Question

What is the orthocenter of a triangle with corners at `(2 ,8 )`, `(3 ,4 )`, and (6 ,3 )#?

Answer 1

`(-2/11, 16/11)`

Explanation:

Let the vertices of triangle ABC be `A(2, 8)`, `B(3,4)` & `C(6, 3)`

Now, the equation of altitude drawn from vertex `A(2, 8)` which is perpendicular to the opposite side BC

`y-8=\frac{-1}{\frac{4-3}{3-6}}(x-2)`

`y-8=3(x-2)`

`3x-y=-2\ .........(1)`

Similarly, the equation of altitude drawn from vertex `B(3, 4)` which is perpendicular to the opposite side AC

`y-4=\frac{-1}{\frac{8-3}{2-6}}(x-3)`

`y-4=4/5(x-3)`

`4x-5y=-8\ .........(2)`

Multiplying (1) by `5` & then subtracting from (2) we get

`4x-5y-(15x-5y)=-8-(-10)`

`-11x=2`

`x=-2/11`

`\implies y=3x+2=3(-2/11)+2=16/11`

The ortho-center is the point of intersection of altitudes drawn from vertices to the opposite sides of a triangle hence the orthocenter of given triangle is

`(-2/11, 16/11)`