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 + 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.


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