The point on x-axis which is equidistant from (5,9) and (-4,6) is

(4,1)

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(1,0)

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(2,0)

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(3,0)

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Solution

The correct option is D (3,0)
Let P(x,0) be the point on x-axis which is equidistant from A(5,9) and B(-4,6).

Then, PA = PB

Using the distance formula for PA and PB

(5 – x)² + (9)² = (-4 – x)² + (6)²

x = 3

Hence the required point is (3,0)

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