# Orthocenter of a triangle

In a triangle *ABC* the **orthocenter** *H* is the intersection point of the three altitudes of the triangle.

Every triangle has three altitudes (or heights) and three sides (or bases).

An altitude of a triangle (h_{a}, h_{b} y h_{c}) is a perpendicular line segment from a vertex to the opposite side. This line containing the opposite side is called the extended base of the altitude.

Altitude can also be understood as the distance between the base and the vertex.

**Where is the Orthocenter of a Triangle Located?**

- If it’s an obtuse triangle the orthocenter is located outside the triangle (as we see in the picture above).
- If it’s an acute triangle the orthocenter is located inside the triangle.
- If it’s a right triangle the orthocenter lies on the vertex of the right angle.

## Euler Line

In any **non-equilateral triangle** the **orthocenter** (*H*), the centroid (*G*) and the circumcenter (*O*) are aligned. The line that contains these three points is called the Euler line.

In an equilateral triangle all three centers are in the same place.

The relative distances between the triangle centers remain constant.

**Distances between centers**:

It is true that the distance from the **orthocenter** (*H*) to the centroid (*G*) is twice that of the centroid (*G*) to the circumcenter (*O*). Or put another way, the *HG* segment is twice the *GO* segment:

When the triangle is equilateral, the barycenter, orthocenter, circumcenter, and incenter coincide in the same interior point, which is at the same distance from the three vertices.

This distance to the three vertices of an equilateral triangle is equal to from one side and, therefore, to the vertex, being *h* its altitude (or height).

## Practice Exercise

Find the coordinates of the **orthocenter** *H* of a triangle *ABC* with the vertex coordinates *A* (3, 5), *B* (4, -1), and *C* (-4, 1).

**SOLUTION:**

In order to find the orthocenter we have to solve the equations for the altitudes *h _{a}* (starting from vertex

*A*) and

*h*(from vertex

_{b}*B*) and see where is located the intersection point (

*H*) of both altitudes.

**STEP 1: Find equation for h_{a}:**

Firstly we will find the equation of the line that passes through side *BC*, which is the opposite of vertex *A*. This equation is obtained knowing that it passes through points *B* (4, -1) and *C* (-4, 1). The general equation of the line that passes through two known points is:

The equation of the line that contains side *BC* and its slope *m* will be:

The slope of the line that contains the altitude *h _{a}*, being perpendicular to the side

*BC*, is the inverse and of the opposite sign to the slope of the line that contains the side. So, we have that:

Knowing that the height *h _{a}* passes through the vertex

*A*(3, 5), we can obtain the equation of its line. The equation of a line, knowing a point and its slope (inverse slope and opposite sign to that found for

*BC*, that is,

*m*= 4):

_{p} So, the equation for *h _{a}* is

*y*-4

*x*+7 = 0.

**STEP 2: Find equation for h_{b}:**

Now we proceed in the same way to find the equation of the line that contains the height *h _{b}*, that is, to which, starting from vertex

*B*, is perpendicular to side

*AC*.

First, the slope of the line *AC*:

And, now, the equation of the line that contains the height *h _{b}*, that is, the one that starts from vertex

*B*and is perpendicular to side

*AC*. Therefore, its slope

*m*will be –7/4.

_{p} Then, the equation of hb is 4*y* +7*x* -24 = 0.

We solve this system of two first degree equations with two unknowns. Its roots will be the coordinates of its intersection, that is, the location of the **orthocenter** *H* of the triangle *ABC*.

Solving this system of two first-degree equations we have that:

Therefore, the altitudes cross at (2.26, 2.04).

And the coordinates of the **orthocenter** are ** H (2.26, 2.04)**.