Question

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

Answer 1

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x₁, y₁) 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)`

=>x² + 2x + 1 + y² -4y + 4 = x² + 9  -6x + y² + 4 - 4y

⇒ 8x = 8

⇒ x = 1

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

In ΔABC,

AB² + BC² = AC²

`=> (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 + y² + 4 − 4y + 4 + y² − 4y + 4 =16

⇒ 2y² + 16 − 8 y =16

⇒ 2y² − 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+x₁=2

x1 =1

and

` (y + y_1)/2 = 2`

⇒ y + y₁ = 4

If y = 0,

y₁ = 4

If y = 4,

y₁ = 0

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