Question

Find all points on the line $x+y=4$ that lie at a unit distance from the line $4x+3y=10$

Answer 1

To take any point on the line $x+y=4$ ......$\left(1\right)$

Let $x=\alpha ,$ then from $\left(1\right),y=4-\alpha$

$\therefore P(\alpha ,4-\alpha )$ is any point on the line $\left(1\right)$

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

$4x+3y-10=0$ is $1$ unit

$\Rightarrow \frac{|4\alpha +3\times (4-\alpha )-10|}{\sqrt{4^2+3^2}}=1$

$\Rightarrow \frac{|4\alpha +12-3\alpha -10|}{\sqrt{16+9}}=1$

$\Rightarrow \frac{|\alpha +2|}{\sqrt{25}}=1$

$\Rightarrow \alpha +2=\pm 5$

$\Rightarrow \alpha =-2\pm 5$

$\therefore \alpha =3,-7$

Hence, the required points are $(3,4-3)$ and $(-7,4+7)$ or $(3,1)$ and $(-7,11)$