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 $\mathrm{C}$).
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 $$R = \begin{pmatrix}cos &amp; sin\\ -sin &amp; cos\end{pmatrix};$$ 
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.
Assuming your triangle is oriented so that the adjacent side is on the $x$-axis we have that slope is $\frac{rise}{run} = \frac{opposite}{adjacent}$