Ta có: $\frac{6}{3}$ = $\frac{8}{4}$ $\neq$ $\frac{-13}{-27}$ $\Rightarrow$ $\Delta$ // $\Delta'$
Lấy điểm A(0; $\frac{13}{8}$) $\in$ $\Delta$.
Ta có: d($\Delta$, $\Delta'$) = d(A; $\Delta'$) = $\frac{|4.\frac{13}{8}-27|}{\sqrt{3^{2} + 4^{2}}}$ = $\frac{41}{10}$