Question 1164035
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 3y\ =\ -5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{3}x\ -\ \frac{5}{3}]


Slopes of parallel lines are equal, so the slope of the desired line is *[tex \Large -\frac{1}{3}]


An equation of a line with slope *[tex \Large -\frac{1}{3}] that passes through the point *[tex \Large (-7,2)] is, using the Point-Slope form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 2\ =\ -\frac{1}{3}(x\ +\ 7)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{3}x\ -\frac{1}{3}]


And in *[tex \Large Ax\ +\ By\ =\ C] form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x\ -\ 3y\ =\ 1]


But we want a form of the equation where the constant term is 3, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3x\ -\ 9y\ =\ 3]


You can do


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -9\ -\ (-3)]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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