Given the line $y = \dfrac{3}{4}x + 6$ and a line $L$ parallel to the given line and $4$ units from it. A possible equation for $L$ is:
$\textbf{(A)}\ y =\frac{3}{4}x+1\qquad \textbf{(B)}\ y =\frac{3}{4}x\qquad \textbf{(C)}\ y =\frac{3}{4}x-\frac{2}{3}\qquad  \\ \textbf{(D)}\ y = \dfrac{3}{4}x -1 \qquad \textbf{(E)}\ y = \dfrac{3}{4}x + 2$
If a point on line $L$ is four units away from another line, the perpendicular distance between the two lines is $4$.  Since a line perpendicular to the original has a slope of $-\tfrac{4}{3}$, a point on that line is $5d$ units away and can be derived from subtracting $4d$ from the y-coordinate and adding $3d$ to the x-coordinate of a point on the original line.
We want the perpendicular distance to be $4$, so $d = \tfrac{4}{5}$.  Since one point on the original line is $(0,6)$, a point on line $L$ will be $(0+3 \cdot \tfrac{4}{5},6 - 4 \cdot \tfrac{4}{5})$, which is simplified as $(\tfrac{12}{5},\tfrac{14}{5})$.  Using point-slope form, the equation of line $L$ is
\[y - \frac{14}{5} = \frac{3}{4} (x - \frac{12}{5})\]
\[y = \frac{3}{4} x + 1\]
The answer is $\boxed{\textbf{(A)}}$.