The distance between the parallel lines $3\mathrm{x}-4\mathrm{y}+9=0$  and $6x-8y-15=0$ ,is

Putting  $\mathrm{y}=0$ in $3\mathrm{x}-4\mathrm{y}+9=0$ , we get  $\mathrm{x}=-3.$
Thus(-3, 0)  is a point on the line $3\mathrm{x}-4\mathrm{y}+9=0$ .
Now, length of the $\perp$  from $\left(-3,0\right)$  to   $6x-8y-15=0$
 $=\frac{\left|6\left(-3\right)-8\left(0\right)-15\right|}{\sqrt{6^2+{\left(-4\right)}^2}}=\frac{33}{10}$