Let triangleABC " be the triangle with corners at"
A(4,1), B(1,3) and C(5,2)
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
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Slope of bar(AB) =(1-3)/(4-1)=-2/3
bar(AB)_|_bar(CN)=>slope of bar(CN)=3/2 ,
bar(CN) passes through C(5,2)
:.The equn. of bar(CN) is :y-2=3/2(x-5)
=>2y-4=3x-15
i.e. color(red)(3x-2y=11.....to (1)
Slope of bar(BC) =(2-3)/(5-1)=-1/4
bar(AL)_|_bar(BC)=>slope of bar(AL)=4 , bar(AL) passes through A(4,1)
:.The equn. of bar(AL) is :y-1=4(x-4)
=>y-1=4x-16
i.e. color(red)(y=4x-15.....to (2)
Subst. y=4x-15 into (1) ,we get
3x-2(4x-15)=11=>3x-8x+30=11
-5x=-19
=>color(blue)( x=19/5
From equn.(2) we get
y=4(19/5)-15=>y=(76-75)/5=>color(blue)(y=1/5
Hence, the orthocenter of triangle is (19/5,1/5)=(3.8,0.2)
