Formula for centroid and mid-point: 1 Mark each
Let A$(x_1,y_1)$,B$(x_2,y_2)$ and C$(x_3,y_3)$ be the vertices of triangle ABC.
Let P(1,1), Q(2,-3), R(3,4) be the mid-points of sides AB, BC and CA respectively.
Then, P is the mid-point of BC
$\Rightarrow \frac{x_1+x_2}{2}=1,\frac{y_1+y_2}{2}=1$
$\Rightarrow x_1+x_2=2$ and $y_1+y_2=2\dots \dots \left(1\right)$
Q is the mid-point of AB
$\Rightarrow \frac{x_2+x_3}{2}=2,\frac{y_2+y_3}{2}=-3$
$\Rightarrow x_2+x_3=4$ and $y_2+y_3=-6\dots \dots \left(2\right)$
R is the mid-point of AC
$\Rightarrow \frac{x_1+x_3}{2}=3,\frac{y_1+y_3}{2}=4$
$\Rightarrow x_1+x_3=6$ and $y_1+y_3=8\dots \dots \left(3\right)$
Adding (1),(2) and (3) we get,
$x_1+x_2+x_2+x_3+x_1+x_3=2+4+6$
and $y_1+y_2+y_2+y_3+y_1+y_3=2-6+8$
$\Rightarrow x_1+x_2+x_3=6\dots \dots \left(4\right)$ 
and $y_1+y_2+y_3=2\dots \dots \left(5\right)$
The coordinates of the centroid of $△ABC$ are
$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=\frac{6}{3},\frac{2}{3}=2,\frac{2}{3}$ [Using (4) and (5)]