Answer
$(4,7); (0,-1); (10,-3)$
Work Step by Step
We know that the vertices must be the same distance apart from each other as the corresponding set of vertices. Thus, since $(7,2)$ is 2 to the left and 4 units above $(5,-2)$, we find that another set of vertices must either be 2 to the left and 4 below $(2,3)$ or 2 to the right and 4 units above this point. Thus, two options are $(4,7)$ and $(0,-1)$. Finally, using the same logic, the last point must be 3 to the right and 5 units below $(7,2)$, making this point $(10,-3)$.
