The distance between the line and the point is:
$$ d = 2 \sqrt{ 5 } $$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
$$\begin{aligned} y & = 2 x + 3 \\2x-y+3&=0\end{aligned}$$After substituting: $ A = 2 $ , $ B = -1 $ , $ C = 3 $ , $ x_0 = 6 $ and $ y_0 = 5 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 2\cdot6 +\left(-1\right)\cdot5 + 3 \right| }{\sqrt{ 2^2 + (-1)^2}} = \\ d =& \frac{ \left| 12 -5 + 3 \right| }{\sqrt{ 4 + 1}} = \\ d =& \frac{ \left| 10 \right| }{\sqrt{ 5}} = \\ d =& \frac{ 10 }{ \sqrt{ 5 } } = \\ d =& 2 \sqrt{ 5 } \end{aligned} $$