Question

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

Answer 1

`(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.