The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.                                                                                                                                           (Ans: (1, 0), (1, 4))

Ans :  AB = BC ⇒ AD² = BC²

(x + 1)² + (y-2⁾² = (x-3)² + (y-2)²

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

2x - 4y + 5 = - 6x - 4y + 13

8x = 13 - 5      8x =     8  ⇒   x = 1

On substituting in (x-3)² + (y-2)² + (x+1)² + (y-2)²

= (-1 -3)² + (2 - 2)²

We get y = 4 or 0.

∴B (1, 4) or (1, 0)

AD = DC ⇒ AD² = DC²

(x1 + 1)² + (y1-2)² = (x1-3)² + (y1-2)²

∴ x = 1.

On substituting in (x1 + 1)² + (y1-2)² + (x1-3)² + (y1-2)² = 16

We get y1 = 0 or 4.

∴ D (1, 4) or (1, 0)

∴the opposite vertices are (1, 4) & (1, 0)