# 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?

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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)+164(5)=x^2-14x+49+16x^2-14x+65-20=0x^2-14x+45=0x^2-9x-5x+45=0x(x-9)-5(x-9)=0(x-9)(x-5)=0x=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 $(7,-4)$.

Updated on 10-Oct-2022 13:28:28