Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks
Let ABCD be a square and two opposite vertices of it are A(-1, 2) and C(3, 2). ABCD is a square.

Since ABCD is a square.
$\Rightarrow AB=BC$
$\Rightarrow A{B}^2=B{C}^2$
[Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$]
$\Rightarrow {(x+1)}^2+{(y-2)}^2={(x-3)}^2+{(y-2)}^2$
$\Rightarrow x^2+2x+1=x^2-6x+9$
$\Rightarrow 2x+6x=9-1=8$
$\Rightarrow 8x=8\Rightarrow x=1$
ABC is right $\Delta$ at B, then
$A{C}^2=A{B}^2+B{C}^2$ (Pythagoras theorem)
$\Rightarrow (3+1{)}^2+(2-2{)}^2=(x+1{)}^2+(y-2{)}^2+(x-3{)}^2+(y-2{)}^2$
$\Rightarrow 16=2(y-2{)}^2+(1+1{)}^2+(1-3{)}^2$
$\Rightarrow 16=2(y-2{)}^2+4+4\Rightarrow 2(y-2{)}^2=16-8=8$
$\Rightarrow (y-2{)}^2=4\Rightarrow y-2=\pm 2\Rightarrow y=4$ and 0
i.e when x = 1 then y = 4 and 0
Co-ordinates of the opposite vertices are: B(1,0) or D(1,4)