Correct. The slope of $\displaystyle{y}=\frac{{1}}{{3}}{x}-{2}\usepackage{cmbright}\Huge\$ is $\displaystyle\frac{{1}}{{3}}\usepackage{cmbright}\Huge\$ , so any line perpendicular to it will have slope -3 (the opposite reciprocal of $\displaystyle\frac{{1}}{{3}}\usepackage{cmbright}\Huge\$ ). The equation $\displaystyle{y}-{5}=-{3}{\left({x}+{1}\right)}\usepackage{cmbright}\Huge\$ is the point slope form of the line with slope -3 that goes through (-1, 5).