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Question

Find the point on x-axis which is equi-distance from (2,-5)and (-2,9)

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Solution

Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
That is PA = PB
Hence PA2 = PB2 → (1)
Distance between two points is √[(x2 - x1)2 + (y2 - y1)2]
PA = √[(2 - x)2 + (-5 - 0)2]
PA2 = 4 - 4x +x2 + 25 = x2 - 4x + 29
Similarly, PB2 = x2 + 4x + 85
Equation (1) becomes
x2 - 4x + 29 = x2 + 4x + 85
- 8x = 56
x = -7
Hence the point on x-axis is (-7, 0)

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