The distance between the lines $$\frac{x}{3} + \frac{y}{4} = 1$$ and 8x + 6y = 5 in units, is

we know the fourmula of distance between two line 

$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$ = $$\pm$$ $$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$

then we have here  $$(A_{1}$$ = 1\3  $$(B_{1}$$ = 1\4 and $$(c_{1}$$ = -1 and 

                             $$(A_{2}$$ = 8 $$(B_{2}$$ = 8  and $$(c_{2}$$ = -5 

then 

|$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$| = $$\pm$$ |$$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$|

|$$\frac{1\3 + 1\4 - 1}\sqrt{{1\3}^{2}+{1\4}^{2}}$$| = $$\pm$$ |$$\frac{8 +6 -5}\sqrt{{8}^{2}+{6}^{2}}$$|

1 + 9\10 = 10\19  Answer