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
\[A\left( x_1 , y_1 \right), B\left( x_2 , y_2 \right)\text{ and } C\left( x_3 , y_3 \right)\]  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:

\[x_1 + x_2 + x_3 = 1 + 2 + 4 = 7\]
\[ y_1 + y_2 + y_3 = - 1 - 1 - 3 = - 5\]

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

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