Perpendicular lines have slopes that are negative reciprocals. First, we write the given equation in slope intercept form by solving for y:

x-3y=-18

-3y=-x-18

\(\displaystyle{y}=\frac{{-{x}-{18}}}{{-{{3}}}}\)

\(\displaystyle{y}=\frac{{1}}{{3}}{x}+{6}\)

So, the slope of the given line is \(\displaystyle\frac{{1}}{{3}}\) which means that the slope of the line perpendicular to it is:

\(\displaystyle-\frac{{\frac{{1}}{{1}}}}{{3}}=-{3}\)

x-3y=-18

-3y=-x-18

\(\displaystyle{y}=\frac{{-{x}-{18}}}{{-{{3}}}}\)

\(\displaystyle{y}=\frac{{1}}{{3}}{x}+{6}\)

So, the slope of the given line is \(\displaystyle\frac{{1}}{{3}}\) which means that the slope of the line perpendicular to it is:

\(\displaystyle-\frac{{\frac{{1}}{{1}}}}{{3}}=-{3}\)