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Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (- 3, 4). |
Answer» Coordinates of the points are A(3, 6) and B(-3, 4) Let the point P(x, y) is equidistant from A and B Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) ⇒ PA = PB \(\sqrt{(x - 3)^2 + (y - 6)^2}\) = \(\sqrt{(x + 3)^2 + (y - 4)^2}\) On squaring both sides, we get (x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2 x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16 3x + y = 5 |
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