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
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