Coordinates of the centroid of a triangle in 3D

Content updated: October 22, 2026

The centroid of a triangle is the intersection of its three medians. Each centroid coordinate is the arithmetic mean of the three corresponding vertex coordinates; the triangle need not be right-angled.

On this page: Idea · Formula · Examples · Practice · Further study

1. Why average the three coordinates?

Let $ABC$ be a nondegenerate triangle, and let $N$ be the midpoint of $BC$. The centroid $G$ lies on the median $AN$ and satisfies $\overrightarrow{AG}=\frac23\overrightarrow{AN}$. Since $N=(B+C)/2$ componentwise,

The same argument for the other two medians yields the same point. Thus the formula determines the intersection of all three medians, rather than merely a point on one median.

2. Formula and conditions

Let the three vertices $A(x_A;y_A;z_A)$, $B(x_B;y_B;z_B)$, and $C(x_C;y_C;z_C)$ be noncollinear. Their centroid is

Check the result using $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec0$. The formula is symmetric in the three vertices, so changing their order does not change $G$.

Given $G,A,B$, recover the remaining vertex using $C=3G-A-B$. Then check that the three vertices are noncollinear if a triangle is required. Their coordinate average still exists when they are collinear, but the resulting configuration is not a nondegenerate triangle.

The centroid is not the circumcenter. In a right triangle, the midpoint of the hypotenuse is the circumcenter; the centroid still divides each median in the ratio $2:1$ measured from the vertex.

3. Worked examples

Example 1. The centroid and checks on the triangle

Let $A(1;2;3)$, $B(2;3;1)$, and $C(4;-4;1)$. Find the centroid, determine whether the triangle is right-angled, and calculate its area.

By the centroid formula,

Set $\vec u=\overrightarrow{AB}=(1;1;-2)$ and $\vec v=\overrightarrow{AC}=(3;-6;-2)$. At $A$, we have $\vec u\cdot\vec v=3-6+4=1\ne0$, so angle $A$ is not a right angle. At $B$ and $C$, the dot products are $5$ and $48$, respectively, both different from $0$. Thus the given triangle is not right-angled.

To calculate the area, use $AB^2=6$, $AC^2=49$, and $\cos A=1/\sqrt{294}$. Since $A$ is an interior angle of the triangle, $\sin A>0$; hence

The radicand is positive, so the three points are noncollinear. We cannot use one-half the product of two side lengths, as for a right triangle, before checking perpendicularity.

Example 2. Recovering a vertex from the centroid

Let $A(0;0;0)$, $B(3;0;0)$, and the centroid $G(1;1;1)$ be given. Then $C=3G-A-B=(0;3;3)$. The vectors $\overrightarrow{AB}=(3;0;0)$ and $\overrightarrow{AC}=(0;3;3)$ are not parallel, so the data define a triangle. Averaging its three vertex coordinates gives exactly $(1;1;1)$.

4. Practice

1. Find the centroid of the triangle with $A(0;0;0)$, $B(6;0;0)$, and $C(0;3;3)$.
2. Given $A(1;0;2)$, $B(-1;3;1)$, and the centroid $G(2;1;0)$, find $C$.
3. Consider the triangle right-angled at $A$ with $A(0;0;0)$, $B(3;0;0)$, and $C(0;3;0)$. Calculate its centroid and the midpoint of its hypotenuse. Do they coincide?

5. Common mistakes

• Mixing coordinate components, for example using $z_C$ in the formula for $x_G$.
• Calculating the y-coordinate $\frac13$ correctly but copying it as $\frac73$ in the conclusion.
• Confusing the centroid with the midpoint of the hypotenuse or with the circumcenter.
• Assuming a right triangle before checking the coordinate data.

6. Related lessons

• Midpoint coordinates (English lesson)
• The dot product and checks on triangles (English lesson)
• The cross product and its applications (English lesson)

Additional material on Scribd from the original lesson.