{7}(x - 5)$$7y - 7 = x - 5$$7y = x + 2$

The angular points of a triangle are $A(-1, -7)$, $B(5, 1)$, and $C(1, 4)$. The equation of the bisector of the angle $\angle ABC$ is

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