Question

What is the orthocenter of a triangle with corners at `(3 ,2 )`, `(4 ,5 )`, and (2 ,7 )#?

Answer 1

Orthocenter of the triangle is at `( 5.5,6.5)`

Explanation:

Orthocenter is the point where the three "altitudes" of a triangle meet. An "altitude" is a line that goes through a vertex (corner point) and is at right angles to the opposite side.

`A = (3,2) , B(4,5) , C(2,7)` . Let `AD` be the altitude from `A` on `BC` and `CF` be the altitude from `C` on `AB` they meet at point `O` , the orthocenter.

Slope of `BC` is `m_1= (7-5)/(2-4)= -1`
Slope of perpendicular `AD` is `m_2= 1 (m_1*m_2=-1)`

Equation of line `AD` passing through `A(3,2)` is `y-2= 1(x-3)` or
`y-2 = x-3 or x-y=1 (1)`

Slope of `AB` is `m_1= (5-2)/(4-3) =3`
Slope of perpendicular `CF` is `m_2= -1/3 (m_1*m_2=-1)`

Equation of line `CF` passing through `C(2,7)` is `y-7= -1/3(x-2)` or
`y-7 = -1/3 x+2/3 or 1/3x+y = 7+2/3 or 1/3x+y = 23/3` or
`x+3y=23 (2)`

Solving equation(1) and (2) we get their intersection point , which is the orthocenter.

`x-y=1 (1) ; x+3y=23(2)` Subtracting (1) from (2) we get,
`4y=22 :. y=5.5 ; x = y+1=6.5`

Orthocenter of the triangle is at `( 5.5,6.5)` [Ans]