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Which point on x – axis is equidistant from (5, 9) and (-4, 6)? |
Answer» Let A (5, 9) and B (-4, 6) be the given points Let the point on x – axis equidistant from the above points be C(x, 0) Now, we have AC = √[(x – 5)2 + (0 – 9)2] = √[x2 – 10x + 25 + 81] = √[x2 – 10x + 106] And, BC = √[(x – (-4))2 + (0 – 6)2] = √[x2 + 8x + 16 + 36] = √[x2 + 8x + 52] As AC = BC (given condition) So, AC2 = BC2 x2 – 10x + 106 = x2 + 8x + 52 18x = 54 x = 3 Therefore, the point on the x-axis is (3, 0) |
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