Question 276638
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First I have two questions for you.  1: Why do you find it necessary to shout?  Typing in all caps is the electronic equivalent of shouting and is both rude and annoying.  2: Why have you not tried to search for a similar question on this site?  This very same question, albeit with different numbers, has been asked and answered hundreds of times on this site.


Start with the fact that the slopes of perpendicular lines are negative reciprocals, that is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \perp\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ -\frac{1}{m_2}\ \text{ and } m_1,\, m_2\, \neq\, 0]


Take your given equation and put it into slope-intercept form, *[tex \Large y\ =\ mx\ +\ b] by solving for *[tex \Large y].  Then determine the slope of the given line by inspection of the coefficient on *[tex \Large x]


Calculate the negative reciprocal to determine the slope of the perpendicular line.


Use the point-slope form of the 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 slope calculated above.


Once you have your point-slope form, solve for *[tex \Large y] to put it into slope-intercept form.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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