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)#
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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.
