Ta có: $\frac{3}{6}$ = $\frac{4}{8}$ $\neq$ $\frac{-10}{-1}$ $\Rightarrow$ $\Delta$ // $\Delta'$
Lấy điểm M(2; 1) $\in$ $\Delta$
$\Rightarrow$ d($\Delta$; $\Delta'$) = d(M; $\Delta'$) = $\frac{|6. 2 + 8. 1-1|}{\sqrt{6^{2} + 8^{2}}}$ = $\frac{19}{10}$