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 $A B, B C$ and $C A$.
$ \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 $
and $ \frac{x_{1}+x_{2}}{2} =11, \frac{y_{1}+y_{2}}{2}=9$
$\therefore x_{2}+x_{3}=4, x_{3}+x_{1}=4, x_{1}+x_{2}=22$
$\therefore 2\left(x_{1}+x_{2}+x_{3}\right)=30$
$\Rightarrow x_{1}+x_{2}+x_{3}=15$
and $y_{1}+y_{3}=+2, y_{2}+y_{3}=-2 y_{1}+y_{2}=18$
$\therefore 2\left(y_{1}+y_{2}+y_{3}\right)=18$
$\Rightarrow y_{1}+y_{2}+y_{3}=9$
$\therefore $ The centroid is
$=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right) $
$=\left(\frac{15}{3}, \frac{9}{3}\right)=(5,3)$
