If triangle has vertexes at $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, is the intersection points of feet of altitudes

$$x_h = \frac{x_1x_2(y_2-y_1) + x_2x_3(y_3-y_2) + x_3x_1(y_1-y_3) + y_1^2(y_3-y_2) + y_2^2(y_1-y_3) + y_3^2(y_2-y_1)}{x_1y_2 + x_2y_3 + x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3}$$

$$y_h = \frac{y_1y_2(x_1-y_2) + y_2y_3(x_2-x_3) + y_3y_1(x_3-y_1) + x_1^2(x_2-x_3) + x_2^2(x_3-x_1) + x_3^2(x_1-x_2)}{x_1y_2 + x_2y_3 + x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3}$$

I mean, my computation were quite messy and I would like to have a verification of this result.

  • $\begingroup$ When I was a schoolboy I calculated coordinates of similar points in a triangle. The results were similar to yours. They should be known. Also them looks much more nicer if they are represented as a quotient of determinants of some $3x3$ matrices. $\endgroup$ – Alex Ravsky Dec 17 '14 at 21:43

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