Orthocenter is the intersection point of the three lines through a vertex perpendicular to the opposite side. For example, through A, perpendicular to BC; through B, perpendicular to AC; and through C, perpendicular AB.
Line through C perpendicular to AB:
m = (6-6) / (4-0) = 0
The line is horizontal, so a line perpendicular to that one would be vertical and have an undefined slope.
However, since it's going through C, whose x coordinate is 1, the equation for the line through C perpendicular to AB is:
Line perpendicular to BC through A:
m of BC = (6-3)/(4-1) = 3/3 = 1
m perpendicular to BC is therefore -1.
Line perp. to BC through A ... 6 = -1(0) + b
b = 6
Equation of line = y = -1(x) + 6.
Line perpendicular to AC through B:
m of AC = (6-3)/(0-1) = 3/-1 = -3.
m perp to AC = 1/3
Substituting in y=mx+b gives 6 = (1/3)(4) + b
b = 14/3
Line equation -> y = (1/3)x + (14/3)
Since x must be 1 for all three altitudes,
plug in 1 for the other two equations.
y = -(x) + 6
y = -(1) + 6 = 5
(1,5) works for the first two equations.
Verify in the 3rd.
y = (1/3)(1) + (14/3)
y = 15/3 = 5.
The orthocenter of this triangle is at (1,5).