Question

find the point on the x axis which is equidistance from (2,-5)and (-2,9).
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Answer 1

Answer:

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Answer 2

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION}}}$

Let A(2,-5) & B(-2,9) be the given points.

Also let P(x,0) be the points on x-axis such that,

PA= PB

PA^2 = PB^2

Then,

$(x - 2) {}^{2} + (0 + 5) {}^{2} = (x + 2) {}^{2} + (0 - 9) {}^{2} \\ \\ = > (x - 2) {}^{2} - (x + 2) {}^{2} = 81 - 25 \\ \\ = > (x - 2 + x + 2)(x - 2 - x - 2) = 56 \\ \\ = > (2x)( - 4) = 56 \\ \\ = > - 8x = 56 \\ \\ = > x = - \frac{56}{8} = 7 \\ \\ = > x = - 7$

Hence, the required point is (-7,0).

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