Question 1125228
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Using the slope formula, *[tex \Large m\ =\ \frac{y_1\,-\,y2}{x_1\,-\,x_2}] where *[tex \Large (x_1,\,y_1)] and *[tex \Large (x_2,\,y_2)] are the coordinates of two points:


Calculate the slopes of the six lines that contain the following segments: *[tex \Large \overline{AB}\ \in\ L_1], *[tex \Large \overline{BC}\ \in L_2], *[tex \Large \overline{CD}\ \in\ L_3], *[tex \Large \overline{DA}\ \in\ L_4], *[tex \Large \overline{AC}\ \in\ L_5], and *[tex \Large \overline{BD}\ \in\ L_6].


If ABCD is a parallelogram, then *[tex \Large \overline{AB}] will be parallel to *[tex \Large \overline{CD}] so the line segments containing those segments will have equal slopes.  Further, *[tex \Large \overline{BC}] will be parallel to *[tex \Large \overline{AD}] and those segments will lie in lines that have equal slopes.


If ABCD was a rhombus, then the lines containing segments *[tex \Large \overline{AC}] and *[tex \Large \overline{BD}] would be perpendicular. But if they are not perpendicular, but do intersect, then the parallelogram is NOT a rhombus.  Hence the slopes of the diagonals must not be equal to each other and also not be negative reciprocals of each other.


So, in summary, you need to show (where *[tex \Large m_i] is the slope of *[tex \Large L_i])


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_1\ =\ m_3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_2\ =\ m_4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_5\ \not=\ m_6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_5\ \not=\ -\frac{1}{m_6}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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