What is the orthocenter of a triangle with corners at #(9 ,5 )#, #(4 ,4 )#, and (8 ,6 )#?

1 Answer
Oct 18, 2017

Ortho center (23/3, 23/3)

Explanation:

Slope of AB #=(4-5)/(4-9) = 1/5#
Slope of CO perpendicular to AB is -5 (O is the ortho center)
Eqn of OC is
#y-6=-5(x-8)#
#5x+y = 46color(white)(aaa) Eqn (1)#

Slope of AC#=(6-5)/8-9) = -1#
Slope of OB perpendicular to AC is #-(1/-1)=1#
Eqn of OB is
#y-4= 1*(x- 4)#
#-x+y=0 color(white)(aaa) #Eqn (2)

Solving Eqns (1) & (2),
#6x = 46, x= 23/3#
#y= 23/3#
Coordinates of O (8,6)#

Verification :
Slope of BC # = (6-4)/(8-4) = 1/2#
Slope of OA #=-2#
Eqn of OA is
#y - 5 = -2(x - 9)#
#2x+y = 23 color (white)(aaa)# Eqn (3)

Solving Eqn (2) & (3),
#3x = 23, x=23/3, y = 23/3#

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