The distance between the line and the point is:
$$ d = \frac{ 4 \sqrt{ 5}}{ 5 } $$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
$$\begin{aligned} y & = - \frac{ 1 }{ 2 } x + \frac{ 3 }{ 2 } \\2 \cdot y &= 2 \cdot \left( - \frac{ 1 }{ 2 } x + \frac{ 3 }{ 2 } \right) \\2 \cdot y &= - x + 3 \\x+2y-3&=0\end{aligned}$$After substituting: $ A = 1 $ , $ B = 2 $ , $ C = -3 $ , $ x_0 = 1 $ and $ y_0 = -1 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 1\cdot1 +2\cdot\left(-1\right) + \left( -3\right) \right| }{\sqrt{ 1^2 + 2^2}} = \\ d =& \frac{ \left| 1 -2 -3 \right| }{\sqrt{ 1 + 4}} = \\ d =& \frac{ \left| -4 \right| }{\sqrt{ 5}} = \\ d =& \frac{ 4 }{ \sqrt{ 5 } } = \\ d =& \frac{ 4 \sqrt{ 5}}{ 5 } \end{aligned} $$