Find the point on the x-axis which is equidistant from $(3,\ -5)$ and $(-2,\ 4)$.

Given: A point on the x-axis which is equidistant from $( 3,\ -5)$ and $( 2,\ -4)$.

To do: To find the point.

Solution:

Let $P( x,\ 0)$ be the point on the x-axis which is equidistant from $A( 3,\ -5)$ and $B( 2,\ -4)$.

$\Rightarrow PA=PB$

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

$\Rightarrow \sqrt{( 3-x)^2+(-5-0)^2}=\sqrt{( 2-x)^2+( -4-0)^2}$

$\Rightarrow \sqrt{9-6x+x^2+25}=\sqrt{4-4x+x^2+16}$

$\Rightarrow 34-6x+x^2=x^2-4x+20$

$\Rightarrow -6x+4x=20-34$

$\Rightarrow -2x=-14$

$\Rightarrow x=\frac{-14}{-2}=7$

Therefore, point $( 7,\ 0)$ is equidistant from $( 3,\ -5)$ and $( 2,\ -4)$.

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Updated on: 10-Oct-2022

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