Question 656888
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I believe it is an error to refer to an intercept as a single number.  An intercept is properly defined as a point, which, in *[tex \LARGE \mathbb{R}^2] is defined by an ordered pair.  The proper way to refer to the *[tex \LARGE x]-intercept in your problem is *[tex \LARGE (2,0)]


The slope of the given line, the equation of which is presented in slope-intercept form, is simply the coefficient on *[tex \LARGE x].  The slope of any perpendicular to any line where the slope is neither zero nor undefined is the negative reciprocal of the slope of the first line, thus:


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


Now that you have the slope of the desired line and one point on that line, use the point-slope form to derive an equation that represents the desired 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.


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