Why did this method gave the orthocentre of the triangle?

Question

Coordinates of vertices of a triangle are given(say $A$, $B$, $C$). I assumed coordinates of orthocenter 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 $m_1=-1/m_2$ (where $m$ 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? edit: $A(1,0)$, $B(-2,1)$, $C(5,2)$. Let $H(a,b)$. Slope of $AH=b/(a-1)$, slope of $BH = (b-1)/(a+2)$, slope of $CH = (b-2)/(a-5)$. I assumed $AH, BH, CH$ to be mutually perpendicular. Slopes of perpendicular lines have relation $p*q=-1$ where p and q are slopes of the lines. Using the above relation I got $a^2+b^2-a-b-2=0$, $a^2+b^2-6a-2b+5=0$ and $a^2+b^2-3a-3b-8=0$. On solving the three equations i got $a=2$ and $b=-7$ which are the correct coordinates.

Answer 1

The subtraction step (which we identified in the comments as the step where you move from incorrect to correct) effectively replaces two of the lines, e.g., $BH$ and $CH$, with their "difference" (in this example $BC$) which is one of the sides of the triangle. Thus, though you start with equations requiring $AH$,$BH$,$CH$ mutually perpendicular, you end up with an equation describing the weaker condition that $AH$ is perpendicular to $BC$ (and similarly for other sides) -- which is exactly what defines the orthocenter.

To flesh this out a bit: denote the slope of $AH$ by $m_{AH}=\frac{\text{rise}_{AH}}{\text{run}_{AH}}$ (and similarly for other lines). You start with the equation $$m_{AH}m_{BH}=\frac{\text{rise}_{AH}\text{rise}_{BH}}{\text{run}_{AH}\text{run}_{BH}}=-1$$ and multiply out to get $$\text{rise}_{AH}\text{rise}_{BH} = - \text{run}_{AH}\text{run}_{BH}$$ Similarly you get $$\text{rise}_{AH}\text{rise}_{CH} = - \text{run}_{AH}\text{run}_{CH}$$ and then you subtract the two to get $$\text{rise}_{AH}(\text{rise}_{BH}-\text{rise}_{CH}) = - \text{run}_{AH}(\text{run}_{BH}-\text{run}_{CH})$$ But $\text{rise}_{BH}-\text{rise}_{CH} = \text{rise}_{BC}$, and similarly for the run. Thus your equation says $$\text{rise}_{AH}\text{rise}_{BC} = - \text{run}_{AH}\text{run}_{BC}$$ i.e. $$m_{AH}m_{BC}=-1$$