Let #triangleABC " be the triangle with corners at"#
#A(1,3), B(6,9) and C(2,4)#
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) =(3-9)/(1-6)=6/5#
#bar(AB)_|_bar(CN)=>#slope of # bar(CN)=-5/6# ,
# bar(CN)# passes through #C(2,4)#
#:.#The equn. of #bar(CN)# is #:y-4=-5/6(x-2)#
#6y-24=-5x+10#
#i.e. color(red)(5x+6y=34.....to (1)#
Slope of #bar(BC) =(9-4)/(6-2)=5/4#
#bar(AL)_|_bar(BC)=>#slope of # bar(AL)=-4/5# , # bar(AL)# passes through #A(1,3)#
#:.#The equn. of #bar(AL)# is #:y-3=-4/5(x-1)#
#=>5y-15=-4x+4#
#=>color(red)(4x+5y=19.....to (2)#
Taking #eqn.(1)xx5-eqn.(2)xx(-6)# and adding
#color(white)(....)25x+30y=170#
#ul(-24x-30y=-114#
#:.1x-0=56#
#=>color(blue)( x=56#
From equn.#(2)# we get
#4(56)+5y=19=>5y=19-224=-205=>color(blue)(y=-41#
Hence, the orthocenter of triangle is #(56,-41)#
