Найдём координаты вектора \(\overrightarrow {AB} \): \(\overrightarrow {AB} = \left( {x-4;1-6} \right) = \left( {x-4;-5} \right).\)
Длина вектора \(\vec d\left( {x;y} \right)\) равна: \(\left| {\vec d} \right| = \sqrt {{x^2} + {y^2}} \).
Тогда: \(\left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( {x-4} \right)}^2} + {{\left( {-5} \right)}^2}} = \sqrt {{{\left( {x-4} \right)}^2} + 25} .\)
Следовательно: \(\sqrt {{{\left( {x-4} \right)}^2} + 25} = 13\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{\left( {x-4} \right)^2} + 25 = 169\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\left( {x-4} \right)^2} = 144\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-4 = 12,}\\{x-4 = -12}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 16,}\\{x = -8.}\end{array}} \right.} \right.\)
Так как \(x > 0\), то \(x = 16\).
Ответ: 16.