What is the orthocenter of a triangle with corners at #(2 ,8 )#, #(3 ,4 )#, and (6 ,3 )#?

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1 Answer
Harish Chandra Rajpoot
Jun 29, 2018

#(-2/11, 16/11)#

Explanation:

Let the vertices of triangle ABC be #A(2, 8)#, #B(3,4)# & #C(6, 3)#

Now, the equation of altitude drawn from vertex #A(2, 8)# which is perpendicular to the opposite side BC

#y-8=\frac{-1}{\frac{4-3}{3-6}}(x-2)#

#y-8=3(x-2)#

#3x-y=-2\ .........(1)#

Similarly, the equation of altitude drawn from vertex #B(3, 4)# which is perpendicular to the opposite side AC

#y-4=\frac{-1}{\frac{8-3}{2-6}}(x-3)#

#y-4=4/5(x-3)#

#4x-5y=-8\ .........(2)#

Multiplying (1) by #5# & then subtracting from (2) we get

#4x-5y-(15x-5y)=-8-(-10)#

#-11x=2#

#x=-2/11#

#\implies y=3x+2=3(-2/11)+2=16/11#

The ortho-center is the point of intersection of altitudes drawn from vertices to the opposite sides of a triangle hence the orthocenter of given triangle is

#(-2/11, 16/11)#

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