Introduction to centroid — lesson. Mathematics State Board, Class 9.

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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.

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3. Introduction to centroid

Theory: