Is there is a formula to calculate the coordinates of the orthocenter of a triangle?

Question

I'm trying to find the coordinates of the orthocenter (the intersection point of all altitudes) of a triangle given its vertices' coordinates $A=(x_1, y_1), \ B=(x_2, y_2) , \ C=(x_3, y_3)$.

I searched for a general formula but couldn't find one by googling, while formulas for the incenter and centroid seem readily available. This suggests there might not be a simple formula for the orthocenter and could be quite complex.

I attempted to derive the formula myself, but long calculations lead to errors. I formulated and tested an expression three times, but it yielded incorrect results.

Answer 1

There is a formula that is easier to remember, using complex numbers.

If the complex coordinates of the points are $z_1$, $z_2$, $z_3$ then we have the center of the circumscribed circle $o$ and the center of the nine point circle $e$

$$o= \frac{ \left | \begin{matrix} 1 & z_1 & |z_1|^2 \\ 1 & z_2 & |z_2|^2 \\ 1 & z_3 & |z_3|^2 \end{matrix} \right | } { \left | \begin{matrix} 1 & z_1 & \bar z_1 \\ 1 & z_2 & \bar z_2 \\ 1 & z_3 & \bar z_3 \end{matrix} \right | }$$

and

$$e = \frac{1}{2}\cdot \frac{ \left | \begin{matrix} 1 & z_1^2 & \bar z_1 \\ 1 & z_2^2 & \bar z_2 \\ 1 & z_3^2 & \bar z_3 \end{matrix} \right | } { \left | \begin{matrix} 1 & z_1 & \bar z_1 \\ 1 & z_2 & \bar z_2 \\ 1 & z_3 & \bar z_3 \end{matrix} \right | }$$

Now use

$$h = 2 e - o$$

Answer 2

For info on tens of thousands of triangle centers, consult Clark Kimberling's Encyclopedia of Triangle Centers; in particular, the orthocenter entry $X(4)$ gives the following barycentric coordinates: $$u:v:w =\frac1{-a^2+b^2+c^2}:\frac1{-b^2+c^2+a^2}:\frac1{-c^2+a^2+b^2}$$ (As the components involve squares of side-lengths, it's straightforward to re-write them in terms of $(x_i,y_i)$.) These allow you to express the coordinates of the orthocenter as a weighted average of the coordinates of the vertices: $$\frac{uA+vB+wC}{u+v+w}$$

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As a user of GeoGebra, OP should be pleased to know of the app's built-in support for this stuff via the command TriangleCenter(A, B, C, n), where n is the ETC index (incenter, $n=1$; centroid, $n=2$; circumcenter, $n=3$; orthocenter, $n=4$; nine-point center, $n=5$; etc, etc, ETC ...). I believe GeoGebra currently supports up to just-over $n=3000$; I actually used the feature to search for centers that could answer this question, finding $n=1138$ as a result.

(You can also have GeoGebra calculate barycentric $u$-$v$-$w$ coordinates individually, and then define the corresponding point as (u A + v B + w C)/(u + v + w). I have to do that from time to time.)

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Mathematica also has a TriangleCenter[tri, type] function, but (a little surprisingly) it supports only a few named types. As Orthocenter is supported, the app can generate an explicit coordinate formula for you:

Factor[TriangleCenter[{{x1, y1}, {x2, y2}, {x3, y3}}, "Orthocenter"]] yields:

{-(x1 x2 y1 - x1 x3 y1 - x1 x2 y2 + x2 x3 y2 + y1^2 y2 - y1 y2^2 + x1 x3 y3 - x2 x3 y3 - y1^2 y3 + y2^2 y3 + y1 y3^2 - y2 y3^2)/(-x2 y1 + x3 y1 + x1 y2 - x3 y2 - x1 y3 + x2 y3), ( x1^2 x2 - x1 x2^2 - x1^2 x3 + x2^2 x3 + x1 x3^2 - x2 x3^2 + x1 y1 y2 - x2 y1 y2 - x1 y1 y3 + x3 y1 y3 + x2 y2 y3 - x3 y2 y3)/(- x2 y1 + x3 y1 + x1 y2 - x3 y2 - x1 y3 + x2 y3)}

which is a bit of a mess. With a bit of effort, cleverness, and luck, you could re-arrange the above into the form

{x1,y1} p/s + {x2,y2} q/s + {x3,y3} r/s

where expressions p, q, r are obtained from each other by cycling elements: x1$\to$x2$\to$x3$\to$x1 and y1$\to$y2$\to$y3$\to$y1, and s is fully-symmetric. As it happens, p/s, q/s, r/s (with s = p+q+r) are equivalent to the $u$-$v$-$w$ expressions above, with lengths expanded using Cartesian-coordinates.

This highlights another benefit of barycentric coordinates (and similar representations): notational tidiness.

Answer 3

Yet an other answer... Let $(x,y)$ be the vector of coordinates of the orthocenter $H$ of the triangle $A_1A_2A_3$, where $A_j=(x_j,y_j)$ is (seen as a vector) with given components. Then the conditions $A_1H\perp A_2A_3$ and $A_2H\perp A_3A_1$ (and the third is superfluous) lead to a system of equations using the scalar products of the corresponding vectors: $$ \left\{ \begin{aligned} 0 &= (x-x_1)(x_2-x_3)+(y-y_1)(y_2-y_3)\ ,\\ 0 &= (x-x_2)(x_3-x_1)+(y-y_2)(y_3-y_1)\ , \end{aligned} \right. $$ and we can write it as a system in the usual form by isolating the $x,y$ vector: $$ \begin{bmatrix} x_2-x_3 & y_2-y_3\\ x_3-x_1 & y_3-y_1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} x_1(x_2-x_3) + y_1(y_2-y_3)\\ x_2(x_3-x_1) + y_2(y_3-y_1) \end{bmatrix} \ . $$ Now apply Cramer's rule for instance.

Answer 4

In figure O is circumcenter and K is its reflection over BC.Since we have coordinates of vertices we can calculate length of the sides a, b and c and also R the radius of circum circle.We use the fact that the circle passing two vertices and orthicenter H has the equal radius as circumcircle.The coordinates of O and K is the solution of the following system of equations:

$\begin{cases}(X_B-Y_O)^2+(Y_B-Y_O)^2=R^2\\(X_C-X_O)^2+(Y_C-Y_O)^2=R^2\end{cases}$

Now we use this fact that $AH=OK=R_1$. So the coordinates of H is the solution of the following system of equations:

$\begin{cases}(X_K-X_H)^2+(Y_K-Y_H)^2=R^2\\(X_A-X_H)^2+(Y_A-Y_H)^2=R_1^2\end{cases}$