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

Question

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

Maverick Smith · Accepted Answer

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

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]

Parker Allen · Answer

undefined To find the orthocenter of a triangle with vertices at (4, 9), (3, 7), and (1, 1), follow these steps:

1. Calculate the slopes of the lines passing through each pair of vertices to find the perpendicular bisectors.
2. Find the equations of the perpendicular bisectors.
3. Find the intersection point of any two perpendicular bisectors. This point will be the orthocenter of the triangle.