Question 1033047
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Intercepts are points, NOT scalar values.  Saying "x-intercept of 3" is as ridiculous as saying a banana is an elephant.  If 3 is the x-coordinate of the x-intercept and -1 is the y-coordinate of the y-intercept, then you have two ordered pairs that are on the graph of your line, namely (3,0) and (0,-1).


You also cannot find "the" equation of a line.  For any given line there are infinite equations that represent that line.  So, you can find "an" equation of a line, or the slope-intercept form of the equation of a line, or the standard form of the equation of a line, and so on.


If you are given two points use the Two-Point Form of an equation of a line:


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


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


Plug in your coordinate values and do the indicated arithmetic.  Then put it in whatever form you deem appropriate.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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