What is the orthocenter of a triangle with corners at #(1 ,3 )#, #(5 ,7 )#, and (9 ,8 )#?

1 Answer
Rui D.
Mar 25, 2016

#(-10/3,61/3)#

Explanation:

Repeating the points:
#A(1,3)#
#B(5,7)#
#C(9,8)#

The orthocenter of a triangle is the point where the line of the heights relatively to each side (passing through the opposed vertex) meet. So we only need the equations of 2 lines.

The slope of a line is #k=(Delta y)/(Delta x)# and the slope of the line perpendicular to the first is #p=-1/k# (when #k!=0#).

#AB-> k_1=(7-3)/(5-1)=4/4=1# => #p_1=-1#
#BC-> k=(8-7)/(9-5)=1/4# => #p_2=-4#

Equation of line (passing through #C#) in which lays the height perpendicular to AB
#(y-y_C)=p(x-x_C)# => #(y-8)=-1*(x-9)# => #y=-x+9+8# => #y=-x+17# [1]

Equation of line (passing through #A#) in which lays the height perpendicular to BC
#(y-y_A)=p(x-x_A)# => #(y-3)=-4*(x-1)# => #y=-4x+4+3# => #y=-4x+7#[2]

Combining equations [1] and [2]
#{y=-x+17#
#{y=-4x+7# => #-x+17=-4x+7# => #3x=-10# => #x=-10/3#

#->y=10/3+17=(10+51)/3# => #y=61/3#

So the orthocenter #P_"orthocenter"# is #(-10/3,61/3)#

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