Distance between two straight lines

First, obviously we have two parallel straight lines to $r$ and at a distance of $10$. One will be placed on one side of $r$ and the other one to the other side.

If the straight lines that we are looking for result in $Ax + By + C = 0$, the parallelism condition with $r$ means that $A =-3$ and $B = 4$. In this way we have, $$-3x + 4y + C = 0$$

If now we put in the distance condition, that is to say, $d(r,s)=10$, we have: $$d(r,s)=10=\dfrac{|C'-C|}{\sqrt{A^2+B^2}}=\dfrac{|C'-(-1)|}{\sqrt{(-3)^2+4^2}}=\dfrac{|C'+1|}{\sqrt{25}}=\dfrac{|C'+1|}{5}$$ $$|C'+1|=50$$ From which we derive $2$ solutions due to the presence of the absolute value: $$C'+1=50 \rightarrow C' = 49$$ $$C' + 1 = - 50 \rightarrow C' =-51$$ This way, the straight lines $s$ are: $$s:-3x + 4y + 49 = 0$$ $$s':-3x + 4y - 51 = 0$$