What is the orthocenter of a triangle with corners at #(2 ,3 )#, #(5 ,7 )#, and (9 ,6 )#?

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Answer 1 · 2018-01-13T10:58:11

Orthocenter of the triangle is at #( 71/19,189/19) #

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(2,3) , B(5,7) , C(9,6) # . 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= (6-7)/(9-5)= -1/4#

Slope of perpendicular #AD# is #m_2= 4; (m_1*m_2=-1) #

Equation of line #AD# passing through #A(2,3)# is

#y-3= 4(x-2) or 4x -y = 5 (1)#

Slope of #AB# is #m_1= (7-3)/(5-2)= =4/3#

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

Equation of line #CF# passing through #C(9,6)# is

#y-6= -3/4(x-9) or y-6 = -3/4 x+27/4# or

#4y -24= -3x +27 or 3x+4y=51 (2)#

Solving equation(1) and (2) we get their intersection point , which

is the orthocenter. Multiplying equation (1 )by #4# we get

#16x -4y = 20 (3)# Adding equation (3 ) and equation (2 )

we get, #19x =71 :. x=71/19 ; y =4x-5or y=4*71/19-5 # or

#y=189/19#. Orthocenter of the triangle is at #(x,y)# or

#( 71/19,189/19) # [Ans]