[tex]\bf \begin{cases} A=(4,0)\\ B=(-6,10) \end{cases}\qquad B-A\implies (-6-4~~,~~10-0)\implies \stackrel{\textit{component form}}{(-10,10)=AB} \\\\[-0.35em] ~\dotfill\\\\ ||AB||\implies \sqrt{(-10)^2+(10)^2}\implies \sqrt{100+100}\implies \sqrt{2(100)} \\\\\\ \sqrt{2(10^2)}\implies 10\sqrt{2} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{\textit{unit vector}}{\cfrac{AB}{||AB||}}\implies \cfrac{(-10,10)}{10\sqrt{2}}\implies \left( \cfrac{\stackrel{-1}{~~\begin{matrix} -10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{2}}~,~\cfrac{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{2}} \right) \implies \left(\cfrac{-1}{\sqrt{2}}~,~\cfrac{1}{\sqrt{2}} \right)[/tex]
[tex]\bf \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{and rationalizing the denominator}}{\left( -\cfrac{\sqrt{2}}{2}~,~\cfrac{\sqrt{2}}{2} \right)}~\hfill[/tex]
