Question

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

Answer 1

Step 1: Define point on $x$-axis

Let $A$ be the point that is equidistant from the given points on the $x$-axis.

Let the point $(2,-5)$ be denoted as $B$ and point $(-2,9)$ be denoted as $C$

Since the point will lie on $x$-axis, its $y$-coordinate will be $0$

Thus the coordinates of point $A$ will be $(x,0)$

Step 2: Determine distance from $A$ to $B$

We can find the distance between any two points with the help of the distance formula which is given as,

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

where $d$ is the distance between the two points, $x_2$ and $x_1$ are the $x$-coordinate of the two points, and $y_2$ and $y_1$ are the $y$-coordinate of the two points.

Thus, the distance ($d_1$) between points $A$ and $B$ is,

$$\begin{gathered} d_1=\sqrt{{\left(x-2\right)}^2+{(0-\left(-5\right))}^2} \\ {d}_1=\sqrt{{\left(x-2\right)}^2+25} \end{gathered}$$

Step 3: Determine distance from $A$ to $C$

The distance ($d_2$) between points $A$ and $C$ is,

$$\begin{gathered} d_2=\sqrt{{\left(x-\left(-2\right)\right)}^2+{(0-9)}^2} \\ {d}_2=\sqrt{{\left(x+2\right)}^2+81} \end{gathered}$$

Step 4: Determine coordinates of point $A$

Given, $A$ is equidistant from $B$ and $C$,

$$\begin{gathered} \Rightarrow d_1=d_2 \\ \Rightarrow \sqrt{{(x-2)}^2+25}=\sqrt{{(x+2)}^2+81} \\ \Rightarrow {(x-2)}^2+25={(x+2)}^2+81 \\ \Rightarrow x^2-4x+29=x^2+4x+85 \\ \Rightarrow 8x=-56 \\ \Rightarrow x=-7 \end{gathered}$$

Therefore, the point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$ is $(-7,0)$.