Question 486001
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Better than that, let *[tex \Large (x_1,y_1)\ =\ (a,c)] and *[tex \Large (x_2,y_2)\ =\ (b,d)] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_2\ -\ y_1}{x_2\ -\ x_1}\ =\  \frac{d\ -\ c}{b\ -\ a}].


Factor -1 from *[tex \Large (d\ -\ c)]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ -\ c\ =\ (-1)(-d\ +\ c)\ =\ (-1)(c\ -\ d)]


Likewise:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ -\ a\ =\ (-1)(-b\ +\ a)\ =\ (-1)(a\ -\ b)]


So, by substitution:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_2\ -\ y_1}{x_2\ -\ x_1}\ =\  \frac{(-1)(c\ -\ d)}{(-1)(a\ -\ b)}].


But since *[tex \Large \frac{-1}{-1}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(-1)(c\ -\ d)}{(-1)(a\ -\ b)}\ =\ \frac{c\ -\ d}{a\ -\ b}].


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}\ =\  \frac{c\ -\ d}{a\ -\ b}].


also.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}\ =\ \frac{y_2\ -\ y_1}{x_2\ -\ x_1}\ \forall x_i,\,y_i\ \in\ \mathbb{R},\ x_1\ \neq\ x_2\]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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