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Which point on y - axis is equidistant from (2, 3) and (-4, 1)? |
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Answer» Since the point is on y-axis, therefore coordinate of x-axis is zero. Therefore the point is P(0, k) which is equidistance from A(2, 3) and B(- 4, 1) PA = PB \(\sqrt{(2 - 0)^2 + (3 - k)^2}\) = \(\sqrt{(- 4)^2 + (1 - k)^2}\) \(\sqrt{4 + 9 + k^2 - 6k}\) = \(\sqrt{16 + 1 + k^2 - 2k}\) On squaring both sides, we get - 4k = 4 k = - 1 Therefore coordinate is (0, -1) |
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