The distance between the lines $3x–4y+9=0$ and $6x–8y–15=0$  is

$ax+by+c_1=0$  ,$2\left(3x+4y-\frac{15}{2}\right)=0$

                           $3x-4y-\frac{15}{2}=0$

                           $ax+by+c_2=0$

Here $c_1=9,c_2=\frac{-15}{2},a=3,b=-4$

Distance between $3x-4y+9=0$ , $6x-8y-15=0$

$=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$

$=\left|\frac{\frac{-15}{2}-9}{\sqrt{3^2+{\left(-4\right)}^2}}\right|$

$=\left|\frac{-15-18}{2\times 5}\right|$

$=\left|\frac{-33}{10}\right|$

$=\frac{33}{10}$