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