Find the cartesian equation of the locus of the set of points of $P$ problem. 
Find the cartesian equation of the locus of the set of points of $P$. $P$ is at a constant distance of five units from the line $4x-3y=1$

I don't have much intuition on how to solve this one. However, I've done a bit of research online and have found a formula which can apparently help me solve this type of problem:
$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$
Problem is I don't quite understand this formula yet cause my textbook hasn't got any chapters on it. And I don't think the textbook would assume prior knowledge of this formula either. This makes me think that there must be a way to solve the problem without the formula. 
The book gives the answers as $4x-3y=26$ and $4x-3y+24=1$
Hints/guidance on how to solve this problem especially ones without using the formula would be much appreciated. 
 A: $$4x-3y=1.$$ Let $P(u,v)$ be at distance of $5$ from this line.
Then $$\frac{∣4u-3v-1∣}{\sqrt{9+16}}=5.$$ Thus $(u,v)$ satisfies either of the equations $4u-3v=26$ or $4u-3v=-24:\;$ two lines.
A: Another way to approach this problem without using the line equation is:
$4x - 3y = 1$ is $y = \frac{4}{3}x - \frac{1}{3}.$
We can add a vertical offset to this line equation such that the distance from the old line to the shifted line equals 5. We call this offset o. The slope of the line is $m = \frac{4}{3}.$
Notice that when the slope is zero, the offset 5 is added. As the slope increases, the angle $\theta$ between the line perpendicular to $y = \frac{4}{3}x - \frac{1}{3}$ and the line $v$ parallel to the $y$-axis increases.
$\theta = \tan^{-1} m,$ as it is the same angle that the line forms with the $x$-axis. $\theta = 53.13.$
$$\cos\theta = \frac{a}{h},\quad h = \frac{5}{\cos\theta},\quad h = 8.333.$$
d can be +5 or -5. Taking the positive case:
$$y = \frac{4}{3}x - \frac{1}{3} + 8.333,$$
$$3y = 4 - 1 + 24.999,$$
$$3y = 4 + 24.$$

A: The locus is clearly given by two lines, parallel to $l:4x-3y=1$, with distance $5$ from $l$.
The line $4x-3y=1$ goes through $(1,1)$. The squared distance of $(1,1)$ from the line $4x-3y=c$ is given by:
$$ \frac{(4-3-c)^2}{4^2+3^2} $$
so we just need to solve $|1-c|=25$ to find the values of $c$ corresponding to our two lines.
