Question

Let the opposite angular points of a square be (3, 4) and (1, – 1), Find the co-ordinates of the remaining angular points.

Answer 1

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 ⇒ \(x\) = \(\frac{23-10y}{4}\)                      .....(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

\(\frac{23-10y}{4}\) + 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 = \(\frac{1}{2}\) or \(\frac{5}{2}\)

Putting y = \(\frac{1}{2}\) and \(\frac{5}{2}\) respectively in (i), we get \(x\) = \(\frac{9}{2}\) and \(x\) = \(-\frac{1}{2}\).

∴ The required vertices of the square are \(\bigg(\)\(\frac{9}{2}\),\(\frac{1}{2}\)\(\bigg)\)and \(\bigg(\)\(-\frac{1}{2}\), \(\frac{5}{2}\)\(\bigg)\)