Question 188735
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We know that the orthocenter is where the interestion of the altitudes of the triangle.


Let's see,
{{{drawing(400,400,-7,7,-7,7,graph(400,400,-7,7,-7,7),green(triangle(0,4,6,-4,0,-2)),red(locate(.5,4,a)),red(locate(6,-3.5,b)),red(locate(-.5,-2,c)))}}}

Extend the altitude lines passing thru the vertex and make perpendicular to the opposite side.
In this case is an obtuse triangle, the orthocenter lies outside the triangle.


-->{{{drawing(400,400,-7,7,-7,7,grid(1),graph(400,400,-7,7,-7,7,-4,(2/3)x-2,(8/3)x+4,(-1/3)x-2),green(triangle(0,4,6,-4,0,-2)),red(locate(.5,4,a)),red(locate(6,-3.5,b)),red(locate(-.5,-2,c)),green(arrow(0,-2,0,-7)),red(locate(-2,-1.5,90^o)),red(locate(2.7,-.5,90^o)),red(locate(.2,-3,90^o)),circle(-3,-4,.12),circle(-3,-4,.10)),green(line(-3,0,-3,-4)),green(line(-3,-4,0,-4)),green(circle(-3,-4,.25)))}}} ----> ORTHOCENTER (-3,-4)


Thank you,
Jojo</font>