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Which point on the y-axis is equidistant from (2, 3) and (-4, 1)? |
Answer» Let A (2, 3) and B (-4, 1) be the given points Let the point on y – axis equidistant from the above points be C (0, y) Now, we have AC = √[(0 – 2)2 + (y – 3)2] = √[y2 – 6y + 9 + 4] = √[y2 – 6y + 13] And, BC = √[(0 – (-4))2 + (y – 1)2] = √[y2 – 2y + 1 + 16] = √[y2 – 2y + 17] As AC = BC (given condition) So, AC2 = BC2 y2 – 6y + 13 = y2 – 2y + 17 -4y = 4 y = -1 Therefore, the point on the y-axis is (0, -1). |
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