Find the point on the x-axis which is equidistant from $(2,\ -4)$ and $(-2,\ 6)$.

Given: A point on the x-axis which is equidistant from $( 2,\ -4)$ and $( -2,\ 6)$.

To do: To find the point.

Solution:

Let $P( x,\ 0)$ be the point on the x-axis which is equidistant from $A( 2,\ -4)$ and $B( -2,\ 6)$.

$\Rightarrow PA=PB$

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

$\Rightarrow \sqrt{( 2-x)^2+(-4-0)^2}=\sqrt{( -2-x)^2+( 6-0)^2}$

$\Rightarrow \sqrt{4-4x+x^2+16}=\sqrt{4+4x+x^2+36}$

$\Rightarrow 20-4x+x^2=x^2+4x+40$

$\Rightarrow 20-4x=4x+40$

$\Rightarrow -4x-4x=-40-20$

$\Rightarrow -8x=20$

$\Rightarrow x=-\frac{20}{8}=-\frac{5}{2}$

Therefore, point $( -\frac{5}{2},\ 0)$ is equidistant from $( 2,\ -4)$ and $( -2,\ 6)$.

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Updated on: 10-Oct-2022

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