Q. 125.0( 1 Vote )

# If two points (2,

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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