What is the orthocenter of a triangle with corners at (1, 3), (6, 2), and (5, 4)?

1 Answer
Apr 22, 2018

(x, y)=(47/9, 46/9)

Explanation:

Let: A(1, 3), B(6, 2) and C(5, 4) be the vertices of triangle ABC:

Slope of a line through points: (x_1, y_1), (x_2, y_2):
m=(y_2-y_1)/(x_2-x_1)

Slope of AB:
=(2-3)/(6-1)=-1/5
Slope of perpendicular line is 5.
Equation of the altitude from C to AB:
y-y_1=m(x-x_1) =>m=5, C(5,4):
y-4=5(x-5)
y=5x-21

Slope of BC:
=(4-2)/(5-6)=-2
Slope of perpendicular line is 1/2.
Equation of the altitude from A to BC:
y-3=1/2(x-1)
y=(1/2)x+5/2

The intersection of the altitudes equating y's:
5x-21=(1/2)x+5/2
10x-42=x+5
9x=47
x=47/9

y=5*47/9- 21
y=46/9

Thus the Orthocenter is at (x, y)=(47/9, 46/9)

To check the answer you can find the equation of altitude from B to AC and find the intersection of that with one of the other altitudes.