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

• A

  $(-5,-3)$
• B

  $(5,-3)$
• C

  $(3,5)$
• Correct answer
  D

  $(5,3)$

Solution

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 $\quad=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$ $\quad=\left(\frac{15}{3}, \frac{9}{3}\right)=(5,3)$

Hence, the correct answer is option: 'D'.