Centroid Coordinate Formula

A vector proof that the centroid is the plain average of the vertices, G = (A + B + C)/3, from the midpoint formula and the 2:1 median ratio via the section formula. The simplest of the three center coordinate formulas: equal weight on every vertex, valid from any origin.

Keywords: centroid, median, section formula, barycentric coordinates

Prerequisites: centroid-proof · Difficulty: intermediate

The centroid has the simplest vector address of all the triangle centers: the plain average of the three vertices,

G = \frac{A + B + C}{3},

every vertex weighted equally.

Strategy: read each vertex as a position vector and use the section formula. One median, from $A$ to the midpoint $M$ of $B C$ , is enough: the centroid splits it $2 : 1$ from $A$ , and the algebra collapses to the equal- weight average.

Step 1: Triangle and Centroid

The centroid $G$ is where the three medians meet. We will pin down its position as the plain average of the vertices.

Step 2: Vectors and the Section Formula

Read each vertex as a position vector from any fixed origin. The one tool we need is the section formula: the point dividing segment $P Q$ in the ratio $m : n$ , measured from $P$ , is

\frac{n P + m Q}{m + n},

so the nearer endpoint carries the larger weight.

Step 3: $G = \frac{A + B + C}{3}$

Claim. The centroid is the plain average of the three vertices,

G = \frac{A + B + C}{3},

each weighted equally. One median, split $2 : 1$ , proves it.

Step 4: The Midpoint of $B C$

The median from $A$ runs to $M$ , the midpoint of $B C$ . As the midpoint it splits $B C$ in the ratio $1 : 1$ , so the section formula gives $M = \frac{B + C}{2}$ .

Step 5: Split the Median $2 : 1$

The centroid divides the median $A M$ in the ratio $A G : GM = 2 : 1$ from $A$ (proved in Why Medians Meet at One Point). By the section formula,

G = \frac{1 \cdot A + 2 \cdot M}{3} = \frac{A + ( B + C )}{3} = \frac{A + B + C}{3} .

Step 6: The Centroid Coordinates

So the centroid is the plain average of the vertices,

G = \frac{A + B + C}{3} .

The three weights are equal and sum to one, so the identity holds from any origin at all, with no special reference point required. It is the simplest case of barycentric coordinates.

The centroid is the plain average of the vertices,

G = \frac{A + B + C}{3} .

The three weights are equal and sum to one, so the identity holds from any origin at all, with no special reference point required. It is the simplest case of barycentric coordinates, to be set beside the side-weighted incenter $I = \frac{a A + b B + c C}{a + b + c}$ . ∎