Question 256399
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The formula for the slope is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


And if I understand your question correctly, you are asking what would happen if you did the following calculation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_2\ -\ y_1}{x_1\ -\ x_2} ]


instead.  The answer is to be found by considering the fact that for all real numbers *[tex \Large a] and *[tex \Large b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ -\left(b\ -\ a\right)]


Hence the error would cause the slope to have the incorrect sign.  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_2\ -\ y_1}{x_1\ -\ x_2}\ =\ -m]


On the other hand, if you are asking what happens if you do:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_2\ -\ y_1}{x_2\ -\ x_1}]


instead of


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}]


using the same principle, we can see that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}\ = \frac{-\left(y_2\ -\ y_1\right)}{-\left(x_2\ -\ x_1\right)}\ = \frac{y_2\ -\ y_1}{x_2\ -\ x_1}\ = m]


Proving that it doesn't matter which you call Point 1 and which you call Point 2 so long as you remain consistent throughout.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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