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

Let ABCD is a square where two opposite vertices are A(-1, 2) and C(3, 2). Let B(x, y) and D(x1, y1) be the other two vertices.
In Square ABCD, AB = BC = CD = DA
Hence AB = BC ⇒ √(x + 1)² + (y - 2)² = √(3 - x)² + (2 - y)² [by distance formula]
Squaring both sides ⇒ (x + 1)² + (y - 2)² = (3 - x)² + (2 - y)²
⇒ x² + 2x + 1 + y² + 4 - 4y = 9 + x² - 6x + 4 + y² - 4y
⇒ 2x + 5 = 13 - 6x
⇒ 2x + 6x = 13 - 5
⇒ 8x = 8
⇒ x = 1
In △ABC, ∠B = 90° [all angles of square are 90°]
Then according to the Pythagorean theorem
AB² + BC² = AC²
As AB = BC ∴ 2AB² = AC²
⇒ 2(√(x + 1)² + (y - 2)²)² = (√(3 - (-1))² + (2 - 2)²)²
⇒ 2((x + 1)² + (y - 2)²) = (3 + 1)² + (2 - 2)²
⇒ 2(x² + 2x + 1 + y² + 4 - 4y) = (4)²
Put the value of x = 1
⇒ 2(1 + 2 × 1 + 1 + y² + 4 - 4y) = 16
⇒ 2(y² - 4y + 8) = 16
⇒ 2y² - 8y + 16 = 16
⇒ 2y² - 8y = 0
⇒ 2y(y - 4) = 0
⇒ (y - 4) = 0
Hence y = 0 or 4.
As diagonals of a square are equal in length and bisect each other at 90°
Let P is the midpoint of AC ∴ Co-ordinates of P = ( (3 - 1)/2, (2 + 2)/2 ) = (1, 2)
P is also the midpoint of BD then co-ordinates of mid-point of BD = co-ordinates of P
⇒ (x1, y1) = 1, 2 ∴ x1 = 1, y1 = 2
Then other two vertices of square ABCD are (1, 0) or (1, 4) and (1, 2).