The Single Point a Triangle Balances on
Cut a triangle out of stiff card and try to balance it flat on the tip of a pencil. There is exactly one point where it settles without tipping - not the middle of any side, not a corner, but a specific point inside. Move the pencil a hair off that point and the card falls. That balance point is the centroid, the triangle's centre of mass. The ancient Greek mathematician Archimedes studied exactly this kind of balance point over two thousand years ago, and engineers still compute it today to find where a triangular structure carries its weight.
What Is the Centroid of a Triangle?
The centroid of a triangle is the point where the three medians of the triangle cross. A median is a line segment from a vertex to the midpoint of the opposite side, so every triangle has three of them. Remarkably, all three always pass through one common point - they are concurrent lines - and that shared point is the centroid, usually written $G$.
The centroid is one of the classic centres of a triangle, and it is the one that acts as the balance point, or centre of mass, of a uniform triangular sheet. Because it is defined by the medians of a triangle, it always lies inside the triangle, no matter what shape the triangle has. It is one of several points of concurrency that special sets of lines in a triangle produce.
What Is the Formula for the Centroid of a Triangle?
If the three vertices of a triangle are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, the centroid $G$ is the average of the three vertices:
$$G = \left(\frac{x_1 + x_2 + x_3}{3}, ; \frac{y_1 + y_2 + y_3}{3}\right).$$
The variables mean: $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ are the coordinates of the three corners. You add the three $x$-values and divide by 3 for the centroid's $x$-coordinate, then do the same with the three $y$-values. This makes the centroid one of the easiest triangle centres to compute in coordinate geometry — no perpendiculars or angle bisectors needed, just an average.
In What Ratio Does the Centroid Divide a Median?
The centroid divides each median in the ratio 2:1, measured from the vertex. That is, the distance from a vertex to the centroid is twice the distance from the centroid to the midpoint of the opposite side.
$$AG : GD = 2 : 1$$
So the centroid sits $\dfrac{2}{3}$ of the way along each median from the vertex, and $\dfrac{1}{3}$ of the way from the midpoint. If a median has length $m$, then the vertex-to-centroid piece is $\dfrac{2}{3}m$ and the centroid-to-midpoint piece is $\dfrac{1}{3}m$. This 2:1 split holds for all three medians at once, which is part of why the balance point lands where it does.
Where Does the Centroid Formula Come From?
The averaging formula follows from the 2:1 ratio. Take the median from vertex $A(x_1, y_1)$ to $D$, the midpoint of $BC$. First find $D$:
$$D = \left(\frac{x_2 + x_3}{2}, ; \frac{y_2 + y_3}{2}\right).$$
The centroid $G$ lies $\dfrac{2}{3}$ of the way from $A$ to $D$. Using the section formula for a point dividing $AD$ in ratio $2:1$:
$$G_x = \frac{2 \cdot D_x + 1 \cdot x_1}{2 + 1} = \frac{2 \cdot \frac{x_2 + x_3}{2} + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$$
The same steps give $G_y = \dfrac{y_1 + y_2 + y_3}{3}$. The two midpoint-halves cancel the division by 2, leaving the clean average of all three vertices. That is why the 2:1 ratio and the averaging formula are really the same fact.
What Is the Difference Between the Centroid, Incenter, Circumcenter, and Orthocenter?
All four are triangle centres, but each is built from a different set of lines.
Centre | Built from | Special property | Location |
|---|---|---|---|
Centroid | The three medians | Balance point; divides medians 2:1 | Always inside |
Incenter | The three angle bisectors | Centre of the inscribed circle | Always inside |
Circumcenter | The three perpendicular bisectors | Centre of the circumscribed circle | Inside, on, or outside |
Orthocenter | The three altitudes | Where the heights meet | Inside, on, or outside |
The centroid is the only one that is the centre of mass, and the only one guaranteed to be the average of the vertices. For the full treatment of each, see the incenter, the circumcenter of a triangle, and the orthocenter. In any triangle, the centroid, circumcenter, and orthocenter even line up on a single straight line called the Euler line.
What Are the Properties of the Centroid of a Triangle?
The centroid gathers up several facts that are worth stating together, because problems often turn on one of them.
It is the concurrency point of the medians. All three medians of a triangle pass through it, so it always exists and is unique.
It is the average of the vertices. $G = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$, the mean of the three corner coordinates.
It divides each median $2 : 1$. Measured from the vertex, the centroid sits $\dfrac{2}{3}$ of the way along every median.
It always lies inside the triangle. Unlike the circumcenter or orthocenter, it never falls on or outside the triangle, whatever its shape.
It is the centre of mass. A uniform triangular sheet balances exactly at its centroid.
It splits the triangle into six equal-area pieces. The three medians cut the triangle into six smaller triangles, all of the same area.
It lies on the Euler line. The centroid, circumcenter, and orthocenter are collinear, and the centroid divides the segment from orthocenter to circumcenter in a $2 : 1$ ratio.
It coincides with the other centres only when the triangle is equilateral. In that single case the centroid, incenter, circumcenter, and orthocenter are the same point.
Examples of Centroid of a Triangle
Six examples, from a direct average to recovering a missing vertex.
Example 1
Find the centroid of the triangle with vertices $A(0, 0)$, $B(6, 0)$, and $C(0, 9)$.
Average the $x$-values and the $y$-values.
$G_x = \dfrac{0 + 6 + 0}{3} = \dfrac{6}{3} = 2$
$G_y = \dfrac{0 + 0 + 9}{3} = \dfrac{9}{3} = 3$
Final answer: $G = (2, 3)$. The centroid is just the average of the three corners.
Example 2
A student finds the centroid of $A(0, 0)$, $B(6, 0)$, $C(0, 9)$ by averaging only two vertices, writing "$G = \left(\dfrac{0 + 6}{2}, \dfrac{0 + 9}{2}\right) = (3, 4.5)$." Where does this go wrong?
The tempting move treats the centroid like a midpoint and averages a pair of points.
The result $(3, 4.5)$ is actually the midpoint of side... nothing consistent — it mixes $B$'s $x$ with $C$'s $y$, and it ignores one vertex entirely. A centroid must depend on all three corners, so any answer that drops a vertex cannot be right.
The correct method averages all three vertices:
$G = \left(\dfrac{0 + 6 + 0}{3}, \dfrac{0 + 0 + 9}{3}\right) = (2, 3)$
Final answer: $G = (2, 3)$. Divide by 3, not 2 — the centroid uses every vertex.
Example 3
Find the centroid of the triangle with vertices $A(1, 2)$, $B(3, -4)$, and $C(5, 8)$.
Add each coordinate and divide by 3.
$G_x = \dfrac{1 + 3 + 5}{3} = \dfrac{9}{3} = 3$
$G_y = \dfrac{2 + (-4) + 8}{3} = \dfrac{6}{3} = 2$
Final answer: $G = (3, 2)$. Negative coordinates go into the sum just like positive ones.
Example 4
A median of a triangle has length $18 \text{ cm}$. How far is the centroid from the vertex, and from the midpoint?
The centroid divides the median 2:1 from the vertex, so it is $\dfrac{2}{3}$ of the way along.
Vertex to centroid: $\dfrac{2}{3} \times 18 = 12 \text{ cm}$
Centroid to midpoint: $\dfrac{1}{3} \times 18 = 6 \text{ cm}$
Final answer: $12 \text{ cm}$ from the vertex and $6 \text{ cm}$ from the midpoint, which add back to the full $18 \text{ cm}$.
Example 5
A triangle has vertices $A(2, 3)$, $B(4, 7)$, and an unknown vertex $C(x, y)$. Its centroid is $G(3, 5)$. Find $C$.
Use the centroid formula and solve for the unknown coordinate.
For $x$: $\dfrac{2 + 4 + x}{3} = 3$, so $6 + x = 9$, giving $x = 3$.
For $y$: $\dfrac{3 + 7 + y}{3} = 5$, so $10 + y = 15$, giving $y = 5$.
Final answer: $C = (3, 5)$. Working backward, multiply the centroid coordinate by 3 and subtract the known values.
Example 6
On median $AD$, the vertex-to-centroid distance $AG = 10 \text{ cm}$. Find the centroid-to-midpoint distance $GD$ and the full median length.
Use the 2:1 ratio: $AG : GD = 2 : 1$.
$GD = \dfrac{1}{2} \times AG = \dfrac{1}{2} \times 10 = 5 \text{ cm}$
Full median $AD = AG + GD = 10 + 5 = 15 \text{ cm}$
Final answer: $GD = 5 \text{ cm}$ and $AD = 15 \text{ cm}$. The reasoning step is reading the ratio as "the vertex piece is twice the midpoint piece," not the other way round.
Why the Centroid of a Triangle Matters: It Is the Centre of Mass
The centroid is not just a tidy intersection; it is where a triangle's weight effectively acts, which makes it one of the most useful points in applied geometry.
It is the balance point. A uniform triangular sheet balances exactly at its centroid — the property that makes it the centre of mass in physics and engineering.
It is the easiest centre to compute. Averaging the three vertices needs no perpendiculars or bisectors, so it is the go-to centre in coordinate geometry and computer graphics.
It always lies inside. Unlike the circumcenter or orthocenter, the centroid never leaves the triangle, so it is a reliable reference point for any triangle shape.
What Are the Most Common Mistakes With the Centroid of a Triangle?
Three errors account for most wrong answers, and the first is a division slip.
Mistake 1: Dividing by 2 instead of 3
Where it slips in: Treating the centroid like a midpoint and averaging vertices in pairs.
Don't do this: Writing $G = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.
The correct way: The centroid averages all three vertices, so divide by 3: $G = \left(\dfrac{x_1 + x_2+ x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$. The first-instinct error is to reuse the midpoint habit; the fix is remembering a triangle has three corners and the centroid uses every one.
Mistake 2: Flipping the 2:1 ratio
Where it slips in: Reading the ratio as 1:2 from the vertex instead of 2:1.
Don't do this: Putting the shorter piece next to the vertex.
The correct way: The longer piece (the 2 part) is between the vertex and the centroid; the shorter piece (the 1 part) is between the centroid and the midpoint. So the centroid is $\dfrac{2}{3}$ of the way from the vertex. The second-guesser who is unsure can sketch it: the centroid always sits closer to the side's midpoint than to the vertex it came from.
Mistake 3: Confusing the centroid with the incenter or circumcenter
Where it slips in: Assuming all triangle centres are the same point or computed the same way.
Don't do this: Using the median formula when the problem asks for the incenter (angle bisectors) or circumcenter (perpendicular bisectors).
The correct way: Each centre comes from different lines — medians for the centroid, angle bisectors for the incenter, perpendicular bisectors for the circumcenter. They coincide only in an equilateral triangle. The confusion between these close ideas is the most common source of wrong answers on triangle-centre problems.
Conclusion
The centroid of a triangle is where the three medians meet, and it is the triangle's balance point.
Its coordinates are the average of the vertices: $G = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$.
It divides each median in a 2:1 ratio from the vertex, sitting $\dfrac{2}{3}$ of the way along.
It always lies inside the triangle and is the easiest triangle centre to compute.
To take triangle geometry further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor, or math classes online.
Practice These to Solidify Your Understanding
Work through these, then check your answers:
Find the centroid of $A(2, 4)$, $B(6, 2)$, $C(4, 9)$. (Answer to Question 1: $G = \left(\tfrac{12}{3}, \tfrac{15}{3}\right) = (4, 5)$.)
A median is $24 \text{ cm}$ long. How far is the centroid from the vertex? (Answer to Question 2: $\tfrac{2}{3} \times 24 = 16 \text{ cm}$.)
A triangle has vertices $A(1, 1)$, $B(5, 3)$, $C(x, y)$ and centroid $G(3, 3)$. Find $C$. (Answer to Question 3: $x = 3$, $y = 5$, so $C = (3, 5)$.)
If Question 3 tripped you, revisit Example 5 on finding a missing vertex. Want a trainer to walk triangle centres through with your child? Book a free demo class.
Read More
Area of a triangle in coordinate geometry — another vertex-coordinate calculation for triangles.
Altitude of a triangle — the height segments whose intersection gives the orthocenter.
Angle bisector — the lines whose meeting point is the incenter.
Types of triangles — how the centroid behaves across scalene, isosceles, and equilateral triangles.
Similar triangles — triangles that share the same shape and centre proportions.
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