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)