Let ABCD is a square with A(5, 4) and C(1, -6).

Let the coordinates of B are (x, y)

Using distance formula =

⇒ AB = BC

On squaring both sides, we get

……………………….(1)

In ΔABC, Using Pythagoras theorem

AC² = AB² + BC²

AC² = 2BC² [Since AB = BC]

On squaring both sides, we get

………………..(2)

On substituting value of from equation (1) in equation (2), we get

On solving we get y = -3, -1

Substituting these values of y in eqn 1, we get x = 8, -2

Therefore other coordinates are B(8, -3). And D(-2,1)