If the vertices of a triangle are A(0, -6),B( | vedclass

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(D) The side lengths are calculated as follows:$a = BC = \sqrt{(-6-1)^2 + (0-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$$b = CA = \sqrt{(1-0)^2 + (1+

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$(-4, 6)$

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(D) The side lengths are calculated as follows:$a = BC = \sqrt{(-6-1)^2 + (0-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$$b = CA = \sqrt{(1-0)^2 + (1+6)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$$c = AB = \sqrt{(0+6)^2 + (-6-0)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$The coordinates of the excenter $I_A$ opposite to vertex $A(x_1, y_1)$ are given by:$x = \frac{-ax_1 + bx_2 + cx_3}{-a + b + c} = \frac{-5\sqrt{2}(0) + 5\sqrt{2}(-6) + 6\sqrt{2}(1)}{-5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{-30\sqrt{2} + 6\sqrt{2}}{6\sqrt{2}} = \frac{-24\sqrt{2}}{6\sqrt{2}} = -4$$y = \frac{-ay_1 + by_2 + cy_3}{-a + b + c} = \frac{-5\sqrt{2}(-6) + 5\sqrt{2}(0) + 6\sqrt{2}(1)}{-5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{30\sqrt{2} + 6\sqrt{2}}{6\sqrt{2}} = \frac{36\sqrt{2}}{6\sqrt{2}} = 6$Thus,the coordinates of the excenter are $(-4, 6)$.

If the vertices of a triangle are $A(0, -6)$,$B(-6, 0)$,and $C(1, 1)$,find the coordinates of the excenter opposite to vertex $A$.

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