Find the point on the x-axis which is equidistant from $(2,\ -5)$ and $(-2,\ 9)$.

Given: A point on the x-axis which is equidistant from $( 2,\ -5)$ and $( -2,\ 9)$.

To do: To find the point.

Solution:

Let $P( x,\ 0)$ be the point on the x-axis which is equidistant from $A( 2,\ -5)$ and $B( -2,\ 9)$.

$\Rightarrow PA=PB$

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

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

$\Rightarrow \sqrt{4-4x+x^2+25}=\sqrt{4+4x+x^2+81}$

$\Rightarrow 29-4x+x^2=4x+x^2+85$

$\Rightarrow -4x-4x=85-29$

$\Rightarrow -8x=56$

$\Rightarrow x=-\frac{56}{8}=-7$

Therefore, point $( -7,\ 0)$ is equidistant from $( 2,\ -5)$ and $( -2,\ 9)$.

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Updated on: 10-Oct-2022

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