Q:

How do you find the slope of a secant line on a non-smooth curve?

My background is engineering, not mathematics, so please forgive me if my terminology is not right. On a continuous curve a secant line is one formed by connecting two points on the curve. I'm trying to determine the angle of that line on a non-smooth curve. Think the angle of a bicycle that starts off on a level surface and then gets to a sudden linear down-slope. I'm trying to model the angle of the bicycle as it goes over the slope.
For a regular secant the slope is just 
$$    
m = \frac{y_2-y_1}{x_2-x_1}
$$
My problem is that x2 depends on the angle of rotation, so I'm not sure what to do.
If we supposed the function
$$    
f(x) = \begin{cases}
x/2, & x <   0\\
0    & x \ge 0
\end{cases}
$$
How would we construct the function?
EDIT: I'm sorry. I didn't note that I know the value of x for one of the points, but I only know the distance between the points (the length of the bike) not the x value of the second point.
(Also thank you gt6989b for formatting the math.)
Edit 2: Added image of the problem I'm trying to solve. I know (x1, y1) and I know L, but I can't figure out how to find (x2, y2)

A:

As a hint, you framed the problem incorrectly, the question if your curve is smooth doesn't really matter, you would have exactly the same problem if your whole 'surface' was $y=x^2$, for example.
In your example, if you know $(x_1,y_1)$ , you are looking for a point $(x_2,y_2)$ with $x_2 \ge 0, y_2=0$ and the additional condition that
$$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}=L.$$
Putting the information we have into this equation, we get
$$\sqrt{(x_1-x_2)^2 + y_1^2}=L,$$
which we can transform into
$$(x_1-x_2)^2 + y_1^2=L^2,$$
then
$$(x_1-x_2)^2 =L^2-y_1^2,$$
further to
$$x_1-x_2 =\pm\sqrt{L^2-y_1^2},$$
and finally to
$$x_2=x_1\pm\sqrt{L^2-y_1^2}.$$
Since $x_1$ is negative, $x_1-\sqrt{L^2-y_1^2} < 0$, so this solution is not valid.
You need to check the remaining solution
$$x_2=x_1+\sqrt{L^2-y_1^2}.$$
If $L^2-y_1^2<0$, the solution is not real so unusable. It correspondes to the point $(x_1,y_1)$ being 'too far away' from the origin, so distance $L$ is not 'reaching' to the positive $x$-axis. In addition, that $x_2$ might be negative, which is another manifestation of $(x_1,y_1)$ being 'too far away' from the origin.
If $x_2$ as calculated is positive, then you have found your point $(x_2,y_2=0)$ and can calculate the slope between $(x_1,y_1)$ and $(x_2,y_2)$ by the usual formula
$$m =\frac{y_2-y_1}{x_2-x_1}$$.

