Question

find the centroid of a triangle pqr whose vertices are p(1,1)q(2,2)R(-3,-3)​

Answer 1

centroid be (1+2-3/3) and (1+2-3/3) = 0 and 0

Answer 2

We use the concept given below.

$\boxed{\begin{minipage}{11.44cm}\textsf{If the coordinates of the vertices of a triangle are \ (x_1,\ y_1),\ (x_2,\ y_2) \ and \ (x_3,\ y_3), }\\ \\ \textsf{then the coordinates of the centroid of the triangle will be,}\\ \\ \begin{center}\large \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)\end{center}\end{minipage}}$

Coordinates of three vertices of ΔPQR are given.

$\mapsto\ P(1, 1)\\ \\ \mapsto\ Q(2, 2)\\ \\ \mapsto\ R(-3, -3)$

Hence, the coordinates of the centroid will be,

$\large\left(\dfrac{1+2-3}{3},\ \dfrac{1+2-3}{3}\right)\ \ \ \ \ \Longrightarrow\ \ \ \ \ \Large\bold{\left(0,\ 0\right)}$

Hence, origin is the centroid!