<PRE><B>Question 458103</B>
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Solve the given equation for *[tex \Large y] in terms of everything else.  The resulting coefficient on *[tex \Large x] will be the slope of the given line.  Perpendicular lines have slopes that are negative reciprocals, so calculate the negative reciprocal of the slope of the given line.


Then use the point-slope form of an equation of a line:


*[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.


Once you have done all of that you will have &lt;i&gt;&lt;b&gt;an&lt;/b&gt;&lt;/i&gt; equation of the desired line.  From there you can leave it as is, put it into Standard Form, or put it into Slope-Intercept Form.  That is the best you can do.  Since there are infinite representations of the equation whose graph is a given line, you cannot write &lt;i&gt;&lt;b&gt;the&lt;/b&gt;&lt;/i&gt; equation of a line.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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