We have the three vertices of the triangle:
A = (0, 0)
B = (8, 4)
C = (4, 22)
To find the orthocenter, we need to find the intersection point of the altitudes. The altitudes of a triangle are perpendicular lines drawn from each vertex to the opposite side.
First, we need to find the slope of side AB, BC, and CA.
The slope of AB:
m_AB = (4 - 0) / (8 - 0) = 4/8 = 1/2
The slope of BC:
m_BC = (22 - 4) / (4 - 8) = 18 / -4 = -9/2
The slope of AC:
m_AC = (22 - 0) / (4 - 0) = 22/4 = 11/2
Since the slope of the altitude is the negative reciprocal of the slope of the corresponding side, we have:
Slope of altitude from A = -2 (negative reciprocal of AB)
Slope of altitude from B = 2/9 (negative reciprocal of BC)
Slope of altitude from C = -2/11 (negative reciprocal of AC)
Now, let's write the equations of the altitudes using the point-slope form, y - y1 = m(x - x1):
Equation of altitude from A:
y - 0 = -2(x - 0)
y = -2x
Equation of altitude from B:
y - 4 = (2/9)(x - 8)
y = (2/9)x - 16/9 + 4
y = (2/9)x - 16/9 + 36/9
y = (2/9)x + 20/9
Equation of altitude from C:
y - 22 = (-2/11)(x - 4)
y = (-2/11)x + 8/11 + 22
y = (-2/11)x + 8/11 + 242/11
y = (-2/11)x + 250/11
Now, we need to find the point of intersection of these three lines. Solving the system of equations:
-2x = (2/9)x + 20/9
(-2/11)x + 250/11 = (2/9)x + 20/9
From the first equation, multiplying through by 9:
-18x = 2x + 20
-20x = 20
x = -1
Substituting x = -1 into the equation of the altitude from A:
y = -2(-1)
y = 2
Therefore, the orthocenter of the triangle ABC is (-1, 2).