1.

If the point P(a, b) is equidistant from the points (3, 1) and (2, 0) then ____________(a) a + b = -3(b) a – b = -3(c) a + b = 3(d) a – b = 3I have been asked this question in an interview for job.This intriguing question originated from Geometry topic in portion Coordinate Geometry of Mathematics – Class 10

Answer»

Right answer is (a) a + b = -3

Easy explanation: The point is (a, b)

Distance between (3, 1) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-3)^2 + (b-1)^2} \)

= \( \sqrt {a^2-6a + 9 + b^2-2b + 1} \)

= \( \sqrt {a^2-6a + 10 + b^2-2b} \)

Distance between (2, 0) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-2)^2 + (b-0)^2} \)

= \( \sqrt {a^2-4a + 4 + b^2 } \)

Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {a^2-6a + 10 + b^2-2b} = \sqrt {a^2-4a + 4 + b^2 } \)

SQUARING on both SIDES we get,

a^2 – 6a + 10 + b^2 – 2b = a^2 – 4a + 4 + b^2

-6a + 10 – 2b = -4a + 4

-2a – 6 = 2b

-a – b = 3

a + b = -3


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