smc

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 $-\frac{4}{3}$, 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 $d=\frac{4}{5}$. Since one point on the original line is $(0,6)$, a point on line $L$ will be $(0+3\cdot \frac{4}{5},6-4\cdot \frac{4}{5})$, which is simplified as $(\frac{12}{5},\frac{14}{5})$. 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 $\boxed{y=\frac{3}{4}x+1}$.