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

1 Answer
Nov 4, 2017

Answer:

Orthocenter of the triangle is at #( 5.5,6.5) #

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 = (3,2) , B(4,5) , C(2,7) # . 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= (7-5)/(2-4)= -1#
Slope of perpendicular #AD# is #m_2= 1 (m_1*m_2=-1) #

Equation of line #AD# passing through #A(3,2)# is #y-2= 1(x-3)# or
#y-2 = x-3 or x-y=1 (1)#

Slope of #AB# is #m_1= (5-2)/(4-3) =3#
Slope of perpendicular #CF# is #m_2= -1/3 (m_1*m_2=-1) #

Equation of line #CF# passing through #C(2,7)# is #y-7= -1/3(x-2)# or
#y-7 = -1/3 x+2/3 or 1/3x+y = 7+2/3 or 1/3x+y = 23/3 # or
#x+3y=23 (2)#

Solving equation(1) and (2) we get their intersection point , which is the orthocenter.

#x-y=1 (1) ; x+3y=23(2)# Subtracting (1) from (2) we get,
#4y=22 :. y=5.5 ; x = y+1=6.5#

Orthocenter of the triangle is at #( 5.5,6.5) # [Ans]