Question

find the point on x axis which is equidistant from the points (-5,-2)and (3,2)

Answer 1

Step-by-step explanation:

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Answer 2

The point (-1,0) is equidistant from the points (-5,-2)and (3,2).

Step-by-step explanation:

Let (x,0) be the point equidistant from points (-5,-2)and (3,2).

Distance Formula:

$(x_1,y_1), (x_2,y_2)\\\\d = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2}$

Putting the values, we get,

$(x,0),(-5,-2)\\\\d_1 = \sqrt{(-2-0)^2 + (-5-x)^2} = \sqrt{4 + 25 + x^2 + 10x} = \sqrt{x^2 + 10x + 29}\\\\(x,0),(3,2)\\\\d_2 = \sqrt{(2-0)^2 + (3-x)^2} = \sqrt{4 + 9 + x^2 - 6x} = \sqrt{x^2 -6x +13}$

Equating the two distances, we have,

$\sqrt{x^2 + 10x + 29 } = \sqrt{x^2-6x+13}\\x^2 + 10x + 29 = x^2-6x+13\\16x = -6\\x = -1$

Thus, the point (-1,0) is equidistant from the points (-5,-2)and (3,2).

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