Question:
Find a point on x-axis which is equidistant from A (2,-5)and (-2,9)
Answer:
Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
=> 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
=> PA2 = x2 - 4x + 29
Similarly, PB2 = x2 + 4x + 85
From equation 1, we get
=> x2 - 4x + 29 = x2 + 4x + 85
=> -8x = 56
=> x = -7
Hence, the point on x-axis is (-7, 0)
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