Question

Find the equation of two straight lines which are parallel to $x+7y+2=0$ and at unit distance from the point $(1,-1)$.

Answer 1

lines parallel to $x+7y+2=0:x+7y+c=0$ distance from (1,-1) = 1

$\Rightarrow \frac{|1+7(-1)+c|}{\sqrt{1^2}+7^2}=1$

$\Rightarrow |c-6|=\sqrt{50}$

$\Rightarrow c=6\pm \sqrt{50}=6\pm 5\sqrt{2}$

$x+7y+(6+5\sqrt{2})=0$ & $x+7y+(6-5\sqrt{2})=0$