The two opposite verticesof square are(-1,2) …

Let ABCD be a square and B (x, y) be the unknown vertex.
AB = BC
{tex} \Rightarrow {/tex}  AB² = BC² 
 
{tex} \Rightarrow {/tex} (x + 1)² + (y - 2)² = (x - 3)² + (y - 2)²
{tex} \Rightarrow {/tex} x² + 1 + 2x + y² + 4 - 4x = x² - 6x + 9 + y² + 4 - 4x
{tex} \Rightarrow {/tex} 2x + 1 = - 6x + 9
{tex} \Rightarrow {/tex} 8x = 8
{tex} \Rightarrow {/tex} x = 1 ........ (i)
In {tex}\triangle{/tex}ABC, AB² + BC² = AC²
{tex} \Rightarrow {/tex} (x + 1)² + (y - 2)² + (x - 3)² + (y - 2)² = (3 + 1)² + (2 - 2)²
{tex} \Rightarrow {/tex}x² + 1 + 2x + y² - 4x + 4 + x² +  9 - 6x + y² + 4 - 2y = 16 + 0
{tex} \Rightarrow {/tex} 2x² + 2y² + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16
{tex} \Rightarrow {/tex} 2x² + 2y² - 4x - 8y + 2 = 0
{tex} \Rightarrow {/tex} x² + y² - 2x - 4y + 1 = 0 ..... (ii)
Putting the value of x in eq. (ii),
1 + y² - 2 - 4y + 1 = 0 
{tex} \Rightarrow {/tex} y² - 4y = 0 
{tex} \Rightarrow {/tex} y(y - 4) = 0
{tex} \Rightarrow {/tex} y = 0 or 4
Hence the other vertices are (1, 0) and (1, 4).