Question

find the point on the x-axis which is equidistant from (2,-5) and (-2,9)

Answer 1

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Answer 2

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Q. Find the point on the x - axis which is equidistant from (2, -5) and (-2, 9).

Solution :

Since the point on x - axis have its ordinate = 0.

So, P(x, 0) is any point on the x - axis.

Since P(x, 0) is equidistant from A(2, -5) and B(-2, 9).

PA = PB ⇒ $PA^{2} = PB^{2}$

⇒ $(x - 2)^{2} + (0 + 5)^{2} = (x + 2)^{2} + (0 - 9)^{2}$

⇒ $x^{2}$ - 4x + 4 + 25 = $x^{2}$ + 4x + 4 + 81

⇒ -4x - 4x = 81 -25

⇒ -8x = 56

⇒ $x = \frac{56}{-8}$ = -7

∴ The point equidistant from given points on the x - axis is (-7, 0).