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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.

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I think it's even numerically worse because of the squaring and the square root operations. –  user2468 Aug 28 '12 at 2:16
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It tells you that if the slope is very large, then $d/(x_2-x_1)$ is approximately the slope. –  Ragib Zaman Aug 28 '12 at 2:21
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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 $\tan\theta=\sqrt{\sec^2\theta-1}$. –  André Nicolas Aug 28 '12 at 2:22
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@JosephSkelton there's nothing wrong with asking! I was making a remark about something I noticed while thinking of pros/cons. –  user2468 Aug 28 '12 at 2:24
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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. –  Raskolnikov Aug 28 '12 at 7:35

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