# Using distance formula to find slope, any reason to use the concluding equation?

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.

• I think it's even numerically worse because of the squaring and the square root operations.
– user2468
Aug 28, 2012 at 2:16
• It tells you that if the slope is very large, then $d/(x_2-x_1)$ is approximately the slope. Aug 28, 2012 at 2:21
• If you change notation a bit, it is interesting and familiar. Let $d=1$ (we are on a circle), and let $\frac{y_2-y_1}{x_2-x_1}=\tan\theta$. Then we are looking at the fact that $\tan\theta=\sqrt{\sec^2\theta-1}$. Aug 28, 2012 at 2:22
• @JosephSkelton there's nothing wrong with asking! I was making a remark about something I noticed while thinking of pros/cons.
– user2468
Aug 28, 2012 at 2:24
• I think having the slope in terms of the height difference and the distance as measured along the slope would be most interesting. Imagine you are assessing the mean climb percentages in the Tour de France, what is most easy to measure? Distance traveled and height differential. Not horizontal distance travelled. Aug 28, 2012 at 7:35

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 $\frac{3}{4}$.