What is the orthocenter of a triangle with corners at #(9 ,7 )#, #(1 ,1 )#, and (3 ,2 )#?

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Answer 1 · 2016-06-09T15:55:51

#p_o = {2.89005, 2.16812}#

Given a triangle by through its vertices #p_1,p_2,p_3# the orthocenter #p_o# is at the intersection of the corner angles bissectrices.

Taking vertice #p_2# the bissectrix line is given by

#b_2->p = p_2+lambda_2 vec v_2 #

where #p = {x,y}# and

#vec v_2 = (p_1-p_2)/(norm(p_1-p_2))+(p_3-p_2)/(norm(p_3-p_2))#

also for vertice #p_3#

#b_3->p = p_3+lambda_3 vec v_3 #

with

#vec v_3 = (p_1-p_3)/(norm(p_1-p_3))+(p_2-p_3)/(norm(p_2-p_3))#

So the orthocenter is in #b_2 nn b_3# or solving for #lambda_2,lambda_3#

#p_2+lambda_2 vec v_2 = p_3+lambda_3 vec v_3#

In our case we have

#v_2 = {4/5 + 2/(sqrt[5]), 3/5 + 1/(sqrt[5])}#
#v_3 = {-2/(sqrt[5]) + 6/(sqrt[61]), -1/(sqrt[5]) + 5/(sqrt[61])}#

and #{lambda_2 = 1.11545, lambda_3 = 0.871195}#

so #p_o = {2.89005, 2.16812}#

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