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


Answer :

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).

Using Distance Formula and according to given conditions we have:

\(\sqrt{(x - 2)^2 + ( 0 - (-5))^2} \)
\( = \sqrt{ (x - (-2))^2 + ( 0 - 9)^2}\)

=>\(\sqrt {x^2 + 4 - 4x + 25} \)
\( = \sqrt {x^2 + 4 + 4x + 81}\)

Squaring both sides, we get
\(\Rightarrow x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81\)
\(\Rightarrow-4x + 29 = 4x + 85\)
\( \Rightarrow 8x = -56\)
\( \Rightarrow x = -7\)

Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

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