What is the orthocenter of a triangle with corners at #(4 ,7 )#, #(8 ,2 )#, and (5 ,6 )#?

Slope of line segment BC # = m_(BC) = (6-2) / (5-8) = -4/3#

Slope of #m_(AD) = - (1/m_(BC)) = (3/4)#

Equation of altitude passing through A and perpendicular to BC

#y - 7 = (3/4) (x - 4)#

#4y - 3x = 16# Eqn (1)

Slope of line segment AC #m_(AC) = (7-6) / (4-5) = -1#

Slope of altitude BE perpendicular to BC #m_(BE) = -(1/ m_(AC)) = -(1/-1) = 1#

Equation of altitude passing through B and perpendicular to AC

#y - 2 = 1 * (x - 8)#

#y - x = -6# Eqn (2)

Solving Eqns (1), (2) we arrive at the coordinates of orthocenter O

#x = 40, y = 34#

Coordinates of orthocenter #O(40, 34)#

Verification :

Slope of #CF = - (4-8) / (7-2) = (4/5)#

Equation of Altitude CF

#y - 6 = (4/5) (x - 5)#

#5y - 4x = 10# Eqn (3)

Orthocenter coordinates #O ( 40, 34)#

I worked out the semi-general case [here].(https://socratic.org/questions/what-is-the-orthocenter-of-a-triangle-with-corners-at-7-3-4-4-and-2-8)

The conclusion is the orthocenter of the triangle with vertices #(a,b),# #(c,d)# and #(0,0)# is

#(x,y) = { ac + bd }/{ad - bc} (d-b,a-c)#

Let's test it by applying it to this triangle and comparing the result to the other answer.

First we translate (5 ,6) to the origin, giving the two other translated vertices:

#(a,b)=(4,7)-(5,6)=(-1,1)#

#(c,d)=(8,2)-(5,6)=(3,-4) #

We apply the formula in the translated space:

#(x,y) = {-1(3) + 1(-4) }/{-1(-4) - 1 (3) } (-5,-4)= -7(-5,-4)=(35,28)#

Now we translate back for our result:

Orthocenter: #(35,28) + (5,6) = (40,34)#

That matches the other answer!