Respuesta :
find the equation of 2 of the line segment, then find the perpendicular lines to them, then solve for x by making the perpendicular equations equal
A (0,0), B(8,10), C(5,-10)
First, find the slope of AB and AC, slope of AB is 5/4, slope of AC is -2
next, find the negative reciprocals of the slopes: -4/5 and 1/2 respectively
so the altitude of AB is a line with a slope of -4/5 passing through (5,-10)
the altitude of AC is a line with a slope of 1/2 passing through (8,10)
find these two lines:
Plug (5,-10) in y=(-4/5)x+b to find b, b=-6, so y=(-4/5)x -6
plug (8,10) in y=(1/2)x +b to find b, b=6, so y=(1/2) x +6
the point where these two lines meet is the orthocenter.
y=(-4/5)x -6
y=(1/2) x +6
(-4/5)x -6=(1/2) x +6
x=-120/13
y=(1/2)*(-120/13)+6=18/13
so the orthocenter is (-120/13, 18/13)
these numbers look odd, but I've run the vertex through an orthodox calculator, and got the same result.
First, find the slope of AB and AC, slope of AB is 5/4, slope of AC is -2
next, find the negative reciprocals of the slopes: -4/5 and 1/2 respectively
so the altitude of AB is a line with a slope of -4/5 passing through (5,-10)
the altitude of AC is a line with a slope of 1/2 passing through (8,10)
find these two lines:
Plug (5,-10) in y=(-4/5)x+b to find b, b=-6, so y=(-4/5)x -6
plug (8,10) in y=(1/2)x +b to find b, b=6, so y=(1/2) x +6
the point where these two lines meet is the orthocenter.
y=(-4/5)x -6
y=(1/2) x +6
(-4/5)x -6=(1/2) x +6
x=-120/13
y=(1/2)*(-120/13)+6=18/13
so the orthocenter is (-120/13, 18/13)
these numbers look odd, but I've run the vertex through an orthodox calculator, and got the same result.