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in Mathematics by (53.3k points)

Find the locus of the point which is equidistant from the points ( -3, 1) and (7, 5).

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Let A = (-3, 1) and B = (7, 5). Let P = (x, y).

Then AP = PB ⇔ (AP)2 = (PB)2

⇔ (x + 3)2 + (y -1)2 = (x -  7)2 + (y - 5)2

⇔  6x - 2y + 10 = 14x -  10y + 74

⇔ 20 x + 8y - 64 = 0

⇔ 5x + 2y - 16 = 0 

Hence, the equation of  5x + 2y - 16 = 0

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