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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.