Question 806276
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Assume the coordinates of the four vertices are given and that they are *[tex \Large V_1(x_1,y_1)], *[tex \Large V_2(x_2,y_2)], *[tex \Large V_3(x_3,y_3)], and *[tex \Large V_4(x_4,y_4)].  Since coordinate rotation and the labeling of the vertices is arbitrary, we can assume without loss of generality that the following inequalities hold:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ <\ x_2\ <\ x_4\ <\ x_3]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_1\ <\ y_4\ <\ y_2\ <\ y_3]


Now write the following four inequalities using your given coordinates:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ <\ \left(\frac{y_2\ -\ y_1}{x_2\ -\ x_1}\right)\left(x\ -\ x_1\right)\ +\ y_1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ <\ \left(\frac{y_3\ -\ y_2}{x_3\ -\ x_2}\right)\left(x\ -\ x_2\right)\ +\ y_2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ >\ \left(\frac{y_3\ -\ y_4}{x_3\ -\ x_4}\right)\left(x\ -\ x_4\right)\ +\ y_4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ >\ \left(\frac{y_4\ -\ y_1}{x_4\ -\ x_1}\right)\left(x\ -\ x_1\right)\ +\ y_1]


If a point on a side of the parallelogram is considered "in" the parallelogram, then change the inequality operators from strictly less than/greater than, to *[tex \Large \leq] or *[tex \Large \geq]


Then if the coordinates of a point satisfy ALL FOUR inequalities, the point is in the parabola, otherwise it is not.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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