Question

If the points (1, −1), (2, −1) and (4, −3) are the mid-points of the sides of a triangle, then write the coordinates of its centroid.

Answer 1

Let P(1, −1), Q(2, −1) and R(4, −3) be the mid-points of the sides AB, BC and CA, respectively, of $∆$ABC.
Let $A\left(x_1,y_1\right),B\left(x_2,y_2\right)\mathrm{and}C\left(x_3,y_3\right)$ be the vertices of $∆$ABC.
Since, P is the mid-point of AB,

$\frac{x_1+x_2}{2}=1,\frac{y_1+y_2}{2}=-1$ ... (1)

Q is the mid-point of BC.

$\therefore \frac{x_2+x_3}{2}=2,\frac{y_2+y_3}{2}=-1$ ... (2)

R is the mid-point of AC.

$\therefore \frac{x_1+x_3}{2}=4,\frac{y_1+y_3}{2}=-3$ ... (3)

Adding equations (1), (2) and (3), we get:

$$\begin{gathered} x_1+x_2+x_3=1+2+4=7 \\ {y}_1+y_2+y_3=-1-1-3=-5 \end{gathered}$$

$\therefore \mathrm{Centroid}\mathrm{of}∆ABC=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)=\left(\frac{7}{3},\frac{-5}{3}\right)$

Hence, the coordinates of the centroid of the triangle is $\left(\frac{7}{3},\frac{-5}{3}\right)$.