Question 986560
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Exactly.  Any equation of the form *[tex \Large y\ =\ mx] (note the absence of a *[tex \Large b]) passes through the origin because whenever *[tex \Large x\ =\ 0], *[tex \Large y\ =\ 0]


You can also look at it this way.  If you have *[tex \Large y\ =\ mx\ +\ b], then the *[tex \Large y]-coordinate of the *[tex \Large y]-intercept is *[tex \Large b], hence the *[tex \Large y]-intercept is the point *[tex \Large (0,b)].  Now, if *[tex \Large b\ =\ 0], what do you have?


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

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