The distance between the line and the point is:
$$ d = 5 $$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
$$\begin{aligned} y & = \frac{ 4 }{ 3 } x + 3 \\3 \cdot y &= 3 \cdot \left( \frac{ 4 }{ 3 } x + 3 \right) \\3 \cdot y &= 4 x + 9 \\4x-3y+9&=0\end{aligned}$$After substituting: $ A = 4 $ , $ B = -3 $ , $ C = 9 $ , $ x_0 = 4 $ and $ y_0 = 0 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 4\cdot4 +\left(-3\right)\cdot0 + 9 \right| }{\sqrt{ 4^2 + (-3)^2}} = \\ d =& \frac{ \left| 16 + 0 + 9 \right| }{\sqrt{ 16 + 9}} = \\ d =& \frac{ \left| 25 \right| }{\sqrt{ 25}} = \\ d =& \frac{ 25 }{ 5 } = \\ d =& 5 \end{aligned} $$