Question 719280
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In the first place, your line does not have a *[tex \LARGE y]-intercept of 2.  A *[tex \LARGE y]-intercept is a point and a point is designated by an ordered pair.  *[tex \LARGE y]-intercepts have the feature that the *[tex \LARGE x]-coordinate is zero.  So, presuming that you (or your instructor) actually meant to say that the *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept is 2, then the *[tex \LARGE y]-intercept is the point *[tex \LARGE (0,2)]


Now use the two-point form of an equation whose solution set is a set of ordered pairs that comprise 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.


Do the indicated arithmetic and then rearrange the equation so that it is in *[tex \LARGE Ax\ +\ By\ =\ C] form.  Some teachers/professors/instructors/textbook authors demand that *[tex \LARGE A], *[tex \LARGE B], and *[tex \LARGE C] be integers to be proper Standard Form.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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