What is the orthocenter of a triangle with corners at #(1, 3)#, #(6, 2)#, and #(5, 4)#?

1 Answer
Apr 22, 2018

#(x, y)=(47/9, 46/9)#

Explanation:

Let: A(1, 3), B(6, 2) and C(5, 4) be the vertices of triangle ABC:

Slope of a line through points: #(x_1, y_1), (x_2, y_2)#:
#m=(y_2-y_1)/(x_2-x_1)#

Slope of AB:
#=(2-3)/(6-1)=-1/5#
Slope of perpendicular line is 5.
Equation of the altitude from C to AB:
#y-y_1=m(x-x_1)# =>#m=5, C(5,4)#:
#y-4=5(x-5)#
#y=5x-21#

Slope of BC:
#=(4-2)/(5-6)=-2#
Slope of perpendicular line is 1/2.
Equation of the altitude from A to BC:
#y-3=1/2(x-1)#
#y=(1/2)x+5/2#

The intersection of the altitudes equating y's:
#5x-21=(1/2)x+5/2#
#10x-42=x+5#
#9x=47#
#x=47/9#

#y=5*47/9- 21#
#y=46/9#

Thus the Orthocenter is at #(x, y)=(47/9, 46/9)#

To check the answer you can find the equation of altitude from B to AC and find the intersection of that with one of the other altitudes.