Answer :
[tex]A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\----------------\\\sqrt{(1-x)^2+(5-2)^2}=13\\(1-x)^2+3^2=169\\(1-x)^2=169-9\\(1-x)^2=160\\1-x=\pm\sqrt{160}\\-x=\pm\sqrt{16\times10}-1\\-x=\pm4\sqrt{10}-1\\\boxed{x=1\pm4\sqrt{10}}[/tex]
[tex]\sqrt{(1-x)^2+(5-2)^2}=\sqrt{13}\\(1-x)^2+3^2=13\\(1-x)^2=13-9\\(1-x)^2=4\\1-x=\pm\sqrt4\\-x=\pm2-1\\-x=-2-1\ or\ -x=2-1\\-x=-3\ or\ -x=1\\\boxed{x=3\ or\ x=-1}[/tex]
[tex]\sqrt{(1-x)^2+(5-2)^2}=\sqrt{13}\\(1-x)^2+3^2=13\\(1-x)^2=13-9\\(1-x)^2=4\\1-x=\pm\sqrt4\\-x=\pm2-1\\-x=-2-1\ or\ -x=2-1\\-x=-3\ or\ -x=1\\\boxed{x=3\ or\ x=-1}[/tex]
