Question

Find all points on the line $x+y=3$ that lie at a distance of $2$ units from the line $3x+4y=5$

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

$3x+4y-5=0$ is $2$ units

$\Rightarrow \frac{|3\alpha +4\times (3-\alpha )-5|}{\sqrt{3^2+4^2}}=2$

$\Rightarrow \frac{|4\alpha +12-3\alpha -10|}{\sqrt{9+16}}=2$

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

$\Rightarrow \alpha +2=\pm 10$

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

$\therefore \alpha =3,-7$

Hence, the required points are $(3,3-3)$ and $(-7,3+7)$ or $(3,0)$ and $(-7,10)$