Question

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

Answer 1

Orthocenter of the triangle is at `( 16,-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 perpendicular to the opposite side.

`A = (5,7) , B(2,3) , C(4,5)` . 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 line `BC` is `m_1= (5-3)/(4-2)= 1`

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

Equation of line `AD` passing through `A(5,7)` is

`y-7= -1(x-5) or y-7 =-x+5 or x+y =12 ; (1)`

Slope of line `AB` is `m_1= (3-7)/(2-5)=4/3`

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

Equation of line `CF` passing through

`C(4,5)` is `y-5= -3/4(x-4) or 4 y - 20 = -3 x +12` or

`3 x+4 y =32; (2)` Solving equation(1) and (2) we get their

intersection point , which is the orthocenter. Multiplying

equation (1) by `3` we get, `3 x+3 y =36 ; (3)` Subtracting

equation (3) from equation (2) we get,

`y = -4 :. x=12-y= 12+4=16 :. (x,y) = (16 , -4)`

Hence Orthocenter of the triangle is at `( 16,-4)` [Ans]