How do you get the orthocenter?

Question

of the triangle with vertices A(-3,4) B(1,-8) C(9,0)

For some reason I get 25 for X when I do a system of eqts with the two lines

Line1: perpendicular slope of AB and vertex C
Line2: perpendicular slope of AC and vertex B

What am I doing wrong? There is no way X is 25; that's out of the triangle! I foudn the orthocenter graphically, and it gave me about (3,-2)

PLEASE HELP MAX points to whoever can explain this to me thoroughly.

Answer 1

The point-slope form of the equation of a line is:

y-y₀ = m(x-x₀)

So the slope of AB is (-8-4)/(1-(-3)) = -12/4 = -3, so the perpendicular to it has slope 1/3. It passes through C, so line 1 has the equation:

y-0 = 1/3 (x-9)

Or simplifying:

y = x/3 - 3

For line 2, the slope of AC is (0-4)/(9-(-3)) = - 4/12 = -1/3, so the slope of the perpendicular is 3. The line passes through B, so it has the equation:

y-(-8) = 3(x-1)

Simplifying:

y+8 = 3x-3
y = 3x - 11

Putting the two equations in standard form:

y - x/3 = -3
y - 3x = -11

Subtracting equation 1 from equation 2:

-8x/3 = -8

Multiplying by -3/8:

x = 3

Substituting and solving for y:

y = 3/3 - 3
y = 1 - 3
y = -2

So the orthocenter is at (3, -2), which is what you found by graphing.

Answer 2

Yes, it actually is (3,-2). I used the altitude from B to AC (equation --> y=3x-11) and the altitude from A to BC (equation --> y = -x+1). The slopes I got for the 3 sides were:

AB = -3
AC = -1/3
BC = 1

You seem like you know how to do a system of equations, so I won't explain that to you, but it's clear with the two equations above. You must have had one of your equations wrong, but realize in the future, it IS possible for the orthocenter to be outside the triangle (when it's obtuse).