Let ABCD be the given square and let A ≡ (3, 4) and C ≡ (1, –1). Also let B ≡ (x, y). ABCD being a square,AB = BC ⇒ AB2 = BC2, ∠ABC = 90º⇒ \(\big(\sqrt{(x-3)^2+(y-4)^2}\big)^2\) = \(\big(\sqrt{(x-1)^2+(y+1)^2}\big)^2\)⇒ x2 – 6x + 9 + y2 – 8y + 16 = x2 – 2x + 1 + y2 + 2y + 1⇒ 4x + 10y - 23 = 0 ⇒ \(x\) = \(rac{23-10y}{4}\) .....(i)Also, in ΔABC, ∠B = 90º ⇒ AB2 + BC2 = AC2⇒ (x – 3)2 + (y – 4)2 + (x – 1)2 + (y + 1)2 = (3 – 1)2 + (4 + 1)2 ⇒ x2 – 6x + 9 + y2 – 8y + 16 + x2 – 2x + 1 + y2 + 2y + 1 = 4 + 25 ⇒ 2x2 + 2y2 – 8x – 6y + 27 = 29 ⇒ 2x2 + 2y2 – 8x – 6y – 2 = 0 ⇒ x2 + y2 – 4x – 3y – 1 = 0 ...(ii) Now substitute the value of x from (i), we have\(rac{23-10y}{4}\) + y2 – (23 – 10y) – 3y – 1 = 0 ⇒ (23 – 10y)2 + 16y2 – 16 (23 – 10y) – 48y – 16 = 0 ⇒ 529 – 460y + 100y2 + 16y2 – 368 + 160y – 48y – 16 = 0 ⇒ 116y2 – 348y + 145 = 0 ⇒ 4y2 – 12y + 5 = 0 ⇒ (2y – 1) (2y – 5) = 0 ⇒ y = \(rac{1}{2}\) or \(rac{5}{2}\)Putting y = \(rac{1}{2}\) and \(rac{5}{2}\) respectively in (i), we get \(x\) = \(rac{9}{2}\) and \(x\) = \(-rac{1}{2}\).∴ The required vertices of the square are \(\bigg(\)\(rac{9}{2}\),\(rac{1}{2}\)\(\bigg)\)and \(\bigg(\)\(-rac{1}{2}\), \(rac{5}{2}\)\(\bigg)\)