Each side of a square is of length 4.  The centre of the square is (3, 7) and one of its diagonals is parallel to y = x.  Then the coordinates of its vertices are

side of the square is 4, hypotenuse = $\sqrt{2}s=4\sqrt{2}$ half of $4\sqrt{2 }is 2\sqrt{2}$

slope of the diagonal parallel to y=x is 1, inclination=${45}^0$

 $$\begin{gathered} The point on the diagonal \left(vertices\right) on the line with inclination {45}^{0 }and at a dis\tan ce of 2\sqrt{2} from centre(3,7) u\sin g parametric form is \\ x=x_1\pm r \cos \theta , y=y_1\pm r\sin \theta \end{gathered}$$

$\theta =\frac{\pi }{4}$

$x=3\pm 2\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)$           $y=7\pm \left(2\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}\right)$

 $A\left(5,9\right)C\left(1,5\right)$

As diagonals are  perpendicular so slope of other diagonal is 13$5^0$

$\theta =\frac{3\pi }{4}$

The other two vertices are

$B\left(1,9\right)D\left(5,5\right)$     ,