G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). Why centroid divides median 2:1.

Centroid of Triangle — Formula and Example

Question

Find the centroid of a triangle with vertices $A(1, 4)$ , $B(3, -2)$ , and $C(5, 6)$ .

Solution — Step by Step

The centroid $G$ of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ is:

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

We’re simply averaging the $x$ -coordinates and averaging the $y$ -coordinates separately.

A = (1,\ 4), \quad B = (3,\ -2), \quad C = (5,\ 6)

So $x_1 = 1,\ x_2 = 3,\ x_3 = 5$ and $y_1 = 4,\ y_2 = -2,\ y_3 = 6$ .

G_x = \frac{1 + 3 + 5}{3} = \frac{9}{3} = 3

G_y = \frac{4 + (-2) + 6}{3} = \frac{8}{3}

\boxed{G = \left(3,\ \frac{8}{3}\right)}

Why This Works

The centroid is the point where all three medians of a triangle meet. A median connects a vertex to the midpoint of the opposite side.

When we average the three $x$ -coordinates, we’re finding the “balance point” horizontally. Same logic applies vertically. The arithmetic mean of all three vertices gives the physical centre of mass — if you cut a triangle out of cardboard, it would balance perfectly on its centroid.

This is also why the centroid always lies inside the triangle — an average of three coordinates can never land outside the range of those coordinates.

Why the Centroid Divides Each Median in 2:1

This is the follow-up that CBSE and JEE both love to test.

Take median from $A(1, 4)$ to midpoint $M$ of $BC$ .

M = \left(\frac{3+5}{2},\ \frac{-2+6}{2}\right) = (4,\ 2)

Now check: does $G = (3,\ 8/3)$ divide $AM$ in the ratio $2:1$ from vertex $A$ ?

Using the section formula for internal division in ratio $2:1$ :

x = \frac{2(4) + 1(1)}{2+1} = \frac{9}{3} = 3 \checkmark

y = \frac{2(2) + 1(4)}{2+1} = \frac{8}{3} \checkmark

The centroid always sits two-thirds of the way from any vertex to the opposite midpoint. This 2:1 property appears repeatedly in PYQs — worth internalising.

Alternative Method — Verification Using a Different Median

We can verify using the median from $B(3, -2)$ to midpoint $N$ of $AC$ .

N = \left(\frac{1+5}{2},\ \frac{4+6}{2}\right) = (3,\ 5)

Check if $G = (3,\ 8/3)$ divides $BN$ in ratio $2:1$ from $B$ :

x = \frac{2(3) + 1(3)}{3} = \frac{9}{3} = 3 \checkmark

y = \frac{2(5) + 1(-2)}{3} = \frac{8}{3} \checkmark

Same point — both medians confirm our answer. This cross-check takes 30 seconds in an exam and eliminates calculation errors.

Common Mistake

Sign error with negative coordinates. In this problem, $y_2 = -2$ . Students often write $\frac{4 + 2 + 6}{3}$ instead of $\frac{4 + (-2) + 6}{3}$ , getting $G_y = 4$ instead of $8/3$ . Always substitute the sign along with the number.

In CBSE Class 10 board exams, centroid questions are 2-3 mark questions and almost always give you three clean coordinates. If your answer has messy fractions for both $x$ and $y$ , double-check your addition before proceeding.