Q15 of 47 Page 1

Find the coordinates of a point on the x-axis which is equidistant from the points A(2, –5) and B(–2, 9).

Given:

*A = (2, –5)


B = (–2, 9)


Let the point be P (x, 0), ∵ it is on x-axis ∴ y- co-ordinate be 0.


∵ P (x, 0) is equidistant from A (2, –5) and B (–2, 9)


∴ PA = PB


⇒ PA2 = PB2


We know that by Distance Formula,


PA2 = (x – 2)2 + (0 + 5)2


⇒ PA2 = x2 + 4 – 4x + 25


⇒ PA2 = x2 – 4x + 29


And,


PB2 = (x + 2)2 + (0 – 9)2


⇒ PB2 = x2 + 4 + 4x + 81


⇒ PB2 = x2 + 4x + 85


∵ PA2 = PB2


⇒ x2 – 4x + 29 = x2 + 4x + 85


⇒ x2 – 4x + 29 – x2 – 4x – 85 = 0


⇒ – 4x + 29 – 4x – 85 = 0


⇒ – 8x – 56 = 0


⇒ – 8x = 56



Hence, the point = (-7, 0)


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