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

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)