Question

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

Answer 1

To take any point on the line $x-y=5$ ......$\left(1\right)$

Let $x=\alpha ,$ then from $\left(1\right),y=5-\alpha$

$\therefore P(\alpha ,5-\alpha )$ 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 $3$ units

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

$\Rightarrow \frac{|3\alpha +12-20+4\alpha -5|}{\sqrt{9+16}}=3$

$\Rightarrow \frac{|7\alpha -13|}{\sqrt{25}}=3$

$\Rightarrow 7\alpha -13=\pm 15$

$\Rightarrow 7\alpha =-13\pm 15$

$\therefore 7\alpha =2,-28$

$\therefore \alpha =\frac{2}{7},-4$

Hence, the required points are $(\frac{2}{7},5-\frac{2}{7})$ and $(-4,5+4)$

or $(\frac{2}{7},\frac{35-2}{7})$ and $(-4,9)$

or $(\frac{2}{7},\frac{33}{7})$ and $(-4,9)$