well, 2/5 of the distance AB is simply their product, thus 2/5 of AB is just (2/5)AB.
[tex]\bf \stackrel{A}{-7}\stackrel{\textit{7 units}}{\rule[0.35em]{17em}{0.25pt}}0\stackrel{\textit{3 units}}{\rule[0.35em]{10em}{0.25pt}}\stackrel{B}{3} \\\\\\ AB=7+3\implies AB=10~\hspace{7em}\cfrac{2}{5}AB\implies \cfrac{2}{5}(10)\implies 4 \\\\\\ -7\rule[0.35em]{2em}{0.25pt}-6\rule[0.35em]{2em}{0.25pt}-5\rule[0.35em]{2em}{0.25pt}-4\rule[0.35em]{2em}{0.25pt}\boxed{-3}[/tex]
