Let #triangleABC " be the triangle with corners at"#
#A(9,7), B(2,4) and C(8,6)#
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) =(7-4)/(9-2)=3/7#
#bar(AB)_|_bar(CN)=>#slope of # bar(CN)=-7/3# ,
# bar(CN)# passes through #C(8,6)#
#:.#The equn. of #bar(CN)# is #:y-6=-7/3(x-8)#
#3y-18=-7x+56#
#i.e. color(red)(7x+3y=74.....to (1)#
Slope of #bar(BC) =(6-4)/(8-2)=2/6=1/3#
#bar(AL)_|_bar(BC)=>#slope of # bar(AL)=-3# , # bar(AL)# passes through #A(9,7)#
#:.#The equn. of #bar(AL)# is #:y-7=-3(x-9)=>y-7=-3x+27#
#=>3x+y=34#
#i.e. color(red)(y=34-3x.....to (2)#
Subst. #color(red)(y=34-3x# into #(1)# ,we get
#7x+3(34-3x)=74=>7x+102-9x#=#74=>-2x=-28#
#=>color(blue)( x=14#
From equn.#(2)# we get
#y=34-3(14)=34-42=>color(blue)(y=-8#
Hence, the orthocenter of triangle is #(14,-8)#
