Correct. Applying the distance formula, you find $\displaystyle{13}=\sqrt{{{{\left({1}-{6}\right)}}^{{2}}+{{\left({y}-{4}\right)}}^{{2}}}}=\sqrt{{{25}+{{\left({y}-{4}\right)}}^{{2}}}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{13}}^{{2}}={{\left(\sqrt{{{25}+{{\left({y}-{4}\right)}}^{{2}}}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . Solving this equation for y, you find $\displaystyle{169}={25}+{{\left({y}-{4}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{144}={{\left({y}-{4}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . Take the square root of both sides: $\displaystyle{12}={y}-{4}\usepackage{cmbright}\Huge\$ or $\displaystyle{12}=-{\left({y}-{4}\right)}\usepackage{cmbright}\Huge\$ . If $\displaystyle{y}-{4}={12}\usepackage{cmbright}\Huge\$ , then y = 16 and if $\displaystyle-{\left({y}-{4}\right)}={12}\usepackage{cmbright}\Huge\$ , then $\displaystyle{y}-{4}=-{12}\usepackage{cmbright}\Huge\$ , so y = -8. Since both of these answers give points (1, y) that are 13 units from (6,4), y can be either 16 or -8.