Find the equations to the straight lines inclined at an angle $\tan^{-1}(m)$

Question

Find the equations to the straight lines which pass through the point $(h, k)$ and are inclined at an angle $\tan^{-1}(m)$ to the straight line $y = mx + c$.

I'm getting $\tan^{-1}(m)=\frac{M-m}{1+mM}$ or $\tan^{-1}(m)=-\frac{M-m}{1+mM}$, where $M$(say) is the slope of the required lines. However the value of $M$ seems to be in a very complicated form. Is there any simple way to express $M$?

Answer 1

You have two ways to understand "are inclined" (positive and negative angles). This gives two solutions.

Let $\theta$ such that $\tan \theta=m$. The two solutions are $$y=k\text{ for negative inclination }$$ and $$\frac{y-k}{x-h}=\tan 2\theta\iff y=\frac{2m}{1-m^2}(x-h)+k\text{ for positive inclination }$$

Answer 2

Let $m=\tan \theta$, then the line will be \begin{align*} y-k &= (x-h)\tan 2\theta \\ (1-m^2)(y-k) &= 2m(x-h) \end{align*}

provided "your inclination" is in anti-clockwise sense.

If it's just concerned with the angle $\tan^{-1} m$ made, the other line is

$$y-k=0$$

Note that in general, $\theta=\tan^{-1} m+180n^{\circ}$