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

1 Answer
Binayaka C.
Jun 19, 2016

#(40/7,30/7)# is the intersection point of altitudes and is the orthcenter of the triangle.

Explanation:

Orthocenter of a triangle is the point of intersection of all the altitudes of the triangle. Let A(4,3) , B(5,4) and C(2,8,) are the vertices of the triangle.
Let AD be the altitude drawn from A perpendiclar to BC and CE be the altitude drawn fromC on AB.
Slope of the line BC is #(8-4)/(2-5)= -4/3 :. #Slope of AD is #-1/(-4/3) = 3/4#The equation of altitude AD is #y-3= 3/4(x-4) or 4y-12=3x-12 or 4y-3x=0 (1)#
Now Slope of the line AB is #(4-3)/(5-4)=1 :. #Slope of CE is #-1/1 = -1#The equation of altitude CE is #y-8= -1(x-2) or y+x=10(2)#
Solving #4y-3x =0 (1) #and #y+x=10 (2)# we get #x = 40/7 ; y =30/7 :. (40/7,30/7)# is the intersection point of two altitudes and is the orthcenter of the triangle.[Ans]

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