Consider the straight line $$r:-3x+4y-1=0$$, and find the straight lines $$s$$ parallel to $$r$$, and placed at a distance of $$10$$ from $$r$$.
Development:
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: $$$s:-3x + 4y + 49 = 0$$$ $$$s':-3x + 4y - 51 = 0$$$
Solution:
$$s:-3x + 4y + 49 = 0$$
$$s':-3x + 4y - 51 = 0$$