If the vertices of a triangle are (2, -2),(8, | vedclass

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Question

(C) Let the vertices be $A(2, -2)$,$B(8, -2)$,and $C(8, 6)$.First,calculate the side lengths:$c = AB = \sqrt{(8-2)^2 + (-2 - (-2))^2} = \sqrt{6^2 + 0^

Vedclass · Accepted Answer

$(14, 4)$

Vedclass · Answer

(C) Let the vertices be $A(2, -2)$,$B(8, -2)$,and $C(8, 6)$.First,calculate the side lengths:$c = AB = \sqrt{(8-2)^2 + (-2 - (-2))^2} = \sqrt{6^2 + 0^2} = 6$.$a = BC = \sqrt{(8-8)^2 + (6 - (-2))^2} = \sqrt{0^2 + 8^2} = 8$.$b = AC = \sqrt{(8-2)^2 + (6 - (-2))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.Since $a^2 + c^2 = 8^2 + 6^2 = 64 + 36 = 100 = b^2$,the triangle is a right-angled triangle at $B(8, -2)$.The excenter $I_A$ opposite to vertex $A(x_1, y_1)$ is given by the formula:$I_A = (\frac{-ax_1 + bx_2 + cx_3}{-a + b + c}, \frac{-ay_1 + by_2 + cy_3}{-a + b + c})$.Here,$x_1=2, y_1=-2$ (vertex $A$); $x_2=8, y_2=-2$ (vertex $B$); $x_3=8, y_3=6$ (vertex $C$).Denominator: $-a + b + c = -8 + 10 + 6 = 8$.$x$-coordinate: $\frac{-8(2) + 10(8) + 6(8)}{8} = \frac{-16 + 80 + 48}{8} = \frac{112}{8} = 14$.$y$-coordinate: $\frac{-8(-2) + 10(-2) + 6(6)}{8} = \frac{16 - 20 + 36}{8} = \frac{32}{8} = 4$.Thus,the excenter is $(14, 4)$.

If the vertices of a triangle are $(2, -2)$,$(8, -2)$,and $(8, 6)$,what is the excenter opposite to the vertex $(2, -2)$?

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