Example 1.18. Form the equations of lines parallel to the line $3 x+4 y-1=0(l)$ and at a distance of 1 from it.
Solution. We will seek the equation of each line in the form $y=k x+b$. Since the desired line is parallel to the line $l$, its slope $k=-\frac{3}{4}$, and therefore, its equation takes the form:

$$
y=-\frac{3}{4} x+b
$$

or

$$
3 x+4 y-4 b=0
$$

To find the parameter $b$, we will use the fact that the distance from any point on the line $l$, for example, from the point $A(3 ;-2)$ to the line $(*)$ according to the condition is 1. However, this distance can also be calculated directly. For this, we will write the equation of the line $h$ drawn from point $A$ perpendicular to the line $l$:

$$
y+2=\frac{4}{3}(x-3) \text { or } 4 x-3 y-18=0 \text {. }
$$

Solving the system of equations of the lines $h$ and $l$

$$
\begin{aligned}
& 4 x-3 y-18=0 \\
& 3 x+4 y-4 b=0
\end{aligned}
$$

we will find the coordinates of the point $B$ of their intersection:

$$
x_{B}=\frac{72+12 b}{25}, \quad y_{B}=\frac{16 b-54}{25}
$$

Then the desired distance is equal to the length of the segment $A B$:

$$
\begin{aligned}
& A B=\sqrt{\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}}= \\
& =\sqrt{\left(\frac{72+12 b}{25}-3\right)^{2}+\left(\frac{16 b-54}{25}+2\right)^{2}}= \\
& =\frac{\sqrt{(12 b-3)^{2}+(16 b-4)^{2}}}{25}
\end{aligned}
$$

Equating this expression to one, we get an equation in terms of $b$:

$$
\frac{\sqrt{(12 b-3)^{2}+(16 b-4)^{2}}}{25}=1
$$

or

$$
(12 b-3)^{2}+(16 b-4)^{2}=25^{2}
$$

and finally

$$
2 b^{2}-b-3=0
$$

The solutions to this equation are: $b_{1}=\frac{3}{2}, b_{2}=-1$. Substituting the obtained values of $b$ into the equation $(*)$, we write the equations of the desired lines:

$$
3 x+4 y-6=0 \quad \text { and } \quad 3 x+4 y+4=0 .
$$