Answer
$r\displaystyle \cos(\theta-\frac{\pi}{2})=-5$
or
$r\displaystyle \cos(\theta+\frac{\pi}{2})=5$
Work Step by Step
Conversion formulas
$\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
Using the conversion formula, $ y=r\sin\theta$, we get:
$r\sin\theta=-5$
But we want one of the form with cosine.
Apply the symmetry identity about $\displaystyle \frac{\pi}{2},\ \displaystyle \quad \cos(\frac{\pi}{2}-\theta)=\sin\theta$
and the fact that cosine is an even function $\displaystyle \cos[-(\frac{\pi}{2}-\theta)]=\cos(\theta-\frac{\pi}{2})$
So, a parametric equation for the line can be
$r\displaystyle \cos(\theta-\frac{\pi}{2})=-5$
Note: $\displaystyle \cos(\alpha+\frac{\pi}{2})=-\sin\alpha$
so a paramateric equation could also be
$r\displaystyle \cos(\theta+\frac{\pi}{2})=5$
