Question

Find all points on the line $x+y=3$ that lie at a distance of $5$ units from the line $x-y=5$

Answer 1

To take any point on the line $x+y=3$ ......$\left(1\right)$

Let $x=\alpha ,$ then from $\left(1\right),y=3-\alpha$

$\therefore P(\alpha ,3-\alpha )$ is any point on the line $\left(1\right)$

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

$x-y=5$ is $5$ units

$\Rightarrow \frac{|\alpha -(3-\alpha )-5|}{\sqrt{1^2+1^2}}=5$

$\Rightarrow \frac{|\alpha -3+\alpha -5|}{\sqrt{2}}=5$

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

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

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

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

Hence, the required points are $(\frac{13}{2},3-\frac{8+5\sqrt{2}}{2})$ and $(\frac{3}{2},3-\frac{8-5\sqrt{2}}{2})$

or $(\frac{13}{2},\frac{6-8-5\sqrt{2}}{2})$ and $(\frac{3}{2},\frac{6-8+5\sqrt{2}}{2})$

or $(\frac{13}{2},\frac{-2-5\sqrt{2}}{2})$ and $(\frac{3}{2},\frac{-2+5\sqrt{2}}{2})$