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