The distance between the line and the point is:
$$ d = \frac{ 12 \sqrt{ 61}}{ 61 } $$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
$$\begin{aligned} y & = - \frac{ 5 }{ 6 } x - 1 \\6 \cdot y &= 6 \cdot \left( - \frac{ 5 }{ 6 } x - 1 \right) \\6 \cdot y &= - 5 x - 6 \\5x+6y+6&=0\end{aligned}$$After substituting: $ A = 5 $ , $ B = 6 $ , $ C = 6 $ , $ x_0 = 6 $ and $ y_0 = -4 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 5\cdot6 +6\cdot\left(-4\right) + 6 \right| }{\sqrt{ 5^2 + 6^2}} = \\ d =& \frac{ \left| 30 -24 + 6 \right| }{\sqrt{ 25 + 36}} = \\ d =& \frac{ \left| 12 \right| }{\sqrt{ 61}} = \\ d =& \frac{ 12 }{ \sqrt{ 61 } } = \\ d =& \frac{ 12 \sqrt{ 61}}{ 61 } \end{aligned} $$