The distance between the line and the point is:
$$ d = \dfrac{ 6 \sqrt{ 5}}{ 5 } $$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
$$\begin{aligned} y & = \frac{ 1 }{ 2 } x + 1 \\2 \cdot y &= 2 \cdot \left( \frac{ 1 }{ 2 } x + 1 \right) \\2 \cdot y &= x + 2 \\x-2y+2&=0\end{aligned}$$After substituting: $ A = 1 $ , $ B = -2 $ , $ C = 2 $ , $ x_0 = -2 $ and $ y_0 = 3 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 1\cdot\left(-2\right) +\left(-2\right)\cdot3 + 2 \right| }{\sqrt{ 1^2 + (-2)^2}} = \\ d =& \frac{ \left| -2 -6 + 2 \right| }{\sqrt{ 1 + 4}} = \\ d =& \frac{ \left| -6 \right| }{\sqrt{ 5}} = \\ d =& \frac{ 6 }{ \sqrt{ 5 } } = \\ d =& \frac{ 6 \sqrt{ 5}}{ 5 } \end{aligned} $$