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:

x=1

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).