$\begin{array}{1 1}(a)\;(-1,1),(1,-1),y=-x\\(b)\;(1,1),(1,1),y=x\\(c)\;(-2,1),(1,-2),y=-2x\\(d)\;(2,1),(1,-2),y=-3x\end{array}$

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Here $(1,1)$ & $(-1,-1)$ are vertices of diagonal of square which seems to be lying on the line y=x hence the line perpendicular to this y=x is y=-x.

Hence equation of the other diagonal is $y=-x$

Let $D(x_1,y_1)$ & $B(x_2,y_2)$

These points lie on the line $y=-x$ hence $y_1=-x_1,y_2=-x_2$

Now $OA=OD$ where O(0,0) midpoint of AC.

$OA=\sqrt 2$

$OD=\sqrt 2$

$\sqrt{x_1^2+y_1^2}=2$

$x_1^2+y_1^2=2$

$x_1^2+(-x_1)^2=2$

$2x_1^2=2$

$x_1=\pm 1$

So $x_1$ has two values $\pm 1$ in which $x_1=-1$ & $x_2=1$

So $y_1=1$ & $y_2=-1$

Hence $D(-1,1)$ & $B(1,-1)$

Hence (a) is the correct answer.

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