Question

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

Answer 1

$\to$ points on x - axis are of

form $(\alpha ,0)$ and at distance

of $2\sqrt{5}$ would give :

$\sqrt{(\alpha -7{)}^2+(4{)}^2}=2\sqrt{5}$

$\Rightarrow {\alpha }^2+49-14\alpha +16=20$

$\Rightarrow {\alpha }^2-14\alpha +45=0$

$\Rightarrow {\alpha }^2-5\alpha -9\alpha +45=0$

$\Rightarrow \alpha (\alpha -5)-9(\alpha -5)=0$

$\Rightarrow (\alpha -5)(\alpha -9)=0$

$\Rightarrow \alpha =5$ or $\alpha =9$

$\Rightarrow (5,0)$ or $(9,0)$