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

Question

1768 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-18T12:17:19

Orthocenter of the triangle is at #( 16,-4) #

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]