Find the point on the x-axis which is equidistant from $( 5,\ -5)$ and $( 0,\ 9)$.

Given: A point on the x-axis which is equidistant from $( 5,\ -5)$ and $( 0,\ 9)$.

To do: To find the point.

Solution:

Let $P( x,\ 0)$ be the point on the x-axis which is equidistant from $A( 5,\ -5)$ and $B( 0,\ 9)$.

$\Rightarrow PA=PB$

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

$\Rightarrow \sqrt{( 5-x)^2+(-5-0)^2}=\sqrt{( 0-x)^2+( 9-0)^2}$

$\Rightarrow \sqrt{25-50x+x^2+25}=\sqrt{x^2+81}$

$\Rightarrow 50-50x+x^2=x^2+81$

$\Rightarrow 50-50x=81$

$\Rightarrow -50x=81-50$

$\Rightarrow -50x=31$

$\Rightarrow x=-\frac{31}{50}$

Therefore, point $( -\frac{31}{50},\ 0)$ is equidistant from $( 5,\ -5)$ and $( 0,\ 9)$.

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Updated on: 10-Oct-2022

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