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 AB

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

$\Rightarrow x_1+x_2=2and{y}_1+y_2=2$ ....(1)

Q is the mid point of BC

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

$\Rightarrow \frac{x_2+x_3}{2}=4and\frac{y_2+y_3}{2}=-3$

$\Rightarrow x_2+x_3=4and{y}_2+y_3=-6$ .....(2)

R is the mid point of AC

$\Rightarrow x_1+x_3=6and{y}_1+y_3=8$ ......(3)

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

$(x_1,y_1)\equiv (2,8),(x_2,y_2)\equiv (0,-6)$

and $(x_3,y_3)\equiv (4,0)$

Then the coordinates of the centroid

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

$=(\frac{2+0+4}{3},\frac{8-6+0}{3})=(2,\frac{2}{3})$