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We all know that a triangle has $$3$$ medians.
The point at which all the $$3$$ medians intersect is a centroid.
Let us look at the following figure:

Let the points $$A$$, $$B$$ and $$C$$ be $$(x_1$$, $$y_1)$$, $$(x_2$$, $$y_2)$$ and $$(x_3$$, $$y_3)$$ respectively.

The point $$G(x$$, $$y)$$ is the intersection point of all $$3$$ medians.

Here, $$G$$ is the centroid.

The point $$G(x$$, $$y)$$ divides a median internally in the ratio $$2 : 1$$.

Consider the median $$AD$$.

To find the point $$G$$, we should know the value of $$D$$ which is the mid-point of $$BC$$.

To know the value of $$D$$, we should use the mid-point formula.
$$\text{Mid-point} =$$ $$(\frac{x_1 + x_2}{2}$$, $$\frac{y_1 + y_2}{2})$$
When we apply the mid-point formula for $$BC$$, we get:

Mid-point of $$BC = D =$$ $$(\frac{x_2 + x_3}{2}$$, $$\frac{y_2 + y_3}{2})$$

We also know that $$G$$ divides the median $$AD$$ in the ratio $$2 : 1$$. Hence, we should use the section formula to find the value of $$G$$.
$$P(x$$, $$y) =$$ $$(\frac{mx_2 + nx_1}{m + n}$$, $$\frac{my_2 + ny_1}{m + n})$$
When we apply section formula for $$A(x_1$$, $$y_1)$$, and $$D$$$$(\frac{x_2 + x_3}{2}$$, $$\frac{y_2 + y_3}{2})$$ we get:

$$G(x$$, $$y) =$$ $$(\frac{2(\frac{x_2 + x_3}{2}) + x_1}{2 + 1}$$, $$\frac{2(\frac{y_2 + y_3}{2}) + y_1}{2 + 1})$$

[Note: We are considering the values of $$m$$ and $$n$$ as $$2$$ and $$1$$ respectively, as $$G$$ divides the median $$AD$$ in the ratio $$2 : 1$$.]

$$G(x$$, $$y) =$$ $$(\frac{x_1 + x_2 + x_3}{3}$$, $$\frac{y_1 + y_2 + y_3}{3})$$

which is the centroid formula.