Given vertices A(1, 1),B(4, -2),and C(5, 5) o | vedclass

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(B) The internal bisector of $\angle A$ divides the opposite side $BC$ at point $D$ in the ratio of the lengths of the adjacent sides $AB$ and $AC$.Ca

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$x - 5 = 0$

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(B) The internal bisector of $\angle A$ divides the opposite side $BC$ at point $D$ in the ratio of the lengths of the adjacent sides $AB$ and $AC$.Calculate lengths: $AB = \sqrt{(4-1)^2 + (-2-1)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.$AC = \sqrt{(5-1)^2 + (5-1)^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.Using the section formula,point $D$ divides $BC$ in the ratio $3\sqrt{2} : 4\sqrt{2} = 3:4$.$D = \left( \frac{3(5) + 4(4)}{3+4}, \frac{3(5) + 4(-2)}{3+4} \right) = \left( \frac{15+16}{7}, \frac{15-8}{7} \right) = \left( \frac{31}{7}, 1 \right)$.The slope of the internal bisector $AD$ is $m_{AD} = \frac{1-1}{31/7 - 1} = 0$.Since the bisector $AD$ is a horizontal line $(y=1)$,the line perpendicular to it passing through $C(5, 5)$ must be a vertical line.$A$ vertical line passing through $x=5$ is given by $x - 5 = 0$.

Given vertices $A(1, 1)$,$B(4, -2)$,and $C(5, 5)$ of a triangle,find the equation of the line perpendicular to the internal angle bisector of $\angle A$ and passing through vertex $C$.

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