Using slope of lines to determine orthocenter

[Guest]

A(1,0) B(-2,1) C(5,2) are the vertices of a triangle. I assumed coordinates of orthocentre to be H(a, b). I calculated slope of AH, BH, CH in terms of a and b. For some reason I treated these three as mutually perpendicular lines and related their slopes using [tex]x_{1 } = -\frac{1}{x_{2}}[/tex] (where x is the slope of line). I got three equations in a and b, and on solving them i got the correct coordinates.
How did the answer came out to be true despite the fact that the three lines are not perpendicular. I tried another example just to be sure and it also gave the right answer.

[Guest]

The "orthocenter" of a triangle is the point where the three "altitudes" of the triangle intersect. An "altitude" of a triangle is a line, through a vertex of the triangle, perpendicular to the opposites side. There is NO requirement that the altitudes be perpendicular to each other!

The three vertices of this triangle are A= (1, 0), B= (-2, 1), C= (5,2).

Side AB has slope (1- 0)/(-2-1)= -1/3. The altitude perpendicular to that side has slope 3 and passes through (5, 2) so has equation y= 3(x- 5)+ 2= 3x- 15+ 2= 3x- 13.

Side BC has slope (2- 1)/(5-(-2))= 1/7. The altitude perpendicular to that side has slope -7 and passes through (1, 0) so has equation y= -7(x- 1)+ 0= -7x+ 7.

Since a point is determined by two lines, that is sufficient to determine the orthocenter: y= 3x- 13= -7x+ 7. 10x- 20, x= 2, y= 3(2)- 13= 6- 13= -7. Note that y= -7(2)+ 7= -7. The orthocenter is (2, -7).

As a check, the third side, AC, has slope (2- 0)/(5- 1)= 2/4= 1/2. The altitude perpendicular to that side has slope -2 and passes through (-2, 1) so has equation y= -2(x-(-2))+ 1= -2x- 4+ 1= -2x- 3. When x= 2, y= -2(2)- 3= -7. Yes, that altitude also passes through (2, -7), confirming that the orthocenter is (2, -7).