The points \((11,9),(2,1)\) and \((2,-1)\) are the mid-points of the sides of the triangle. Then, the centroid is

Let the vertices of a triangle be \(A\left(x_{1}, y_{1}\right)\), \(B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\). Since, \((11,9),(2,-1)\) and \((2,1)\) are the mid-points of \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\).
\[
\begin{array}{ll}
\frac{x_{2}+x_{3}}{2}=2, & \frac{y_{2}+y_{3}}{2}=-1 \\
\frac{x_{3}+x_{1}}{2}=2, & \frac{y_{3}+y_{1}}{2}=1
\end{array}
\]
and \(\frac{x_{1}+x_{2}}{2}=11, \frac{y_{1}+y_{2}}{2}=9\)



\(\therefore x_{2}+x_{3}=4, \quad x_{3}+x_{1}=4, \quad x_{1}+x_{2}=22\)
\(\therefore \quad 2\left(x_{1}+x_{2}+x_{3}\right)=30\)
\(\Rightarrow \quad x_{1}+x_{2}+x_{3}=15\)
and \(y_{1}+y_{3}=+2, y_{2}+y_{3}=-2, y_{1}+y_{2}=18\)
\(\therefore \quad 2\left(y_{1}+y_{2}+y_{3}\right)=18\)
\(\Rightarrow \quad y_{1}+y_{2}+y_{3}=9\)
\(\therefore\) The centroid is
\(\quad=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
\(\quad=\left(\frac{15}{3}, \frac{9}{3}\right)=(5,3)\)