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

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Answer 1 · 2016-10-18T18:19:21

Let the coordinates of three vertices of the triangle ABC be

#A->(7,8)" "B->(3,4)" "C->(8,3)#

Let the coordinate of the#color(red)(" Ortho center O"->(h,k))#

#m_(AB)->"Slope of AB"=((8-4))/((7-3))=1#

#m_(BC)->"Slope of BC"=((4-3))/((3-8))=-1/5#

#m_(CO)->"Slope of CO"=((k-3))/((h-8))#

#m_(AO)->"Slope of AO"=((k-8))/((h-7))#

O being orthocenter the straight line passing through C and O will be perpendicular to AB,

So #m_(CO)xxm_(AB)=-1#

#=> ((k-3))/((h-8))xx 1=-1#

#=>k=-h+11....(1)#

O being orthocenter the straight line passing through A and O will be perpendicular to BC,

So #m_(AO)xxm_(BC)=-1#

#=> ((k-8))/((h-7))xx(- 1/5)=-1#

#=>k=5h-27....(2)#

Comparing (1) and (2)

#5h-27=-h+11#

#=>6h=38#

#=>h=6 1/3#

Inserting the value of h in(1)

#k=-6 1/3+11=4 2/3#

Hence coordinate of the orthocenter is

#color(green)((6 1/3 "," 4 2/3))#