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?

Given:

The point \( (7,-4) \).

To do:

We have to find the points on the \( x \)-axis which are at a distance of \( 2 \sqrt{5} \) from the point \( (7,-4) \). 

Solution:

A point on the X-axis is of the form $(x, 0)$.

Let $P(x, 0)$ be the point on the X-axis which is at a distance of $2\sqrt5$ from the point $Q(7, -4)$.

We know that,

The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The distance between $P(x,0)$ and $Q(7,-4) is,

$PQ =\sqrt{(7-x)^{2}+(-4-0)^{2}}$

Squaring on both sides, we get,

$PQ^2=(7-x)^{2}+(-4)^{2}$

$(2\sqrt5)^2=7^2+x^2-2(7)(x)+16$

$4(5)=x^2-14x+49+16$

$x^2-14x+65-20=0$

$x^2-14x+45=0$

$x^2-9x-5x+45=0$

$x(x-9)-5(x-9)=0$

$(x-9)(x-5)=0$

$x=9$ or $x=5$

Hence, there are two points on the axis, $(5, 0)$ and $(9, 0)$ which are at a distance of \( 2 \sqrt{5} \) from the point \( (7,-4) \).

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Updated on: 10-Oct-2022

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