Solution:

The bisector points along the sum of the unit vectors of the two lines: $\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}.$ Its slope is $$k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}.$$

Multiplying numerator and denominator by $\sqrt5 - 1$ gives $\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4} = \dfrac{2 + 2\sqrt5}{4} = \dfrac{1 + \sqrt5}{2}.$

Thus, the correct answer is A.