The coordinates of the feet of the perpendicu | vedclass

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(C) Let the triangle be $ABC$ and the feet of the perpendiculars be $D(20, 25)$,$E(8, 16)$,and $F(8, 9)$. The triangle $DEF$ is the orthic triangle of

Vedclass · Accepted Answer

$(10, 15)$

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(C) Let the triangle be $ABC$ and the feet of the perpendiculars be $D(20, 25)$,$E(8, 16)$,and $F(8, 9)$. The triangle $DEF$ is the orthic triangle of $\Delta ABC$.The orthocenter of $\Delta ABC$ is the incenter of its orthic triangle $DEF$.Let the orthocenter be $O(h, k)$. The side lengths of $\Delta DEF$ are:$EF = \sqrt{(8-8)^2 + (16-9)^2} = \sqrt{0^2 + 7^2} = 7$$FD = \sqrt{(20-8)^2 + (25-9)^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$$DE = \sqrt{(20-8)^2 + (25-16)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$The incenter $(h, k)$ of $\Delta DEF$ with vertices $D(x_1, y_1)$,$E(x_2, y_2)$,$F(x_3, y_3)$ and opposite side lengths $a=EF=7$,$b=FD=20$,$c=DE=15$ is given by:$h = \frac{a x_1 + b x_2 + c x_3}{a + b + c} = \frac{7(20) + 20(8) + 15(8)}{7 + 20 + 15} = \frac{140 + 160 + 120}{42} = \frac{420}{42} = 10$$k = \frac{a y_1 + b y_2 + c y_3}{a + b + c} = \frac{7(25) + 20(16) + 15(9)}{7 + 20 + 15} = \frac{175 + 320 + 135}{42} = \frac{630}{42} = 15$Thus,the orthocenter is $(10, 15)$.

The coordinates of the feet of the perpendiculars from the vertices of a triangle to the opposite sides are $(20, 25)$,$(8, 16)$,and $(8, 9)$. Find the orthocenter of the triangle.

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