What is the orthocenter of a triangle with corners at (4 ,9 ), (3 ,7 ), and (1 ,1 )#?

1 Answer
Jul 12, 2017

Orthocenter of the triangle is at ( -53,28)

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 = (4,9) , B(3,7) , C(1,1) . 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= (1-7)/(1-3)= 3
Slope of perpendicular AD is m_2= -1/3 (m_1*m_2=-1)
Equation of line AD passing through A(4,9) is y-9= -1/3(x-4) or
y-9 = -1/3 x+4/3 or y +1/3x = 9+4/3 or y +1/3x = 31/3 (1)

Slope of AB is m_1= (7-9)/(3-4)= =2
Slope of perpendicular CF is m_2= -1/2 (m_1*m_2=-1)
Equation of line CF passing through C(1,1) is y-1= -1/2(x-1) or
y-1 = -1/2 x+1/2 or y +1/2x = 1+1/2 or y +1/2x = 3/2 (2)

Solving equation(1) and (2) we get their intersection point , which is the orthocenter.
y +1/3x = 31/3 (1)
y +1/2x = 3/2 (2) Subtracting (2) from (1) we get,

-1/6x = (31/3-3/2) = 53/6 or x = - 53/cancel6*cancel6 or x=-53
Putting x= -53 in equation (2) we get y-53/2 =3/2 or y=53/2+3/2 or 56/2=28 :. x= -53 ,y = 28

Orthocenter of the triangle is at ( -53,28) [Ans]