A is (3, 3)
C is (5, 7)

ABCD is a square, so AC and BD are diagonals having the same length.

Midpoint of AC (let’s call it M) = (4, 5)

There is the Long and tedious way of finding the equation of BD and using lengths to solve for the coordinates, and there is an easy way. You can do the long one on your own, but I will do the short one (which may not be in your syllabus).

A to M is a movement of +1 unit in the x-direction and +2 units in the y-direction (a concept linked to gradients and vectors). Because ABCD is a square, AM must be perpendicular to and equal in length as MB. As such, M to B is a movement +2 units in the x-direction and -1 unit the y-direction OR -2 units in the x-direction and +1 in the y-direction. In the context of perpendicular lines, these movements basically lead to the products of the gradients of AM and MC being -1.

M to D is similar, and will account for the other case.

M is (4, 5) and there is a movement of EITHER +2 units in the x-direction and -1 unit in the y-direction OR -2 units in the x-direction and +1 unit in the y-direction. These cases account for the coordinates of B and D.

Thus, B and D are (6, 4) and (2, 6).

The standard way of solving requires you to find equation BD and using lengths.

Gradient of AC = (5 - 3) / (7 - 3) = 1/2
so gradient of BD = -2.

M is (4, 5).

Since M is on BD,

y = -2x + c
5 = -2 (4) + c
c = 13

BD: y = -2x + 13

Length of BD
= Length of AC
= sqrt [(5 - 3)^2 + (7 - 3)^2]
= sqrt 20

so length of BM and DM is 1/2 * sqrt 20 each. This simplifies to sqrt 5.

Let B be (a, b).

Then, length of BM is sqrt 20.

sqrt [(4 - a)^2 + (5 - b)^2] = sqrt 5
(4 - a)^2 + (5 - b)^2 = 5 (this is Eq. 1)

Since B lies along the line y = -2x + 13,

b = -2a + 13 (this is Eq. 2)

You will then solve a simul for this. I am currently not available to do the simul for you so you can proceed from here on your own.