How does one go about finding the slopes of the lines making a given angle with a line of a given slope ? Say, we have a line of slope m and are said to find the slopes of lines making angle of $\theta$ with said line. How does one find the slopes?
PS: It is obvious that there must be two lines. But my question is how can one find out both slopes from a single concise formula? Derivation of said formula will be appreciated.
You have that $m=\tan \alpha$ where $\alpha$ is the angle that forms the line with the $OX$-axis. Now, if we have a line that forms an angle $\theta$ with the given line then its slope is $\tan(\alpha+\theta)$ or $\tan(\alpha-\theta).$ In the first case we have:
$$\tan(\alpha+\theta)=\dfrac{\tan\alpha+\tan\theta}{1-\tan \alpha\tan\theta}=\dfrac{m+\tan\theta}{1-m\tan\theta}.$$ In the second case
$$\tan(\alpha-\theta)=\dfrac{\tan\alpha-\tan\theta}{1+\tan \alpha\tan\theta}=\dfrac{m-\tan\theta}{1+m\tan\theta}.$$
Edit
If we call $m'=\tan (\alpha+\theta)$ then we have $$\tan\theta=\dfrac{m'-m}{1+mm'}.$$ If we call $m'=\tan (\alpha-\theta)$ then we have $$\tan\theta=\dfrac{m-m'}{1+mm'}.$$ So, you have, $$\tan\theta=\left|\dfrac{m-m'}{1+mm'}\right|$$ When you solve this you get the values above $$\dfrac{m+\tan\theta}{1-m\tan\theta},\quad\dfrac{m-\tan\theta}{1+m\tan\theta}.$$