Question 576464
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Since in the given equation the *[tex \LARGE x] and *[tex \LARGE y] terms are on the same side of the equals sign, take the opposite of the coefficent on *[tex \LARGE x] and divide by the coefficient on *[tex \LARGE y] to find the slope of the line represented by the given equation.


Use the point-slope form of an equation of a line to write the desired equation:


*[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.


Put the derived equation into either Standard or Slope-Intercept Form as desired by your teacher/professor/instructor.0


However, please note that you cannot write "the" equation of any line.  You can only write "an" equation of a line.  In the first place, you could just substitute the values in the point-slope form and have an equation of the desired line, or you could solve for *[tex \Large y] to put it into slope-intercept form, or you could rearrange it into standard form, namely *[tex \Large Ax + By = C].  Furthermore, for a given *[tex \Large A], *[tex \Large B], and *[tex \Large C], *[tex \Large kAx + kBy = kC] where *[tex \Large k\ \in\ \R] describes a set of equations with an infinite number of elements, each of which graphs to the same line in *[tex \Large \R^2].


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