If two points (2, 3) and (−6, −5) are equidistant from the point (x, y), show that x + y + 3 = 0.

Formula used:

Let the points be A (x, y), B (2, 3) and C (–6, –5)

Distance of AB

⇒ AB = √ ((2 – x)² + (3 – y)²)

⇒ AB = √ ((4 – 4x + x²) + (9 – 6y + y²))

⇒ AB = √ (4 – 4x + x² + 9 – 6y + y²)

⇒ AB = √ x² + y² – 4x – 6y + 13

Distance of BC

⇒ BC = √ ((–6 – x)² + (–5 – y)²)

⇒ BC = √ ((36 + x² + 12x) + (25 + y² + 10y))

⇒ BC = √ (36 + x² + 12x + 25 + y² + 10y)

⇒ BC = √ (x² + y² + 12x + 10y + 61)

i.e. AB = BC (∵ Given)

⇒ √x² + y² – 4x – 6y + 13 = √ x² + y² + 12x + 10y + 61

Squaring both sides

⇒ x² + y² – 4x – 6y + 13 = x² + y² + 12x + 10y + 61

⇒ x² + y² – 4x – 6y + 13 – x² – y² – 12x – 10y – 61 = 0

⇒ –16x – 16 y – 48 = 0

⇒ –4(x + y + 3) = 0

⇒ x + y + 3 = 0

Hence proved.