What is the orthocenter of a triangle with corners at #(9 ,7 )#, #(6 ,9 )#, and (2 ,4 )#?

1 Answer
Feb 17, 2016

Orthocenter is at #(132/23,221/23)~~(5.74,9.61)#
#color(white)("XXX")#...based on the outrageous assumption that my arithmetic is valid

Explanation:

For convenience label the points on this triangle as in the diagram below:
enter image source here
Note that (by definition) the orthocenter of a triangle is the point where the altitudes meet (labelled #O# in this diagram).

Slope(#AB#)#=(7-4)/(9-2)=3/7#
Slope(#CD#)#=-7/3# (since it is perpendicular to #AB#)
Equation of line (#CD#) [using point slope form is]
#color(white)("XXX")y-9=-7/3(x-6)#
#color(white)("XXX")3y-27=-7x+42#
#color(white)("XXX")7x+3y=69#

Slope(#CB#)#=(9-7)/(6-9)=-2/3#
Slope(#AE#)#=3/2#
Equation of line (#AE#)
#color(white)("XXX")y-4=3/2(x-2)#
#color(white)("XXX")2y-8=3x-6#
#color(white)("XXX")3x-2y=-2#

Solving the system of equations:
#{([1],,7x+3y=69),([2],,3x-2y=-2):}#

#rArr {([1]xx2,,14x+6y=138),([2]xx3,,9x-6y=-6):}#

#rarr 23x=132#
#rarr x=132/23#

and
#rArr {([1]xx3,,21x+9y=207),([2]xx7,,21x-14=-14):}#

#rarr 23y=221#
#rarr y=221/23#