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}#