If (-4, 6), (2, 3) and (-2, -5) are the verti | vedclass

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Question

(A) Let the vertices be $A(-4, 6)$,$B(2, 3)$,and $C(-2, -5)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:Side $a = BC = \sqr

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$(-1, 2)$

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(A) Let the vertices be $A(-4, 6)$,$B(2, 3)$,and $C(-2, -5)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:Side $a = BC = \sqrt{(-2 - 2)^2 + (-5 - 3)^2} = \sqrt{(-4)^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.Side $b = AC = \sqrt{(-2 - (-4))^2 + (-5 - 6)^2} = \sqrt{2^2 + (-11)^2} = \sqrt{4 + 121} = \sqrt{125} = 5\sqrt{5}$.Side $c = AB = \sqrt{(2 - (-4))^2 + (3 - 6)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$.The incenter $(I)$ is given by the formula $I = \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)$.$x_I = \frac{4\sqrt{5}(-4) + 5\sqrt{5}(2) + 3\sqrt{5}(-2)}{4\sqrt{5} + 5\sqrt{5} + 3\sqrt{5}} = \frac{\sqrt{5}(-16 + 10 - 6)}{12\sqrt{5}} = \frac{-12}{12} = -1$.$y_I = \frac{4\sqrt{5}(6) + 5\sqrt{5}(3) + 3\sqrt{5}(-5)}{4\sqrt{5} + 5\sqrt{5} + 3\sqrt{5}} = \frac{\sqrt{5}(24 + 15 - 15)}{12\sqrt{5}} = \frac{24}{12} = 2$.Thus,the incenter is $(-1, 2)$.

If $(-4, 6), (2, 3)$ and $(-2, -5)$ are the vertices of a triangle,find its incenter.

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