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

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)#