What is the orthocenter of a triangle with corners at #(1 ,3 )#, #(5 ,7 )#, and (2 ,3 )#?

Let the triangle be ABC with corners at

#A(1,3),B(5,7) and C(2,3).#

so,the slope of #"line "(AB)=(7-3)/(5-1)=4/4=1#

Let , #bar(CN) _|_bar( AB)#

#:.# The slope of #"line "CN=-1/1=-1#,and it passes through#C(2,3).#

#:.#The equn. of #"line "CN# ,is :

#y-3=-1(x-2)=>y-3=-x+2#

#i.e. x+y=5...to(1)#

Now , the slope of #"line "(BC)=(7-3)/(5-2)=4/3#

Let , #bar(AM) _|_bar( BC)#

#:.# The slope of #"line "AM=-1/(4/3)=-3/4#,and it passes through#A(1,3).#

#:.#The equn. of #"line "AM# ,is :

#y-3=-3/4(x-1)=>4y-12=-3x+3#

#i.e. 3x+4y=15...to(2)#

The intersection of #"line "CN and"line "AM# is the orthocenter of #triangleABC#.

So we solve equn. #(1) and (2)#

Multiply equn #(1)# by #3# and subtracting from #(2)# we get

#3x+4y=15...to(2)#

#ul(-3x-3y=-15)...to(1)xx(-3)#

#=>y=0#

From #(1)#,

#x+0=5=>x=5#

Hence, orthocentre of #triangle ABC# is #H(5,0)#
.......................................................................................

Note:

If #"line " l# passes through #P(x_1,y_1) and Q(x_2,y_2),then#

#(1)#slope of #l# is #=m=(y_2-y_1)/(x_2-x_1)#

#(2)#The equn. of #l# (passes thr' #P(x_1,y_1)# ,is :
#y-y_1=m(x-x_1)#

#(3)# If #l_1_|_l_2 ,then, m_1*m_2=-1=>m_2=-1/m_1#

#(4)# Orthocentre is the point,where three altitudes of triangle intersect.