InterviewSolution
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InterviewSolution
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What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?(a) (10, 0)(b) (13, 0)(c) (11, 0)(d) (12, 0)I had been asked this question during an interview.My doubt stems from Geometry in section Coordinate Geometry of Mathematics – Class 10 |
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Answer» RIGHT ANSWER is (c) (11, 0) The EXPLANATION: Let the point on x-axis be (x, 0) Distance between (9, 8) and (x, 0) = \( \SQRT {(x_2-x_1)^2 + (y_2-y_1)^2} \) = \( \sqrt {(x-9)^2 + (0-8)^2} \) = \( \sqrt {x^2-18x + 81 + (-8)^2} \) = \( \sqrt {x^2-18x + 81 + 64} \) = \( \sqrt {x^2-18x + 145} \) Distance between (3, 2) and (x, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \) = \( \sqrt {(x-3)^2 + (0-2)^2} \) = \( \sqrt {x^2-6x + 9 + (-2)^2} \) = \( \sqrt {x^2-6x + 9 + 4} \) = \( \sqrt {x^2-6x + 13} \) Since, the point ( x, 0) is EQUIDISTANT to (3, 2) and (9, 8) The distances will be equal ∴ \( \sqrt {x^2-18x + 145} = \sqrt {x^2-6x + 13} \) Squaring on both sides we get, x^2 – 18x + 145 = x^2 – 6x + 13 -18x + 145 = -6x + 13 -18x + 6x = -145 + 13 -12x = -132 x = \( \frac {132}{12} \) = 11 The point is (11, 0) |
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