Two opposite vertices of a square are (1, –2) and (–5, 6) then the other two vertices are

Two opposite vertices of a square are let A=(1, -2), C=(-5, 6)

 $$\begin{gathered} \mathrm{AC}=\sqrt{{(-5-1)}^2+{(6+2)}^2} \\ =\sqrt{36+64} \\ =\sqrt{100} \\ =10 \end{gathered}$$

Mid point of $\mathrm{AC}=\left(\frac{1-5}{2}, \frac{-2+6}{2}\right)$

 $=(-2, 2)$

In a square diagonals bisect each other and also their lengths are equal.

By verification first option, other two vertices are let B=(2, 5), D=(-6, -1)

Now $\mathrm{BD}=\sqrt{{(-8)}^2+{(-6)}^2}=\sqrt{100}=10$

And mid point of $\mathrm{BD}=\left(\frac{2-6}{2}, \frac{5-1}{2}\right)=(-2, 2)$