Question

find the points on y axis which is equidistant from(-5, - 2 )(- 9 - 2)
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Answer 1

$\textbf{Concept: }$

$\text{The distance between two points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}$

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

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

Let the required point on y axis be P(h,k)

since P lies on y-axis, h=0

Therefore, P is (0,k)

According to given data, PA=PB

$\sqrt{(0+5)^2+(k+2)^2}=\sqrt{(0+9)^2+(k+2)^2}$

$\implies\,25+(k+2)^2=81+(k+2)^2$

$\implies\,25=81$

This is not possible

Hence there is no point y axis which is equidistance from A and B

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Answer 2

No such point exists on y-axis.

Step-by-step explanation:

• We are asked to find out a point on y-axis. So, it's coordinates would be (0, y).

• It is given that it is equidistant from (-5, -2) & (-9, -2).

It means that;

• The distance between (0, y) & (-5, -2) = The distance between (0, y) & (-9, -2)

We know that;

• The distance between $(x_1, y_1) \ and \ (x_2, y_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

• The distance between (0, y) & (-5, -2) = $\sqrt{(0 - (-5))^2 + (y - (-2))^2}$

Similarly,

• The distance between (0, y) & (-9, -2) = $\sqrt{(0 - (-9))^2 + (y - (-2))^2}$

Since, both are equal. So we write as;

• $\sqrt{(0 - (-5))^2 + (y - (-2))^2} = \sqrt{(0 - (-9))^2 + (y - (-2))^2}$

Squaring both the sides, we get;

• $(0 - (-5))^2 + (y - (-2))^2 = (0 - (-9))^2 + (y - (-2))^2$

        $25 + (y + 2)^2 = (9)^2 + (y + 2)^2$

        $25 = 81$ This is not possible.

It means, there is no real value of y is possible for which the point having coordinate (0, y) would be equidistant from (-5,-2) & (-9, -2).