Find the orthocenter of a triangle ABC with vertices at A (2, 2), B (0, - 4), and C (6, - 2).

Hello,

H = (3,-1)

(AB) ≡y-2 = (x-2) * (-4-2) / (0-2) = = >y=3x-4

(BC) ≡y+4 = (x-0) * (-2+4) / (6-0) = = >y=x/3-4

(AA') ≡y-2 = (x-2) * (-3) = = >y=-3x+8

(CC') ≡y+2 = (x-6) * (-1/3) = = >y=-x/3

(AA') ∩ (CC'):

x+3y=0 (1)

3x+y=8 (2)

==>-3 * (1) + (2) : - y8=8==>y=-1 and x=3