First find the equation of line of the side of the triangle,

#bar(AB), bar(BC) " and " bar(CA)#

**Equation of line sets for: #color(red)(bar(AB))#:**

#y_(bar(AB)) = m_(bar(AB)) x+b_(bar(AB))# with #m=(3-5)/(6-2)=-1/2#

Use B(2,5) which is on #y_(bar(AB)) = m_(bar(AB)) x+b_(bar(AB))#

Letting #y=5# and #x=2# solve for #b_(bar(AB))#

#5=-1/2*2+b_(bar(AB)); b_(bar(AB)) = 6# thus

#y_(bar(AB)) = -1/2 x+6#

A perpendicular equation of line will have a slope

#m_(bar(AB))=-1/m_(bar(AB)); -1/2=-1/m_(bar(AB));m_(Pr_bar(AB))= 2#

Thus the perpendicular equation of linr to AB is:

#y_(Pr_bar(AB)) = 2x+ b_(Pr_bar(AB))#

use #A(7,9)# to determine #b_(Pr_bar(AB))#

#9 = 2*7+ b_(Pr_bar(AB)); b_(Pr_bar(AB))=-5#

So the pair of equation sets, for #color(red)bar(AB)# are:

#[(y), (color(blue)(y_(pr)))]_(bar(AB)) = [(-1/2, 6) , (color(blue)(2), color(blue)(-5) )]*[(x), (1) ]#

**Equation of line sets for: #color(red)(bar(BC))#:**

#y_(bar(A=BC))=m_(bar(BC)) x+b_(bar(BC))# with #m=(9-5)/(7-2)=4/5#

#5=4/5*2+b; b=17/5# the perpendicular has slope and y-intercept:

#m_(pr) = -1/m; m_(pr)=-5/4#

Using point# A(6,3)# to determine #b_(pr)#:

#3=-5/4*6+b_(pr); b_(pr)=21/2#

#[(y), (color(magenta)(y_(pr)))]_(bar(BC)) = [(4/5, 17/5) , (color(magenta)(-5/4), color(magenta)(21/2))]*[(x), (1) ]#

The Orthocenter is the point where the perpendicular line to AB and BC meet, i.e.: #color(blue)(y_(Pr_bar(AB))) = color(magenta)(y_(Pr_bar(BC))#

#color(blue)(2x-5)=color(magenta)(-5/4x +21/2); x=62/13#

Using x=62/13 solve for y:

#y=2x-5|_(x=(62/13))#, #y=2*62/13-5= 59/13#