Question

Find all points on the line $x-y=2$ that lie at a distance of $5$ units from the line $x+2y=5$

Answer 1

To take any point on the line $x-y=2$ ......$\left(1\right)$

Let $x=\alpha ,$ then from $\left(1\right),y=\alpha -2$

$\therefore P(\alpha ,\alpha -2)$ is any point on the line $\left(1\right)$

It will be a required point if its perpendicular distance from the line

$x+2y=5$ is $5$ units

$\Rightarrow \frac{|\alpha -2(\alpha -2)-5|}{\sqrt{1^2+1^2}}=5$

$\Rightarrow \frac{|\alpha -2\alpha +4|}{\sqrt{2}}=5$

$\Rightarrow \frac{|-\alpha +8|}{\sqrt{2}}=5$

$\Rightarrow \alpha -8=\pm 5\sqrt{2}$

$\Rightarrow \alpha =8\pm 5\sqrt{2}$

$\therefore \alpha =\frac{8+5\sqrt{2}}{2},\frac{8-5\sqrt{2}}{2}$

Hence, the required points are $(\frac{8+5\sqrt{2}}{2},\frac{8+5\sqrt{2}}{2}-2)$ and $(\frac{8-5\sqrt{2}}{2},\frac{8-5\sqrt{2}}{2}-2)$

or $(\frac{8+5\sqrt{2}}{2},\frac{8+5\sqrt{2}-4}{2})$ and $(\frac{8-5\sqrt{2}}{2},\frac{8-5\sqrt{2}-4}{2})$

or $(\frac{8+5\sqrt{2}}{2},\frac{4+5\sqrt{2}}{2})$ and $(\frac{8-5\sqrt{2}}{2},\frac{4-5\sqrt{2}}{2})$