Question

What is the orthocenter of a triangle with corners at `(6 ,3 )`, `(2 ,4 )`, and (7 ,9 )#?

Answer 1

Orthocenter of the triangle is at `(5.6,3.4)`

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 = (6,3) , B(2,4) , C(7,9)` . 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= (9-4)/(7-2)=5/5= 1`

Slope of perpendicular `AD` is `m_2= -1 (m_1*m_2=-1)`

Equation of line `AD` passing through `A(6,3)` is

`y-3= -1(x-6)or y-3 = -x+6 or x +y = 9 (1)`

Slope of `AB` is `m_1= (4-3)/(2-6)= -1/4`

Slope of perpendicular `CF` is `m_2= -1/(-1/4)=4`

Equation of line `CF` passing through `C(7,9)` is

`y-9= 4(x-7) or y-9 = 4x-28 or 4x-y=19 (2)`

Solving equation(1) and (2) we get their intersection point , which

is the orthocenter. Adding equation(1) and (2) we get,

`5x=28 or x = 28/5=5.6 and y= 9-x =9-5.6 =3.4`

Orthocenter of the triangle is at `(5.6,3.4)` [Ans]