Centroid of a triangle

We find the centroid of the triangle with vertices $A(0, 0)$ , $B(4, 0)$ and $C(2, 3)$ ¹. The centroid is the average of the coordinates of the vertices.

G = \left( \frac{0 + 4 + 2}{3}, \ \frac{0 + 0 + 3}{3} \right) = (2, 1)

The $x$ -coordinates are averaged among themselves and the $y$ -coordinates among themselves; nothing more is needed.

Why the average works

The centroid is the intersection of the three medians, the segments joining a vertex to the midpoint of the opposite side. The midpoint of $BC$ is $M\left( 3, \dfrac{3}{2} \right)$ , and the point dividing the median $AM$ in the ratio $2 : 1$ from $A$ lies two-thirds of the way from $A$ toward $M$ .

\left(\frac{2}{3} \times 3, \ \frac{2}{3} \times \frac{3}{2}\right) = (2, 1)

It does coincide with $G$ .

The same from any median

Median	Midpoint of the opposite side	The point two-thirds along
From $A$	$M\left( 3, \dfrac{3}{2} \right)$	$(2, 1)$
From $B$	$N\left( 1, \dfrac{3}{2} \right)$	$(2, 1)$
From $C$	$L(2, 0)$	$(2, 1)$

That the three medians meet at a single point, and that this point divides each of them in the ratio $2 : 1$ from the vertex, is the theorem of the centroid.

The physical meaning

A triangular plate cut from material of uniform thickness balances exactly when supported at this point. Thinking instead of equal weights placed at the three vertices gives the same place, and that is precisely the average of the coordinates. The mean of a set of data is called the center of gravity of its distribution as an extension of the same idea.

Always inside

The centroid always lies inside the triangle. Being an average of the vertices, it has no way of straying outside.

Point	Can it fall outside the triangle
Centroid	no
Incenter	no
Circumcenter	yes, for an obtuse triangle
Orthocenter	yes, for an obtuse triangle

On the graph the three vertices are marked, and the large dot inside at $(2, 1)$ is the centroid.

Centroid, Wikipedia