Question 623756
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Technically what you ask cannot be done.  The *[tex \LARGE x]-intercept is NOT 2, and the *[tex \LARGE y]-intercept is NOT *[tex \LARGE \frac{2}{5}].  Intercepts are not numbers, intercepts are points.  In *[tex \LARGE \mathbb{R}^2], points are designated by ordered pairs.  It is possible that the *[tex \LARGE x]-coordinate of the *[tex \LARGE x]-intercept is 2 and that the *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept is *[tex \LARGE \frac{2}{5}].


Making that assumption (since nothing else seems to make any sense), you have two points which is sufficient to uniquely define a line in *[tex \LARGE \mathbb{R}^2].  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.


Then solve your derived equation for *[tex \LARGE y] in terms of everything else. That is, put it into *[tex \LARGE y\ =\ mx\ +\ b] form, where *[tex \LARGE m] is the slope and *[tex \LARGE b] is the *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept.


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