What Is the Centroid Formula?

The centroid of a triangle in coordinate geometry is defined as the unique point of intersection of its three medians. This point is always situated within the interior of the triangle and divides each median in the fixed ratio $2:1$ from the corresponding vertex.

Formal Statement and Expression for the Centroid of a Triangle

Consider a triangle $\triangle ABC$ with vertices $A\left(x_1,\,y_1\right)$, $B\left(x_2,\,y_2\right)$, and $C\left(x_3,\,y_3\right)$ in the Cartesian plane. The centroid $G$ is the coordinate point given by the arithmetic mean of the respective coordinates of the three vertices, expressed as:

$G = \left( \dfrac{x_1 + x_2 + x_3}{3}\,,\,\dfrac{y_1 + y_2 + y_3}{3} \right)$.

The point $G$ always lies within the triangle, regardless of the triangle’s type (acute, obtuse, right, scalene, equilateral, or isosceles), and serves as its center of mass for uniform density.

Stepwise Derivation of the Centroid Formula Using Coordinate Geometry

Let triangle $ABC$ have vertices $A\left(x_1,\,y_1\right)$, $B\left(x_2,\,y_2\right)$, and $C\left(x_3,\,y_3\right)$. The median from vertex $A$ meets side $BC$ at point $D$, which is the midpoint of $BC$.

The coordinates of $D$, the midpoint of $BC$, are:

$D = \left( \dfrac{x_2 + x_3}{2}\,,\,\dfrac{y_2 + y_3}{2} \right)$.

The centroid $G$ divides the median $AD$ in the ratio $2:1$, with $AG : GD = 2:1$. To determine the coordinates of $G$, apply the internal section formula for a line segment divided in ratio $m:n$:

If $P(x_1,\,y_1)$ and $Q(x_2,\,y_2)$ are points, and point $R$ divides $PQ$ in ratio $m:n$, then $R = \left( \dfrac{mx_2 + nx_1}{m+n}\,,\,\dfrac{my_2 + ny_1}{m+n} \right)$.

Here, $A$ has coordinates $\left(x_1,\,y_1\right)$, $D$ has coordinates $\left( \dfrac{x_2 + x_3}{2},\,\dfrac{y_2 + y_3}{2} \right)$, and $G$ divides $AD$ in $2:1$ starting from $A$. Thus, $m = 2$ and $n = 1$.

Substitute these values into the section formula to obtain the coordinates of $G$:

$G_x = \dfrac{2 \left( \dfrac{x_2 + x_3}{2} \right) + 1 \cdot x_1}{2 + 1}$

$= \dfrac{ \left( x_2 + x_3 \right) + x_1 }{3 }$

$= \dfrac{ x_1 + x_2 + x_3 }{3 }$

Similarly, for the $y$-coordinate:

$G_y = \dfrac{2 \left( \dfrac{y_2 + y_3}{2} \right) + 1 \cdot y_1}{2 + 1}$

$= \dfrac{ \left( y_2 + y_3 \right) + y_1 }{3 }$

$= \dfrac{ y_1 + y_2 + y_3 }{3 }$

Result: The centroid $G$ of triangle $ABC$ is $\left( \dfrac{x_1 + x_2 + x_3}{3}\,,\,\dfrac{y_1 + y_2 + y_3}{3} \right)$.

Geometrical Characteristics of the Centroid

The centroid $G$ is the unique point of concurrency of all three medians of a triangle. Each median is divided by the centroid such that the segment containing the vertex is twice that containing the midpoint of the side. This property holds for every triangle and ensures that $G$ always lies strictly inside the triangle, except in the degenerate case where the three vertices are collinear, in which case all medians are also collinear.

For a triangle of uniform density, the centroid coincides with the center of mass or ‘center of gravity’ of the lamina, which is relevant in problems of statics, mechanical balancing, and integration over triangular shapes. The concept and formula generalize naturally to centroids of polygons and more complex figures, and underpin analytical approaches in calculus and engineering. For further reading, see Centroid Formula.

Worked Examples Using the Centroid Formula

Example 1: Compute the centroid of the triangle with vertices $(4,\,3)$, $(6,\,5)$, and $(5,\,4)$.

Given: $A(4,\,3)$, $B(6,\,5)$, $C(5,\,4)$.

Substitution: Apply $G = \left( \dfrac{x_1 + x_2 + x_3}{3},\dfrac{y_1 + y_2 + y_3}{3} \right )$.

$G_x = \dfrac{4 + 6 + 5}{3} = \dfrac{15}{3} = 5$

$G_y = \dfrac{3 + 5 + 4}{3} = \dfrac{12}{3} = 4$

Final Result: The centroid is located at $(5,\,4)$.

Example 2: The centroid of a triangle is at $(3,\,3)$ and two vertices are $(1,\,5)$ and $(-1,\,1)$. The third vertex is $(k,\,3)$. Find the value of $k$.

Given: $(x_1,\,y_1) = (1,\,5)$, $(x_2,\,y_2) = (-1,\,1)$, $(x_3,\,y_3) = (k,\,3)$, $G = (3,\,3)$.

Substitution: $3 = \dfrac{1 + (-1) + k}{3}$.

$1 + (-1) + k = 3k$

$k = 9$

Final Result: The value of $k$ is $9$.

Example 3: Determine the centroid of a triangle whose vertices are $(1,\,3)$, $(2,\,1)$, and $(3,\,2)$.

Given: $A(1,\,3)$, $B(2,\,1)$, $C(3,\,2)$.

Substitution:

$G_x = \dfrac{1 + 2 + 3}{3} = \dfrac{6}{3} = 2$

$G_y = \dfrac{3 + 1 + 2}{3} = \dfrac{6}{3} = 2$

Final Result: The centroid is at $(2,\,2)$.

Application of the Section Formula in Centroid Derivation

The derivation of the centroid formula employs the section formula, a result fundamental in coordinate geometry. For the median from vertex $A$ to side $BC$, the centroid $G$ divides the segment joining $A$ and $D$ (the midpoint of $BC$) in the ratio $2:1$. This approach is consistent regardless of the specific median considered, resulting in the same point $G$ via analogous calculations for vertices $B$ and $C$.

The centroid formula forms the foundation for higher analytical concepts, such as calculating the centroid of composite shapes, integration in calculus for centroids of laminae, and in solving statics problems involving triangles. For related results on triangle geometry, see Area Of A Triangle Formula.