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 April 11th, 2017, 03:28 PM #1 Senior Member   Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10 Why do we do y1-y2 x1-x2 Was wondering why we don't do the opposite x1-×2 y1-y2? Or does it not matter? Would it lead to undefined answers when they wouldn't be over wise? Thanks. Last edited by skipjack; April 12th, 2017 at 05:47 AM.
 April 11th, 2017, 03:42 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716 wat?
 April 11th, 2017, 03:45 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,477 Thanks: 693 2 * 3 = 6 ....hang on...and..and..and 3 * 2 = 6 !!!!!!!!!!
April 11th, 2017, 05:41 PM   #4
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Quote:
 Why do we do y1-y2 x1-x2
is that an attempt to write the formula for slope?

$m = \dfrac{y_1-y_2}{x_1-x_2}$

the standard definition for slope is $\dfrac{\Delta y}{\Delta x}$ ... that's why.

 April 11th, 2017, 09:16 PM #5 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1386 $\dfrac{y_1-y_2}{x_1-x_2}$ and $\dfrac{y_2-y_1}{x_2-x_1}$ are equivalent. Their reciprocals give the slope of $x$ with respect to $y$. Thanks from GIjoefan1976 Last edited by skipjack; April 12th, 2017 at 05:44 AM.
 April 12th, 2017, 01:28 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,049 Thanks: 680 Math Focus: Physics, mathematical modelling, numerical and computational solutions If you calculated $\displaystyle \frac{x_2 - x_1}{y_2 - y_1}$, you'd just calculate something different... If you wanted some sort of measure which increases the more horizontal a line or curve is (i.e. instead of "steepness" you want "shallowness"... a bigger number means a more shallow line/curve) then I suppose $\displaystyle \frac{x_2 - x_1}{y_2 - y_1}$ could be useful. Otherwise, I wouldn't bother with it.
 April 12th, 2017, 09:41 AM #7 Senior Member   Joined: May 2016 From: USA Posts: 785 Thanks: 312 It is conventional to show y as the dependent variable and x as the independent variable. That is why the conventional notation is y = some formula in x. It is frequently important that we know how the dependent variable (conventionally y) will change in response to a change in the independent variable (conventionally x). Now we could measure the ratio between the change in y and the change in x as $\dfrac{\Delta x}{\Delta y}$ or $\dfrac{\Delta y}{\Delta x}$ where $\Delta$ means change in. It is convenient to measure the ratio as $\dfrac{\Delta y}{\Delta x}$ because that gives a simpler formula for linear equations, namely $y = b + \dfrac{\Delta y}{\Delta x}$ instead of $y = b + \dfrac{x}{\dfrac{\Delta x}{\Delta y}}.$ Everyone prefers to multiply than to divide. So the convention is $\dfrac{\Delta y}{\Delta x}.$ Thanks from GIjoefan1976
April 12th, 2017, 12:44 PM   #8
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 Originally Posted by JeffM1 It is conventional to show y as the dependent variable and x as the independent variable. That is why the conventional notation is y = some formula in x. It is frequently important that we know how the dependent variable (conventionally y) will change in response to a change in the independent variable (conventionally x). Now we could measure the ratio between the change in y and the change in x as $\dfrac{\Delta x}{\Delta y}$ or $\dfrac{\Delta y}{\Delta x}$ where $\Delta$ means change in. It is convenient to measure the ratio as $\dfrac{\Delta y}{\Delta x}$ because that gives a simpler formula for linear equations, namely $y = b + \dfrac{\Delta y}{\Delta x}$ instead of $y = b + \dfrac{x}{\dfrac{\Delta x}{\Delta y}}.$ Everyone prefers to multiply than to divide. So the convention is $\dfrac{\Delta y}{\Delta x}.$
okay, thanks everyone. it just seemed to help me do it left to right.

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