Question

If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid.

Answer 1

Let P(1,1), Q(2,-3), R(3,4) be the mid-points of sides AB, BC and CA respectively of triangle ABC. Let A$(x_1,y_1)$,B$(x_2,y_2)$ and C$(x_3,y_3)$ be the vertices of triangle ABC. Then, P is the mid-point of BC

$\Rightarrow \frac{x_1+x_2}{2}$ = 1, $\frac{y_1+y_2}{2}$ = 1

$\Rightarrow x_1+x_2$ = 2 and $y_1+y_2$ = 2 ........(1)

Q is the mid-point of BC

$\Rightarrow \frac{x_2+x_3}{2}$ = 2, $\frac{y_2+y_3}{2}$ = -3

$\Rightarrow x_2+x_3$ = 3 and $y_2+y_3$ = -6 ........(2)

R is the mid-point of AC

$\Rightarrow \frac{x_1+x_3}{2}$ = 3, $\frac{y_1+y_3}{2}$ = 4

$\Rightarrow x_1+x_3$ = 6 and $y_1+y_3$ = 8 ........(3)

From (1),(2) and (3), we get

$x_1+x_2+x_2+x_3+x_1+x_3$ = 2 + 4 + 6 and $y_1+y_2+y_2+y_3+y_1+y_3$ = 2 - 6 + 8

$\Rightarrow x_1+x_2+x_3$ = 6 and $y_1+y_2+y_3$ = 2 ........(4)

The coordinates of the centroid of $△$ABC are

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

= ($\frac{6}{3},\frac{2}{3}$) [Using (4)]

= ($2,\frac{2}{3}$)