For the triangle with vertices A (6, 7), B (- | vedclass

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(B) According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $\overline{BC}$ in the ratio of the lengths of the ot

Vedclass · Accepted Answer

$\left(\frac{30}{7}, \frac{13}{7}\right)$

Vedclass · Answer

(B) According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $\overline{BC}$ in the ratio of the lengths of the other two sides,$AB$ and $AC$.Step $1$: Calculate the lengths of sides $AB$ and $AC$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AB = \sqrt{(-2 - 6)^2 + (3 - 7)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.$AC = \sqrt{(9 - 6)^2 + (1 - 7)^2} = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.Step $2$: The ratio $m:n$ in which the bisector divides $\overline{BC}$ is $AB:AC = 4\sqrt{5}:3\sqrt{5} = 4:3$.Step $3$: Use the section formula to find the coordinates of the point $D(x, y)$ on $\overline{BC}$ that divides it in the ratio $4:3$ internally.$x = \frac{mx_2 + nx_1}{m + n} = \frac{4(9) + 3(-2)}{4 + 3} = \frac{36 - 6}{7} = \frac{30}{7}$.$y = \frac{my_2 + ny_1}{m + n} = \frac{4(1) + 3(3)}{4 + 3} = \frac{4 + 9}{7} = \frac{13}{7}$.Thus,the coordinates are $\left(\frac{30}{7}, \frac{13}{7}\right)$.

For the triangle with vertices $A (6, 7),$ $B (-2, 3)$ and $C (9, 1),$ find the coordinates of the point on $\overline{BC}$ where the bisector of $\angle A$ intersects $\overline{BC}$.

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