Q. 125.0( 1 Vote )

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

Answer :

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 + x^{2}) + (9 – 6y + y^{2}))

⇒ AB = √ (4 – 4x + x^{2} + 9 – 6y + y^{2})

⇒ AB = √ x^{2} + y^{2} – 4x – 6y + 13

Distance of BC

⇒ BC = √ ((–6 – x)^{2} + (–5 – y)^{2})

⇒ BC = √ ((36 + x^{2} + 12x) + (25 + y^{2} + 10y))

⇒ BC = √ (36 + x^{2} + 12x + 25 + y^{2} + 10y)

⇒ BC = √ (x^{2} + y^{2} + 12x + 10y + 61)

i.e. AB = BC (∵ Given)

⇒ √x^{2} + y^{2} – 4x – 6y + 13 = √ x^{2} + y^{2} + 12x + 10y + 61

Squaring both sides

⇒ x^{2} + y^{2} – 4x – 6y + 13 = x^{2} + y^{2} + 12x + 10y + 61

⇒ x^{2} + y^{2} – 4x – 6y + 13 – x^{2} – y^{2} – 12x – 10y – 61 = 0

⇒ –16x – 16 y – 48 = 0

⇒ –4(x + y + 3) = 0

⇒ x + y + 3 = 0

Hence proved.

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