Question

The points $A$ $(x_1,y_1),B(x_2,y_2)$ and $C\left(x_3{y}_3\right)$ are the vertices of $\Delta$ ABC.
The median $AD$ meets $BC$ at $D$.

What are the coordinates of the centroid of the triangle $ABC$?

• A. $\frac{2x_1+2x_2+x_3}{3},\frac{2y_1+2y_2+y_3}{3}$
• B. $\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$
• C. $x_1+x_2+x_{3}3,y_1+y_2+y_{3}3$
• D. $\frac{x_1+x_2+2x_3}{3},\frac{y_1+y_2+2y_3}{3}$

Answer 1

The correct option is B $\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$

The points $A(x_1,y_1),B(x_2,y_2)$ and $C\left(x_3{y}_3\right)$ are the vertices of $\Delta ABC$.

Let the median $AD$ meet $BC$ at $D$.

To find: Co-ordinates of its centroid $G(X,Y)$ .

$G(X,Y)$ of $\Delta ABC$ will divide its median $AD$ in the ratio $m:n=2:1$

i.e $AG:GD=2:1$ when $D(x_4,y_4)$ is the mid-point of $BC$.

So, the co-ordinates of $D(x_4,y_4)$, by the section formula for mid point, are

$D(x_4,y_4)=(\frac{x_3+x_2}{2},\frac{y_3+y_2}{2})$.

$\therefore$ the co-ordinates of $G(X,Y)$ by the section formula is,

$X=\frac{n{x}_1+n{x}_4}{m+n}=\frac{\frac{x_1+x_3}{2}\times 2+x_2\times 1}{2+1}=\frac{x_1+x_2+x_3}{3}$ and

$Y=\frac{n{y}_1+n{y}_4}{m+n}=\frac{\frac{y_1+y_3}{2}\times 2+y_2\times 1}{2+1}=\frac{y_1+y_2+y_3}{3}$

$\therefore G(X,Y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

The co-ordinates of the centroid $G(X,Y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$