# Centroid of a Triangle

The **centroid of a triangle** (or **barycenter of a triangle**) *G* is the point where the three medians of the triangle meet.

The medians of a triangle are the line segments created by joining one vertex to the midpoint of the opposite side. Since every triangle has three sides and three angles, it has three medians (m_{a}, m_{b} and m_{c}).

**Centroid theorem**: the distance between the centroid and its corresponding vertex is twice the distance between the barycenter and the midpoint of the opposite side. That is, the distance from the centroid to each vertex is 2/3 the length of each median. This is true for every triangle.

In physics, the **centroid of a triangle** (*G*) would be its **center of gravity**.

The centroid is always inside the triangle.

## How to Find the Coordinates of the Centroid

If we know the coordinates of the three vertices of the triangle, the coordinates of the centroid will be the arithmetic mean of those coordinates.

Let’s find the arithmetic mean of these coordinates and we will have the centroid of a triangle in a coordinate system:

Where, *x _{a}*,

*x*and

_{b}*x*are the

_{c}*x*coordinates of the vertices of a triangle,

*y*,

_{a}*y*and

_{b}*y*are the y coordinates and

_{c}*G*is the centroid.

## Practice Exercise

Let ABC be a triangle with the vertex coordinates *A* (-4, 0), *B* (2, -3), *C* (4, 2). The midpoints of the side BC, AC and AB are a’, b’, and c’, respectively. The centroid of a triangle is represented as *G*.

- Find the coordinates of the centroid of the triangle above with the given vertices.
- Find the coordinates of the centroid of the triangle above through the intersection of two of its medians, knowing the equations of the lines to which they belong.

**Solution:**

**1. Knowing the vertex coordinates we calculate the coordinates of the centroid** of the triangle obtaining the arithmetic mean of the three coordinates in the abscissa and the ordinate axis:

**2. Find the coordinates of the centroid of the triangle above through the intersection of two of its medians**.

We can also calculate the coordinates of the centroid using the intersection of two of its medians when we know the equations of the lines to which they belong.

We chose the medians *m _{a}* and

*m*. To find the equations of their lines we have the coordinates of the vertices

_{c}*A*and

*C*. We will need to know the coordinates of the midpoints of the opposite sides of each vertex, which we will call

*a’*and

*c’*.

With two points we can obtain the equation of the two lines.

From vertex *A* to midpoint *a’*, the equation of its line is:

The equation for the first median is:

From vertex *C* to midpoint *c’*, the equation of its line is:

The equation for the second median is:

We solve this system of two linear or first degree equations with two unknowns. Its two roots will be the coordinates of the point where the two lines are cut, that is, the sought-after coordinates of the centroid of this triangle:

First we multiply the two terms of the first equation by 2 so that x has an integer coefficient (1) and then we clear the unknown *x* to solve the system of equations by the **substitution method**:

We substitute the value of *x* in the second equation to find the *y*:

Into the x equation:

So we have obtained the same result for the **centroid** coordinates of this triangle as with the first procedure.

## 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).

Triangles are a fascinating little polygons. I always say to my students: Do not under estimate the little triangle