What is the slope of the adjacent/opposite in a right triangle? Suppose I have the lengths of the adjacent, opposite, and hypotenuse of a right triangle (in this instance $a, b$ and $c$ respectively), and coordinates of point $\mathrm{B}$ and $\mathrm{C}$ ($\mathrm{A}$ is unknown).  My question is, how do I calculate the slope of the opppsite? (the slope made with points $\mathrm{A}$ and $\mathrm{C}$). 

 A: 2 options: 


*

*to simply get the slope you can take the slope of $CB$, convert it to angle using $\tan^{-1}$ (you'd also have to add $\pi$ if $B$ is to the left of $C$), get the angle $ACB$ using $\tan^{-1}$, then add those angle together and take $\tan$ of sum. This can be simplified by using formula for tangent of sum: 
$$\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$
Long story short, if $s_1$ is the slope of $CB$ and $s_2 = \frac{a}{b}$, then the slope of $CA$ is simply
$$\frac{s_1+s_2}{1-s_1s_2}.$$

*If you want to find the coordinates of the point $A$, you may want to use Rational parametrization of a circle. Given $M = \frac{1}{2}(B+C)$, $v = B -M$; $cos = \frac{b^2-a^2}{c^2}$, $sin = \frac{2ab}{c^2}$ and finally $$R = \begin{pmatrix}cos & sin\\ -sin & cos\end{pmatrix};$$ 
you get $A = M + Rv$.


Well, if you do not know how to work with matrices and vectors, you may want to ignore the 2nd option.
A: Assuming your triangle is oriented so that the adjacent side is on the $x$-axis we have that slope is $\frac{rise}{run} = \frac{opposite}{adjacent}$
