We can show the situation as : The centre of mass of square is at point $ D(x,y) $ The position co-ordinate of point D $ (x,y)\equiv \left( \frac{0+1}{2},\frac{1+0}{2} \right) $ $ =\left( \frac{1}{2},\frac{1}{2} \right) $ Hence, position vector or centre of mass D is $ =x\hat{i}+y\hat{j} $ $ =\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j} $ $ =\frac{1}{2}(\hat{i}+\hat{j}) $