Question

Find the points on the $x$-axis which are at a distance of $2\sqrt{5}$ from the point $(7,-4)$. How many such points are there $?$

Answer 1

Find points on the $x$-axis:

Given, the points are $2\sqrt{5}$ units away from $(7,-4)$

Also given, the points lie on the $x$-axis, thus their $y$-coordinate is $0$

Let $x$ be the the $x$-coordinate of the points.

Therefore, the coordinates of the points will be in the form $(x,0)$

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.

Substituting $(x_1,y_1)=(x,0)$ and $\left(x_2,y_2\right)=(7,-4)$ and $d=2\sqrt{5}$

$$\begin{gathered} \Rightarrow \sqrt{{(7-x)}^2+{(-4-0)}^2}=2\sqrt{5} \\ \Rightarrow {(7-x)}^2+16=20 \\ \Rightarrow x^2-14x+49=4 \\ \Rightarrow x^2-14x+45=0 \\ \Rightarrow (x-9)(x-5)=0 \end{gathered}$$

$\Rightarrow x=9$ or $x=5$

Therefore, there are two points that are a distance of $2\sqrt{5}$ away from the point $(7,-4)$ and their coordinates are $(9,0)$ and $(5,0)$.