The ends of a diagonal of a square are (2,-3) and (-1, 1). Another vertex of the square can be

For square ABCD, the vertices of diagonal AD, A(2, -3) and D(-1, 1), are given.
Let one of the other two vertices be B(x,y). Now,

$\begin{array}{l}\mathrm{AB}\perp \mathrm{BD}\\ \therefore \left(\frac{\mathrm{y}+3}{\mathrm{x}-2}\right)\left(\frac{\mathrm{y}-1}{\mathrm{x}+1}\right)=-1\\ \mathrm{or} {\mathrm{y}}^2+2\mathrm{y}-3=-{\mathrm{x}}^2+\mathrm{x}+2\\ \mathrm{or} {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+2\mathrm{y}-5=0-----\left(1\right)\end{array}$

Also, AB= BD

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-2{)}^2+(\mathrm{y}+3{)}^2=(\mathrm{x}+1{)}^2+(\mathrm{y}-1{)}^2\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\mathrm{x}-8\mathrm{y}-11=0----\left(2\right)\end{array}$

Solving (1) and (2), we get two points as  $\left(\frac{-3}{2},\frac{-5}{2}\right)$ and $\left(\frac{5}{2},\frac{1}{2}\right)$