Question 559702
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a. use the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of points P and Q.


b. Presuming you mean "what is AN equation of the line containing the segment PQ", 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 either point P or point Q and *[tex \Large m] is the slope calculated in part a.


c.  Use the midpoint formulas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ = \frac{x_1 + x_2}{2}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{y_1 + y_2}{2}]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of points P and Q.  Then the midpoint is the point *[tex \Large \left(x_m,y_m\right)]


d.  Again, presuming you mean "AN equation": The perpendicular bisector of PQ is the line perpendicular to PQ that passes through the midpoint.  The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line.  Calculate the negative reciprocal of the slope that you calculated in part a, that is calculate *[tex \Large -\frac{1}{m}].  Then use the point-slope form of an equation of a line to derive the desired equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m_\perp(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the midpoint of segment PQ and *[tex \Large m_\perp] is the negative reciprocal of the slope calculated in part a.


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