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 $M$?
Best Answer
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 $$\frac{y-k}{x-h}=\tan 2\theta\iff y=\frac{2m}{1-m^2}(x-h)+k\text{ for positive inclination }$$