Question 595981
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First off, *[tex \LARGE y\ +\ 8\ =\ \frac{2}{3}\left(x\ -\ (-2)\right)]


is incorrect.


Your given point has 2, not -2 as an *[tex \LARGE x]-coordinate.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 8\ =\ \frac{2}{3}\left(x\ -\ 2\right)]


Now, distribute the slope fraction:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 8\ =\ \frac{2}{3}x\ -\ \frac{4}{3}]


Then add -8 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{2}{3}x\ -\ \frac{4}{3}\ -\ 8]


But 8 is the same thing as *[tex \LARGE \frac{24}{3}], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{2}{3}x\ -\ \frac{4}{3}\ -\ \frac{24}{3}]


Now just add the numerators on the fractions with like denominators and you have it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{2}{3}x\ -\ \frac{28}{3}]


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