Let #triangleABC# be the triangle with corners at
#A(2,0), B(3,4) and C(6,3)#.
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 .
#diamond#Slope of #bar(AB)#=#(4-0)/(3-2)#=#4=>#slope of #bar(CN)#=#-1/4[because#altitudes]
Now, #bar(CN)# passes through #C(6,3)#
#:.# Equn. of #bar(CN)# is: #y-3=-1/4(x-6)#
#i.e. color(red)(x+4y=18...to(1)#
#diamond#Slope of #bar(BC)#=#(3-4)/(6-3)#=#-1/3=>#slope of #bar(AL)=3[because#altitudes]
Now, #bar(AL)# passes through #A(2,0)#
#:.# Equn. of #bar(AL)# is: #y-0=3(x-2)#
#i.e. color(red)(3x-y=6...to(2)#
#=>color(red)(y=3x-6...to(3)#
Putting ,#y=3x-6# into #(1)# we get
#x+4(3x-6)=18=>x+12x-24=18#
#=>13x=42#
#=>color(blue)(x=42/13#
From #(3)# we get,
#y=3(42/13)-6=(126-78)/13#
#=>color(blue)(y=48/13#
Hence, **the orthocenter of the triangle is :
** # (42/13,48/13)~~(3.23,3.69)#
Please see the graph.
