Problems from Distance between two straight lines

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: $$d(r,s)=10={\displaystyle \frac{|C^{\prime }-C|}{\sqrt{A^2+B^2}}}={\displaystyle \frac{|C^{\prime }-(-1)|}{\sqrt{(-3{)}^2+4^2}}}={\displaystyle \frac{|C^{\prime }+1|}{\sqrt{25}}}={\displaystyle \frac{|C^{\prime }+1|}{5}}$$ $$|C^{\prime }+1|=50$$ From which we derive $2$ solutions due to the presence of the absolute value: $$C^{\prime }+1=50\to C^{\prime }=49$$ $$C^{\prime }+1=-50\to C^{\prime }=-51$$ This way, the straight lines $s$ are: $$s:-3x+4y+49=0$$ $$s^{\prime }:-3x+4y-51=0$$

Solution:

$s:-3x+4y+49=0$

$s^{\prime }:-3x+4y-51=0$

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