Question

Find all points on the line $x+y=1$ that lie at a unit distance from the line $x+3y=2$

Answer 1

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

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

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

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

$x+3y=2$ is $1$ unit

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

$\Rightarrow \frac{|\alpha +3-3\alpha -2|}{\sqrt{1+9}}=1$

$\Rightarrow \frac{|-2\alpha +1|}{\sqrt{10}}=1$

$\Rightarrow 2\alpha -1=\pm \sqrt{10}$

$\Rightarrow \alpha =\frac{\sqrt{10}}{2},\frac{-\sqrt{10}}{2}$

Hence, the required points are $(\frac{\sqrt{10}}{2},1-\frac{\sqrt{10}}{2})$ and $(\frac{-\sqrt{10}}{2},1+\frac{\sqrt{10}}{2})$

or $(\frac{\sqrt{10}}{2},\frac{2-\sqrt{10}}{2})$ and $(\frac{-\sqrt{10}}{2},\frac{2+\sqrt{10}}{2})$