The distance between the line and the point is:
$$ d = \sqrt{ 5 } $$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
$$\begin{aligned} y & = 2 x + 4 \\2x-y+4&=0\end{aligned}$$After substituting: $ A = 2 $ , $ B = -1 $ , $ C = 4 $ , $ x_0 = 2 $ and $ y_0 = 3 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 2\cdot2 +\left(-1\right)\cdot3 + 4 \right| }{\sqrt{ 2^2 + (-1)^2}} = \\ d =& \frac{ \left| 4 -3 + 4 \right| }{\sqrt{ 4 + 1}} = \\ d =& \frac{ \left| 5 \right| }{\sqrt{ 5}} = \\ d =& \frac{ 5 }{ \sqrt{ 5 } } = \\ d =& \sqrt{ 5 } \end{aligned} $$