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Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9). |
Answer» Let point P(x, 0) is on x-axis and equidistant from A(2, -5) and B (-2, 9) PA = PB or PA2 = PB2 (2 – x)2 + (– 5 – 0)2 = (– 2 – x)2 + (9 – 0)2 (2 – x)2 + (– 5)2 = (– 2 – x)2 + (9)2 29 + x2 – 4x = 85 + x2 + 4x 56 = – 8x or x = – 7 The point on ×-axis is (– 7, 0) |
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