Question

Find all points on the line $x+y=6$ that lie at a distance of $4$ units from the line $2x-4y=5$

Answer 1

To take any point on the line $x+y=6$ ......$\left(1\right)$

Let $x=\alpha ,$ then from $\left(1\right),y=6-\alpha$

$\therefore P(\alpha ,6-\alpha )$ is any point on the line $\left(1\right)$

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

$2x-4y=5$ is $4$ units

$\Rightarrow \frac{|2\alpha -4\times (6-\alpha )-5|}{\sqrt{2^2+4^2}}=4$

$\Rightarrow \frac{|2\alpha -24+4\alpha -5|}{\sqrt{4+16}}=4$

$\Rightarrow \frac{|6\alpha -29|}{\sqrt{20}}=4$

$\Rightarrow \frac{|6\alpha -29|}{2\sqrt{5}}=4$

$\Rightarrow 6\alpha -29=\pm 8\sqrt{5}$

$\Rightarrow 6\alpha =29\pm 8\sqrt{5}$

$\therefore \alpha =\frac{29+8\sqrt{5}}{6},\frac{29-8\sqrt{5}}{6}$

Hence, the required points are $(\frac{29+8\sqrt{5}}{6},6-\frac{29+8\sqrt{5}}{6})$ and $(\frac{29-8\sqrt{5}}{6},6-\frac{29-8\sqrt{5}}{6})$

or $(\frac{29+8\sqrt{5}}{6},\frac{36-29-8\sqrt{5}}{6})$ and $(\frac{29-8\sqrt{5}}{6},\frac{36-29+8\sqrt{5}}{6})$

or $(\frac{29+8\sqrt{5}}{6},\frac{7-29\sqrt{5}}{6})$ and $(\frac{29-8\sqrt{5}}{6},\frac{7-29\sqrt{5}}{6})$