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\(\text{The line has slope }m=\dfrac 1 3 \text{ and lies in the I and III quadrants}\\ \text{As }\sin(\theta)<0,~\theta \text{ must lie in the III quadrant and thus }\cos(\theta) < 0\\ \text{the III quadrant point }(-3,-1) \text{ lies on the line and from this we know that }\\ \cos(\theta) = \dfrac{-3}{\sqrt{(-3)^2+(-1)^2}} = -\dfrac{3}{\sqrt{10}}=-\dfrac{3\sqrt{10}}{10}\)

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