How does one calculate the coordinates of a triangle's orthocentre?
I was surfing through the net and got this formula:
$$x-\rm{coordinate}= \frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
$$y-\rm{coordinate}= \frac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A+\tan B+\tan C} $$
How do you prove this? Is there any method using just the coordinates?
On
Consider $ \Delta ABC $ in the plane. For now take origin at $ A $, and positions of $ B, C $ as $ \vec{b} = (b_1, b_2)$ and $ \vec{c} = (c_1, c_2) $ respectively.
Any vector perpendicular to nonzero vector $ \vec{x} = (x_1, x_2) $ is of the form $t (-x_2, x_1) $ with $ t \in \mathbb{R} $. So altitude line through $ B $ is $ (\vec{b} + t(-c_2, c_1))_{t \in \mathbb{R}}$, and that through $ C $ is $ (\vec{c} + s(-b_2, b_1))_{s \in \mathbb{R}} $.
For their intersection point we should solve $ b_1 - \color{goldenrod}{t}c_2 = c_1 - \color{goldenrod}{s} b_2 $ and $ b_2 + \color{goldenrod}{t}c_1 = c_2 + \color{goldenrod}{s}b_1 $ for $ \color{goldenrod}{s},\color{goldenrod}{t} $. Taking $ \text{Eq1} \times b_1 + \text{Eq2} \times b_2 $ eliminates $ s $ to give $ b_1 ^2 -tb_1 c_2 + b_2 ^2 + tb_2 c_1 = b_1 c_1 + b_2 c_2$, i.e. $ t = \frac{\vec{b} \cdot (\vec{b} - \vec{c})}{\text{det}(\vec{b}, \vec{c})} $. So they intersect at $ \vec{b} + \frac{\vec{b} \cdot (\vec{b} - \vec{c})}{\text{det}(\vec{b}, \vec{c})} (-c_2, c_1) $, which on expanding becomes $ \frac{\vec{b}\cdot\vec{c}}{\det(\vec{b}, \vec{c})} (-(b_2 - c_2), b_1 - c_1) .$
It can be rewritten as $ \frac{\vec{b}\cdot \vec{c}}{\det(\vec{b}, \vec{c})} (\vec{b}-\vec{c})^{\perp} $, where $ \vec{z}^{\perp} := (-z_2, z_1) $. This is clearly perpendicular to $ \vec{b} - \vec{c} $, and hence lies on the altitude through $ A $, proving concurrency.
To summarise,
Th: Consider points $ A(\vec{a}),B(\vec{b}),C(\vec{c}) $ in the plane forming a triangle. Defining $ f(\vec{x}, \vec{y}) := \dfrac{\vec{x}\cdot\vec{y}}{\det(\vec{x}, \vec{y})} (\vec{x}-\vec{y})^{\perp}$ where $ \vec{z}^{\perp} := (-z_2, z_1) $, the altitudes of $ \Delta ABC $ all intersect at $ \vec{a} + f(\vec{b}-\vec{a}, \vec{c}-\vec{a}) .$
Hint: In barycentric coordinates system, coordinates of a point $X$ in the plane of triangle $\Delta ABC$ is determined by the ratios $\lambda_1=\frac{[\Delta XBC]}{[\Delta ABC]},\lambda_2 =\frac{[\Delta XCA]}{[\Delta ABC]}$, and $\lambda_3=\frac{[\Delta XAB]}{[\Delta ABC]}$ where the brackets denote the (signed) area of the enclosed triangles. That is to say $X=\lambda_1 A+\lambda_2 B + \lambda_3 C$. You only need to show that for $X$ being the orthocenter we have $\lambda_1=\frac{\tan A}{\sum \tan A}$, $\lambda_2=\frac{\tan B}{\sum \tan A}$, and $\lambda_3=\frac{\tan C}{\sum \tan A}$.