How this instrument works
The centroid is the point where a triangle balances flat on a pin, and the formula for finding it is disarmingly simple: average the three x-coordinates for cx, average the three y-coordinates for cy. That simplicity is not a coincidence — for a uniform flat plate, the centre of mass of any triangle happens to equal the arithmetic mean of its three corners, a fact that does not hold for most other flat shapes.
The reason traces back to the medians — the three lines from each vertex to the midpoint of the opposite side. Every triangle's medians meet at a single point, and that point always sits two-thirds of the way from each vertex toward the midpoint it aims at. Averaging the vertices lands on exactly that 2:1 point, which is why the mean formula and the median-intersection definition describe the same location.
One property is easy to miss until you compare triangles: the centroid always lies strictly inside the shape, however thin, obtuse, or lopsided it gets. That is not true of the circumcenter or the orthocenter, both of which can wander outside an obtuse triangle entirely. Feed in three collinear points — zero enclosed area — and the formula still returns a sensible answer: the mean of the three points, sitting on the line they share.
- Enter the first corner's coordinates in the x₁ and y₁ fields.
- Enter the second corner in x₂ and y₂, and the third in x₃ and y₃ — order doesn't matter, since the formula only sums the three values.
- Read the answer in Centroid x and Centroid y, the coordinates of the balance point.
- To try a different triangle, change any of the six vertex fields; both centroid outputs update together.
Worked example — a triangle at (0,0), (6,0), (3,6)
Take a triangle with vertices at (0, 0), (6, 0), and (3, 6) — a wide base with the apex pushed off to one side. The centroid: cx = (0 + 6 + 3) ⁄ 3 = 9 ⁄ 3 = 3, and cy = (0 + 0 + 6) ⁄ 3 = 6 ⁄ 3 = 2. The balance point sits at (3, 2) — cut this outline from stiff card and it rests level on a pencil tip placed right there.
Notice that (3, 2) is not the midpoint of any side, nor the centre of the bounding box — it is pulled toward the base because two of the three vertices sit on it. That pull is exactly what averaging does: two votes at y = 0 and one vote at y = 6 settle at y = 2, a third of the way up rather than half.
Questions
What is the centroid of a triangle?
The centroid is the triangle's balance point — the spot where a flat, uniform cutout of the shape would rest level on a single support. It equals the average of the three vertices: add the x-coordinates and divide by three for cx, then do the same with the y-coordinates for cy.
How does the centroid formula relate to the medians?
A median runs from one vertex to the midpoint of the opposite side, and all three medians of any triangle cross at a single point. That crossing point splits each median in a 2:1 ratio, closer to the side than the vertex — and averaging the three vertex coordinates lands on exactly that 2:1 point, which is why the two descriptions agree.
Can the centroid fall outside the triangle?
No. Unlike the circumcenter or the orthocenter, the centroid always sits inside the triangle, however obtuse or stretched the shape is. It is a plain average of three points, and an average of three points always lies within the triangle they form, or on its boundary if the points happen to be collinear.
Is the centroid the same point as the incenter or circumcenter?
Only for an equilateral triangle, where the centroid, incenter, circumcenter, and orthocenter all coincide at one point. For any other triangle they separate: the incenter balances distance to the three sides, the circumcenter is equidistant from the three vertices, and only the centroid is the plain average of the corners.
What happens if the three points are collinear?
The formula still returns a value — the mean of the three coordinates — but a triangle with collinear vertices has zero area, so that value marks a point on the shared line rather than a true balance point. This sheet does not block that input; it simply carries out the averaging, exactly as the formula always does.
Does the order I list the vertices in matter?
No. Addition doesn't depend on order, so cx and cy come out identical whether the vertices are entered clockwise, counter-clockwise, or in any mix. Swapping which corner is labeled first, second, or third relabels the same triangle; it never changes where the centroid lands.