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April 11th, 2017, 03:28 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Why do we do y1y2 x1x2
Was wondering why we don't do the opposite x1×2 y1y2? 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: USA Posts: 1,781 Thanks: 919 
wat?

April 11th, 2017, 03:45 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,817 Thanks: 760 
2 * 3 = 6 ....hang on...and..and..and 3 * 2 = 6 !!!!!!!!!!

April 11th, 2017, 05:41 PM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,719 Thanks: 1374  Quote:
$m = \dfrac{y_1y_2}{x_1x_2}$ 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: 18,689 Thanks: 1522 
$\dfrac{y_1y_2}{x_1x_2}$ and $\dfrac{y_2y_1}{x_2x_1}$ are equivalent. Their reciprocals give the slope of $x$ with respect to $y$.
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: 2,090 Thanks: 701 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: 916 Thanks: 366 
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 $\dfrac{\Delta y}{\Delta x}.$ 
April 12th, 2017, 12:44 PM  #8  
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Quote:
 