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

1 Answer
Oct 8, 2016

Answer:

The orthocenter is at the point #(4/3, -17/3)#

Explanation:

Let's begin by writing the equation of the line that goes through point #(3,1)# and perpendicular to the line going through points #(1,3)# and #(5,2)#.

The slope, m, of the line going through points #(1,3)# and #(5,2)# is:

#m = (3 - 2)/(1 - 5) = -1/4#

The slope, n, of any line perpendicular is

#n = -1/m = 4#

Use point slope form of the equation of a line to obtain the equation of a line that we desire:

#y - y_1 = n(x - x_1)#

#y - 1 = 4(x - 3)#

#y - 1 = 4x - 12#

#y = 4x - 11#

Write the equation of a line that goes through point #(5,2)# perpendicular to the line through the points #(3,1)# and #(1,3)#:

The slope, m, of the line that goes through the points #(3,1)# and #(1,3)# is:

#m = (3 - 1)/(1 - 3) = 2/-2 = -1#

The slope, n, of any line perpendicular is:

#n = -1/m = 1#

Again, Use the point slope form for the point #(5,2)#:

#y - y_1 = n(x - x_1)#

#y - 2 = 1(x - 5)#

#y = x - 7#

The orthocenter is at the intersection of these two lines:

#y = 4x - 11#
#y = x - 7#

#x - 7 = 4x - 11#

#4 = 3x#

#x = 4/3#

#y = 4/3 - 7#

#y = -17/3#

The orthocenter is at the point #(4/3, -17/3)#