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

The two opposite vertices of a square are (-1,2) and (3, 2). Let the two other vertices be B[x, y) and D(x₁, y₁)
The sides of square are equal. Therefore,
∴ AB = BC
⇒ √((x+1)²(y-2)²) = √((x-3)²+(y-2)²
⇒ √(x²+1+2x+y²+4-4y) = √(x²+9-6x+y²+4-4y)
Squaring both sides, we get
x²+1+2x +y + 4 -4y = x²+9-6x + y²+4-4y
⇒ x = 8 ⇒ x = 1
All the angles of square is 90°. Therefore, in AABC,
AB²+BC² = AC²
⇒ (√((x+1)²+(y-2)²))+(√((1-3)²+(y-2)²)) = (√(3+1)²+(2-2)²))
⇒ 4+y²+4-4y+4+y²-4y+4 = 16
⇒ 2y²+16-8y = 0
⇒ y(y-4) = 0
⇒ y = 0 or y = 4
We know that, the diagonals of square bisect each other. Therefore oordinates of mid-points of AC = Coordinates of mid-points of BD
⇒ ((-1+3)/2, (2+2)/2) = ((x+x₁)/2, (y+y₁)/2)
⇒ (1, 2) = ((1+x₁)/2, (y+y₁)/2)
On comparing, we get,
(1+x₁)/2 = 1
⇒ 1+x₁ = 2 ⇒ x₁ = 1
and (y+y₁)/2 ⇒ y+y₁ = 4
If y = 0, y₁ = 4,
If y = 4, y₁ = 0
Hence, the coordinates of other two vertices of square are (1, 0) and (1, 4).

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