Centroid Formula – Derivation -Solved Examples

Centroid Formula

The centroid of a triangle is the point where all three medians of a triangle intersect each other. It is also known as the geometric centre and always lies inside the triangle.

This point divides every median of the triangle in the ratio of 2:1.

The coordinates of all the vertices of △ABC are given below

G is the centroid of △ABC

Centroid Formula for \triangle \mathrm{ABC}=\left(\frac{x_{a}+x_{b}+x_{c}}{3},\frac{y_{a}+y_{b}+y_{c}}{3}\right)

Derivation

In △ABC

A, B and C are the vertices having x coordinates and y coordinates as shown below

P – Midpoint of BC

Q – Midpoint of CA

R – Midpoint of AB 

Using section formula, the coordinates of P, Q and R are calculated as shown in the table below:

The point G divides AP in the ratio of 2:1.

Using section formula,

x coordinate of G =\frac{2\left(\frac{x_{b}+x_{c}}{2}\right)+1\left(x_{a}\right)}{2+1}=\frac{x_{a}+x_{b}+x_{c}}{3}

y coordinate of G =\frac{2\left(\frac{y_{b}+y_{c}}{2}\right)+1\left(y_{a}\right)}{2+1}=\frac{y_{a}+y_{b}+y_{c}}{3}

Hence it is proved that the coordinates of centroid G of a triangle are:

\left(\frac{x_{a}+x_{b}+x_{c}}{3},\frac{y_{a}+y_{b}+y_{c}}{3}\right)

Solved Examples

1. The vertices of a triangle are (0,3), (5,6) and (1,2). Find its centroid.

Solution:

Let the centroid of △ABC be G

Using the centroid formula,

x coordinate of G =\frac{x_{a}+x_{b}+x_{c}}{3}=\frac{0+5+1}{3}=2

y coordinate of G =\frac{y_{a}+y_{b}+y_{c}}{3}=\frac{3+6+2}{3}=\frac{11}{3}

Therefore, the coordinates of the centroid are \left(2, \frac{11}{3}\right).

2. The centroid of △ABC is (0,3). Find the coordinates of point A if the other two points are  (-2,4) and (2,2).

Solution:

Using centroid formula 

x coordinate of G =\frac{x_{a}+x_{b}+x_{c}}{3}=\frac{x-2+2}{3}

\Rightarrow \frac{x-2+2}{3}=0
\Rightarrow x=0

y coordinate of \mathrm{G}=\frac{y_{a}+y_{b}+y_{c}}{3}=\frac{y+4+2}{3}

\Rightarrow y+4=9

\therefore y=9-4=5

Hence the coordinate of point A is (0,5).