The two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$. Find the coordinates of the other two vertices.

(A) Let $A(-1, 2)$ and $C(3, 2)$ be the given opposite vertices of a square $ABCD$. Let the other two vertices be $B(x, y)$ and $D(x_1, y_1)$.
Since $ABCD$ is a square,all sides are equal,so $AB = BC$.
Using the distance formula: $\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x-3)^2 + (y-2)^2}$.
Squaring both sides: $(x+1)^2 + (y-2)^2 = (x-3)^2 + (y-2)^2$.
$x^2 + 2x + 1 = x^2 - 6x + 9 \Rightarrow 8x = 8 \Rightarrow x = 1$.
In $\triangle ABC$,$\angle B = 90^\circ$,so by Pythagoras theorem: $AB^2 + BC^2 = AC^2$.
$AC^2 = (3 - (-1))^2 + (2 - 2)^2 = 4^2 + 0^2 = 16$.
Since $AB = BC$,$2AB^2 = 16 \Rightarrow AB^2 = 8$.
$(x+1)^2 + (y-2)^2 = 8$. Substituting $x=1$: $(1+1)^2 + (y-2)^2 = 8 \Rightarrow 4 + (y-2)^2 = 8 \Rightarrow (y-2)^2 = 4$.
$y-2 = \pm 2 \Rightarrow y = 4$ or $y = 0$.
Thus,$B$ is $(1, 4)$ and $D$ is $(1, 0)$ (or vice versa).
The coordinates of the other two vertices are $(1, 0)$ and $(1, 4)$.