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

1 Answer
Sep 28, 2017

Answer:

#(-6.bar(3),-1.bar(3))#

Explanation:

#Let# #A = (3,1)#
#Let# #B = (1,6)#
#Let# #C = (2, 2)#

Equation for altitude through A:
#x(x_3-x_2)+y(y_3-y_2)=x_1(x_3-x_2)+y1(y_3-y_2)#
#=>x(2-1)+y(2-6)=(3)(2-1)+(1)(2-6)#
#=>x-4y=3-4#
#=>color(red)(x-4y+1=0)#-----(1)

Equation for altitude through B:
#x(x_1-x_3)+y(y_1-y_3)=x_2(x_1-x_3)+y2(y_1-y_3)#
#=>x(3-2)+y(1-2)=(1)(3-2)+(6)(1-2)#
#=>x-y=1-6#
#=>color(blue)(x-y+5=0#-----(2)

Equating (1) & (2):
#color(red)(x-y+5)=color(blue)(x-4y+1#
#=>-y+4=1-5#
#=>color(orange)(y=-4/3#-----(3)

Plugging (3) in (2):
#color(blue)(x-4)color(orange)((-4/3))color(blue)(+1)=0#
#=>color(violet)(x=-19/3#

The orthocenter is at #(-19/3,-4/3)# OR #(-6.333...,-1.333...)#
which is actually outside the #triangle# because the #triangle# is an obtuse #triangle#. Click here to find more.