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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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