If the vertices of a triangle ABC are A(1,7), | vedclass

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

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$7x-9y+26=0$

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(A) Given vertices are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$.Let the internal angle bisector of $\angle ABC$ meet the side $AC$ at point $D$.By the Angle Bisector Theorem,the ratio in which $D$ divides $AC$ is $\frac{AD}{DC} = \frac{BA}{BC}$.Calculating the lengths of sides $BA$ and $BC$:$BA = \sqrt{(1 - (-5))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.$BC = \sqrt{(7 - (-5))^2 + (4 - (-1))^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.Thus,$\frac{AD}{DC} = \frac{10}{13}$.Using the section formula,the coordinates of $D$ are:$D = \left( \frac{10(7) + 13(1)}{10 + 13}, \frac{10(4) + 13(7)}{10 + 13} \right) = \left( \frac{70 + 13}{23}, \frac{40 + 91}{23} \right) = \left( \frac{83}{23}, \frac{131}{23} \right)$.The equation of the line passing through $B(-5, -1)$ and $D\left(\frac{83}{23}, \frac{131}{23}\right)$ is:$y - (-1) = \frac{\frac{131}{23} - (-1)}{\frac{83}{23} - (-5)} (x - (-5))$$y + 1 = \frac{\frac{131 + 23}{23}}{\frac{83 + 115}{23}} (x + 5)$$y + 1 = \frac{154}{198} (x + 5)$$y + 1 = \frac{7}{9} (x + 5)$$9y + 9 = 7x + 35$$7x - 9y + 26 = 0$.

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

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