Square Center from Four Points

Let us assume that the four points each lie on a different side of the square. The edges of the square will then be parallel to the unit vectors $\mathbf{u} = \binom{\cos\theta}{\sin\theta} \hspace{2cm} \mathbf{v} = \binom{-\sin\theta}{\cos\theta}$ for some $\theta$ . We then determine the value of $\theta$ by requiring that $\big(\overrightarrow{P_1} - \overrightarrow{P_3}\big)\cdot \mathbf{u} \;= \; \big(\overrightarrow{P_2} - \overrightarrow{P_4}\big)\cdot \mathbf{v} \;= \; 10$

and we deduce after numerical work that $\theta = 1.221730$ . We can then determine the centre of the square by requiring that $\left[\binom{3.2}{y} - \overrightarrow{P_3}\right]\cdot \mathbf{u} \; = \; 5$ and this tells us that $y = 5.699999$ . Thus we can say that the coordinates of the centre of the square are $\boxed{(3.2,5.7)}$ . It is easy to draw the square so defined, and see we have the answer.