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Download now on Google PlayWe 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.

\(\text{Mid-point} =\) \((\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2})\)

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 \(G\).

\(P(x\), \(y) =\) \((\frac{mx_2 + nx_1}{m + n}\), \(\frac{my_2 + ny_1}{m + n})\)

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

[

**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.