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Find a point on the x-axis which is equidistant from the points (7, 6) and (- 3, 4). |
Answer» points A(7, 6) and B(-3, 4) are equidistance from point P. Let the coordinates of point are P(x, 0) Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) ⇒ PA = PB \(\sqrt{(x - 7)^2 + (0 - 6)^2}\) = \(\sqrt{(x + 3)^2 + (0 - 4)^2}\) On squaring both sides, we get (x - 7)2 + (0 - 6)2 = (x + 3)2 + (0 - 4)2 x2 - 14x + 49 + 36 = x2 + 6x + 9 + 16 5x = 15 x = 3 Therefore coordinates are (3, 0) |
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