Find the slope of a line given a point and an angle

$\tan \left( \tan^{-1}\left(\frac{y}{x}\right) - \theta\right)$ is the slope $m$.

Then use "point slope formula" (if you want an equation of the line, that is...)

$y-y_1 = m(x - x_1)$

---

For variety, I'll explain.

Labeling the origin "$O$" and the point $(x, y)$ "$P$", the segment $\overline{OP}$ makes an angle of $\tan^{-1} \left(\frac{y}{x}\right)$ with the positive x-axis. But this is the sum of $\theta$ and the angle $\phi$ that your line makes with the positive x-axis (since we have opposite interior angles).

So $\tan^{-1}\left(\frac{y}{x}\right) - \theta = \phi$.

Finally, $\tan \phi = m$.

Another method: If you know the exterior angle theorem, you know that The exterior angle is the sum of remote interior angles thus:

$ \tan^{-1}\frac{y}{x} = \theta + $ unknown angle

thus,

$ \tan^{-1}\frac{y}{x} - \theta = $ unknown angle

$ \tan(\tan^{-1}\frac{y}{x} - \theta )= \tan( $unknown angle) $$ \tan(\tan^{-1}\frac{y}{x} - \theta )= Slope$$