The distance between the line and the point is:
$$ d = \frac{ 25 \sqrt{ 13}}{ 13 } $$The distance from the point $ (x_0, y_0) $ to the line $ Ax + By + C = 0 $ is given by:
$$ d = \frac{ \left| Ax_0 + B_yo + C \right| }{ \sqrt{A^2 + B^2}} $$To apply this formula, we first need to express the line in standard form
$$\begin{aligned} y & = \frac{ 2 }{ 3 } x + 4 \\3 \cdot y &= 3 \cdot \left( \frac{ 2 }{ 3 } x + 4 \right) \\3 \cdot y &= 2 x + 12 \\2x-3y+12&=0\end{aligned}$$After substituting: $ A = 2 $ , $ B = -3 $ , $ C = 12 $ , $ x_0 = 8 $ and $ y_0 = 1 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 2\cdot8 +\left(-3\right)\cdot1 + 12 \right| }{\sqrt{ 2^2 + (-3)^2}} = \\ d =& \frac{ \left| 16 -3 + 12 \right| }{\sqrt{ 4 + 9}} = \\ d =& \frac{ \left| 25 \right| }{\sqrt{ 13}} = \\ d =& \frac{ 25 }{ \sqrt{ 13 } } = \\ d =& \frac{ 25 \sqrt{ 13}}{ 13 } \end{aligned} $$