Find the equations of the two straight lines which are parallel to the line $12 x + 5 y + 2 = 0$ and at an unit distance from the point $\left(1, -1\right)$.
Equation of the given line is: $\hspace{1em}$ $12x + 5y + 2 = 0$
i.e. $y = -\dfrac{12}{5}x - \dfrac{2}{5}$
$\therefore$ $\;$ Slope of the given line is $= m = - \dfrac{12}{5}$
Since the required line is parallel to the given line,
$\therefore$ $\;$ Slope of the required line $= m = - \dfrac{12}{5}$ $\;\;\; \cdots \; (1)$
Let the equation of the required line be $\hspace{1em}$ $y = mx + c$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ In view of equation $(1)$, equation $(2)$ becomes
$y = - \dfrac{12}{5} x + c$
i.e. $5y = -12 x + 5 c$ $\implies$ $12x + 5y - 5c = 0$ $\;\;\; \cdots \; (2a)$
Distance of a line $ax + by + c = 0$ from a point $\left(x_1, y_1\right)$ is $d = \left|\dfrac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\right|$
Given: $\left(x_1 , y_1\right) = \left(1, -1\right)$
and the distance of the line from the point $= d = 1$
$\therefore$ $\;$ We have for the required line
$1 = \left|\dfrac{12 \times 1 + 5 \times \left(-1\right) - 5 c}{\sqrt{12^2 + 5^2}}\right|$
i.e. $\pm 1 = \dfrac{7 - 5 c}{13}$
$\implies$ $13 = 7 - 5c$ $\;\;\;$ i.e. $- 5 c = 6$ $\implies$ $c = - \dfrac{6}{5}$ $\;\;\; \cdots \; (3a)$
or, $-13 = 7 - 5c$ $\;\;\;$ i.e. $- 5 c = -20$ $\implies$ $c = 4$ $\;\;\; \cdots \; (3b)$
$\therefore$ $\;$ In view of equation $(3a)$ equation $(2a)$ becomes
$12 x + 5 y - 5 \times \left(- \dfrac{6}{5}\right) = 0$ $\implies$ $12 x + 5 y + 6 = 0$
and in view of equation $(3b)$ equation $(2a)$ becomes
$12 x + 5 y - 5 \times 4 = 0$ $\implies$ $12 x + 5y - 20 = 0$
Therefore, the equations of the required lines are: $\hspace{1em}$ $12x + 5y + 6 = 0$ and $12 x + 5y - 20 = 0$