Question 1091708
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Find orthocentre of triangle with vertices (-2,-1),(6,-1),(2,5)
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The orthocenter is the point where all three altitudes of the triangle intersect. 


Notice that the side of the triangle, connecting the vertices (-2,-1) and (6,-1), is horizontal line parallel to x-axis.

Therefore, the altitude drawn to this side is vertical line x = const, and since it passes through the point (2,5), this constant is equal to 2,
and the equation of this altitude is x = 2.


We will find the orthocenter as the intersection point of this altitude with the other altitude drawn from the vertex (6,-1).

This altitude is perpendicular to the side of the triangle connecting two other points, (-2,-1) and (2,5).

The slope of this side/segment is m = {{{(dy)/(dx)}}} = {{{(5-(-1))/(2-(-2))}}} = {{{6/4}}} = {{{3/2}}}.


Since the altitude from the point (6,-1) is perpendicular to this side, it has the slope {{{-2/3)}}}.
And since it passes through the point (6,-1), its equation is

y - (-1) = {{{(-2/3)*(x-6)}}},   or    

y + 1 = {{{(-2/3)*x + 4}}},   or   y = {{{(-2/3)x + 3}}}.


Its intersection with the line x = 2 has y-coordinate  y = {{{-4/3 + 3}}} = {{{1}}} {{{2/3}}}.


Thus the orthocenter is the point (x,y) = ({{{2}}},{{{1}}}{{{2/3}}}).
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