Question

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

Answer 1

On x axis, the y coordinate is 0

So, let us assume the point to be (x,0)

Distance from (2,-5) = Distance from (-2,9)

(x-2)² + 25 = (x+2)² + 81

-56 = 8x

x = -7

So, the point required is (-7,0)

Answer 2

Step-by-step explanation:

AnswEr:

Let us consider given Points are A(2,-5) and B(-2,9).

If Point is on x - axis than y would be 0.

So, P(x, 0)

$\large\bold{\underline{\underline{\sf{\pink{According\:to\: Question\:Now,}}}}}:-$

Point P is equidistant from the Points A and B.

______________________

By using Distance Formula -

$:\implies\sf\small{\underline{\boxed{\sf{\red{\sqrt{(x_{1} - x_{2})^2\:+\: (y_{1} - y_{2})^2}}}}}}$

Now, Distance Between (x, 0) & (2, -5) is :-

$:\implies\sf\sqrt{(x - 2)^2 + (0 - (-5))^2}$

$:\implies\sf\sqrt{(x -2)^2 + (5)^2}$

Distances Between (x, 0) & (-2, 9) is :-

$:\implies\sf\sqrt{(x -(-2))^2 + (0-(-9))^2}$

$:\implies\sf\sqrt{(x + 2)^2 + (9)^2}$

Both are Equal. So,

$:\implies\sf\sqrt{(x -2)^2 + 5^2} = \sqrt{(x +2)^2 + 9^2}$

$:\implies\sf (x -2)^2 + 25 = (x + 2)^2 + 81$

$:\implies\sf x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81$

$:\implies\sf 8x = 25 - 81$

$:\implies\sf 8x = -56$

$:\implies\sf x = \cancel\dfrac{-56}{\:\:8}$

$:\implies\sf\small{\underline{\boxed{\sf{\red{x = -7}}}}}$

Hence, The Required Point is P(-7,0).

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$\rule{150}2$