Let #triangle ABC# be the triangle with corners at
#A(4,9) ,B(3,4) and C(1,1)#
Let #bar(AL) , bar(BM) and bar(CN) # be the altitudes of sides
#bar(BC) ,bar(AC) ,and bar(AB)# respectively.
Let #(x,y)# be the intersection of three altitudes .
Slope of #bar(AB) =(9-4)/(4-3)=5#
#bar(AB)_|_bar(CN)=>#slope of # bar(CN)#=#-1/5# , # bar(CN)# passes through #C(1,1)#
#:.#The equn. of #bar(CN)# is #:y-1=-1/5(x-1)#
#=>5y-5=-x+1#
#i.e. color(red)(x=6-5y.....to (1)#
Slope of #bar(BC) =(4-1)/(3-1)=3/2#
#bar(AL)_|_bar(BC)=>#slope of # bar(AL)=-2/3# , # bar(AL)# passes through #A(4,9)#
#:.#The equn. of #bar(AL)# is #:y-9=-2/3(x-4)=>3y-27=-2x+8#
#i.e. color(red)(2x+3y=35.....to (2)#
Subst. #x=6-5y# into #(2)# ,we get
#2(6-5y)+3y=35#
#=>-7y=23#
#=>color(blue)( y=-23/7#
From equn.#(1)# we get
#x=6-5(-23/7)=(42+115)/7=>color(blue)(x=157/7#
Hence, the orthocenter of triangle is #(157/7,-23/7)#
