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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.

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#### Solution

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (*x*, *y*), (*x*_{1}, *y*_{1}) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

∴ AB = BC

`=>sqrt((x+1)^2 + (y-2)^2) = sqrt((x-3)^2 + (y-2)^2)`

=>x^{2} + 2x + 1 + y^{2} -4y + 4 = x^{2} + 9 -6x + y^{2} + 4 - 4y

⇒ 8x = 8

⇒ x = 1

We know that in a square, all interior angles are of 90°.

In ΔABC,

AB^{2} + BC^{2} = AC^{2}

`=> (sqrt(((1+1)^2)+(y-2)^2))^2 + (sqrt(((1-3)^2)+(y-2)^2))^2 = (sqrt((3+1)^2+(2-2)^2))^2`

⇒ 4 + *y*^{2} + 4 − 4*y* + 4 + *y*^{2} − 4*y* + 4 =16

⇒ 2*y*^{2} + 16 − 8 *y* =16

⇒ 2*y*^{2} − 8 *y* = 0

⇒ *y* (*y *− 4) = 0

⇒ *y* = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD

Coordinate of point O = ((-1+3)/2, (2+2)/2)

`((1+x_1)/2, (y+ y_1)/2) = (1,2)`

`(1+x_1)/2 = 1`

1+x_{1}=2

x1 =1

and

` (y + y_1)/2 = 2`

⇒ *y* + *y*_{1} = 4

If *y* = 0,

*y*_{1} = 4

If *y* = 4,

*y*_{1} = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

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