Question 618985
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Point D has to be at (-2,3) or (-4,-7) if indeed ADEF is a parallelogram, and it makes no difference which, the quadrilateral is not a rhombus in either case.


Use the mid-point formulas to determine the coordinates of point E and F.


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


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


I'll leave it to you to verify E(-2,3) and F(0,-2)


Then using the properties of a parallelogram and the fact that parallel lines have equal slopes, we can determine that point D is either (-2,3) or (-4,-7).


If ADEF were a rhombus, then either AE = EF or AF = EF (depending on where D is located)


Use the distance formula to disprove each assertion.  ADEF not a rhombus, Q.E.D


Distance formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2}]


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