What is the orthocenter of a triangle with corners at #(4 ,5 )#, #(3 ,7 )#, and (5 ,6 )#?

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Answer 1 · 2017-07-21T07:09:38

The orthocenter of the triangle is #=(13/3,17/3)#

Let the triangle #DeltaABC# be

#A=(4,5)#

#B=(3,7)#

#C=(5,6)#

The slope of the line #BC# is #=(6-7)/(5-3)=-1/2#

The slope of the line perpendicular to #BC# is #=2#

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

#y-5=2(x-4)#...................#(1)#

#y=2x-8+5=2x-3#

The slope of the line #AB# is #=(7-5)/(3-4)=2/-1=-2#

The slope of the line perpendicular to #AB# is #=1/2#

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

#y-6=1/2(x-5)#

#y=1/2x-5/2+6#

#y=1/2x+7/2#...................#(2)#

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

#2x-3=1/2x+7/2#

#2x-1/2x=7/2+3#

#3x=13#, #=>#, #x=13/3#

#y=2*13/3-3=17/3#

The orthocenter of the triangle is #=(13/3,17/3)#