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Answer :

Let ABCD be a square with vertices A( \( x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)\)

Let each side of square be 'a'.

So applying Pythagoras theorem in ∆ABD, we get

\( BD^2 = AB^2 + DA^2 \)

\( \Rightarrow \sqrt{(3+1)^2 \ + \ (2-2)^2} \ = \ a^2 \ + \ a^2 \)

\( \Rightarrow 4 \ = \ 2 a^2 \)

\(\Rightarrow a \ = \ 2 \sqrt{2} \)

Now, DA = AB {\(\because \) ABCD is a square }

\( \Rightarrow \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2} \) {by distance formula}

\(\Rightarrow (x_1 +1)^2 \ = \ (3-x_1)^2 \)

\( \Rightarrow x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1 \)

\(\Rightarrow x_1 \ = \ 1 \)

Now since each side is \( 2 \sqrt{2} \)

\(\therefore \) AB \( = \ 2 \sqrt{2} \)

\( \Rightarrow \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2} \)

\( \Rightarrow(2-y_1)^2 = \ 4 \)

\(\Rightarrow y \ = \ 4 \)

Now, point O is id point of BD . SO coordinates of O are

x = \( \frac{3-1}{2} \) , y = \( \frac{2+2}{2} \)

\(\Rightarrow \) x = 1 , y = 2

Now, since ABCD is a square \(\therefore \) O is mid point of AC also

\( \frac{1+x_2}{2} \ = \ 1 \) , \( \frac{4+ y_2}{2} \ = \ 2 \)

\( \Rightarrow x_2 \ = \ 1 \) , \( y_2 \ = \ 0 \)

\(\therefore \) Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)

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