What is the orthocenter of a triangle with corners at #(5 ,9 )#, #(4 ,3 )#, and (1 ,5 )#?

Repeating the points:
#A(5,9)#
#B(4,3)#
#C(1,5)#

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=(3-9)/(4-5)=(-6)/(-1)=6# => #p=-1/6#
#BC-> k=(5-3)/(1-4)=2/(-3)=-2/3# => #p=3/2#
#CA-> k=(9-5)/(5-1)=4/4=1# => #p=-1#

(It should be obvious that if we choose, for one of the equations the slope #p=-1# our task would be easier. I'll choose indifferently, I'll choose the first and second slopes)

Equation of line (passing through #C#) in which the height perpendicular to AB lays
#(y-5)=-(1/6)(x-5)# => #y=(-x+1)/6+5# => #y=(-x+31)/6#[1]

Equation of line (passing through #A#) in which the height perpendicular to BC lays
#(y-9)=(3/2)(x-5)# => #y=(3x-15)/2+9# => #y=(3x+3)/2# [2]

Combining equations [1] and [2]
#{y=(-x+31)/6#
#{y=(3x+3)/2# => #(-x+31)/6=(3x+3)/2# => #-2x+62=18x+18# => #x=44/20# => #x=11/5#
#-> y=(-11/5+31)/6=(-11+155)/30=144/30# => #y=24/5#

So the orthocenter is #(11/5,24/5)#