Question 1048652
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The angle that the line *[tex \Large 3y\ -\ x\ +\ 1\ =\ 0] makes with the *[tex \Large x] axis is *[tex \Large \arctan\left(\frac{1}{3}\right)] because the slope is *[tex \Large \frac{1}{3}]. A line with a positive slope that is inclined at a 45 degree angle to the given line must have a slope *[tex \Large \tan\left(45\ +\ \arctan\left(\frac{1}{3}\right)\right)]. Using the point-slope form and the given point (2,0):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the calculated slope, we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \tan\left(45\ +\ \arctan\left(\frac{1}{3}\right)\right)(x\ -\ 2) ]


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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