What is the orthocenter of a triangle with corners at #(5 ,4 )#, #(2 ,3 )#, and (7 ,8 )#?

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Answer 1 · 2017-06-17T09:54:25

The orthocenter is #=(10,-1)#

Let the triangle #DeltaABC# be

#A=(5,4)#

#B=(2,3)#

#C=(7,8)#

The slope of the line #BC# is #=(8-3)/(7-2)=5/5=1#

The slope of the line perpendicular to #BC# is #=-1#

The equation of the line through #A# and perpendicular to #BC# is

#y-4=-1(x-5)#

#y-4=-x+5#

#y+x=9#...................#(1)#

The slope of the line #AB# is #=(3-4)/(2-5)=-1/-3=1/3#

The slope of the line perpendicular to #AB# is #=-3#

The equation of the line through #C# and perpendicular to #AB# is

#y-8=-3(x-7)#

#y-8=-3x+21#

#y+3x=29#...................#(2)#

Solving for #x# and #y# in equations #(1)# and #(2)#

#y+3(9-y)=29#

#y+27-3y=29#

#-2y=29-27=2#

#y=-2/2=-1#

#x=9-y=9+1=10#

The orthocenter of the triangle is #=(10,-1)#