Question 1183287
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The line segment, *[tex \Large \overline{MA}], defined by the two given points, given a standard naming convention for the vertices of a quadrilateral, must perforce be one side of the desired rectangle.  Since segments *[tex \Large \overline{AT}] and *[tex \Large \overline{HM}] must, by the definition of a rectangle, be perpendicular to *[tex \Large \overline{MA}], the first step is to calculate the slope of the line that contains *[tex \Large \overline{MA}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{MA}\ =\ \frac{-2\,-\,(-4)}{5\,-\,2}\ =\ \frac{2}{3}]


Perpendicular lines have negative reciprocal slopes, consequently the lines that  contain *[tex \Large \overline{AT}] and *[tex \Large \overline{HM}] must have slope *[tex \Large -\frac{3}{2}].


Using the Point-Slope Form of an equation of a line, we write the equation for the line containing *[tex \Large \overline{AT}] thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ (-2)\ =\ -\frac{3}{2}(x\,-\,5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{3}{2}x\,-\,\frac{11}{2}]


Likewise, the line containing *[tex \Large \overline{HM}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ (-4)\ =\ -\frac{3}{2}(x\,-\,2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{3}{2}x\,-\,1]


Constructing those two lines gives:


*[illustration RectangleStep1.jpg].


Now choose any value you like, except *[tex \Large x\ =\ 2] which would lead us to a degenerate rectangle, and find the value of the function *[tex \Large y(x)\ =\ -\frac{3}{2}x\,-\,1] at that selected value.  I chose *[tex \Large x\ =\ 0] just because it will simplify the arithmetic.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(0)\ =\ -\frac{3}{2}(0)\,-\,1\ =\ -1]


Hence, *[tex \Large (0,-1)] is a point on the line *[tex \Large y\ =\ -\frac{3}{2}x\,-\,1].  This is a suitable point *[tex \Large H].  And a line perpendicular to *[tex \Large y\ =\ -\frac{3}{2}x\,-\,1] through *[tex \Large (0,-1)] would create a suitable fourth side to the desired rectangle.


Finding an equation of the line that contains the segment that comprises the fourth side and then solving the 2X2 system of this fourth side and *[tex \Large  y\ =\ -\frac{3}{2}x\,-\,\frac{11}{2}] to find the coordinates of point *[tex \Large A] are left as an exercise for the student.


 *[illustration RectangleStep2.jpg].

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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