Find the point on the x-axis which is equidistant from $(0,\ -5)$ and $(-2,\ 0)$.

Given: A point on the x-axis which is equidistant from $( 0,\ -5)$ and $( -2,\ 0)$.

To do: To find the point.

Solution:

Let $P( x,\ 0)$ be the point on the x-axis which is equidistant from $A( 0,\ -5)$ and $B( -2,\ 0)$.

$\Rightarrow PA=PB$

On using the distance formula, Distance between the two points $( x_1,\ y_1)$ and $( x_2,\ y_2)=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$

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

$\Rightarrow \sqrt{x^2+25}=\sqrt{4+4x+x^2}$

$\Rightarrow x^2+25=x^2+4x+4$

$\Rightarrow 4x=25-4$

$\Rightarrow 4x=21$

$\Rightarrow x=\frac{21}{4}$

Therefore, point $( \frac{21}{4},\ 0)$ is equidistant from $( 0,\ -5)$ and $( -2,\ 0)$.

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Updated on: 10-Oct-2022

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