TITLE: Using distance formula to find slope, any reason to use the concluding equation?
QUESTION [12 upvotes]: So, today I was observing a class that I will be a TA for this semester and the professor started to talk about the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Well, my mind wandered a little and I started to think about slope. That's when I noticed, with a little bit of algebra we can convert the distance formula into a representation of slope.
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
$$d^2=(x_2-x_1)^2+(y_2-y_1)^2$$
$$\left(\frac{d}{x_2-x_1}\right)^2=1+\left(\frac{y_2-y_1}{x_2-x_1}\right)^2$$
$$\frac{y_2-y_1}{x_2-x_1}=\sqrt{\left(\frac{d}{x_2-x_1}\right)^2-1}$$
I was wondering if anyone knows of any practical reason to use this, or if it's utterly pointless. My first impression is that it's pointless, unless you are given distance and two $x$ values and asked to find slope. But excluding that very unlikely case, I cannot think of a reason.

REPLY [3 votes]: It could be useful if you had the length of the hypotenuse (d or the length of the line segment) and the distance that the segment covers on the x-axis $(x_2 - x_1)$ and you wanted to find the positive  slope.
Ex: You have a segment that is 5 units long and covers 4 units on the x-axis. Using your equation, the positive slope of the line would be $\frac{3}{4}$.