Question

What is the orthocenter of a triangle with corners at `(2 ,7 )`, `(1 ,2 )`, and (3 ,5 )#?

Answer 1

Orthocenter is at `(41/7,31/7)`

Explanation:

Slope of the line AB: `m_1= (2-7)/(1-2)=5`
Slope of CF=perpendicular slope of AB: `m_2= -1/5`
Equation of the line CF is `y-5= -1/5(x-3) or 5y-25 =-x+3 or x+5y=28 (1)`
Slope of the line BC: `m_3= (5-2)/(3-1)=3/2`
Slope of AE=perpendicular slope of BC: `m_4= -1/(3/2)=-2/3`
Equation of the line AE is `y-7= -2/3(x-2) or 3y-21 =-2x+4 or 2x+3y=25 (2)` The intersection of CF & AE is the orthocenter of the triangle, which can be obtained by solving equation(1) & (2)
`x+5y=28 (1)` ; `2x+3y=25 (2)`
`2x+10y=56 (1)` obtained by multiplying 2 on both sides
`2x+3y=25 (2)` subtracting we get `7y=31:. y=31/7; x=28-5*31/7=41/7:.`Orthocenter is at `(41/7,31/7)`[Ans]