Question 202908
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The first thing to realize is that since *[tex \Large Q] and *[tex \Large R] are specified as fixed offsets from the coordinates of *[tex \Large P], you can set *[tex \Large P] at the origin without any loss of generality and then re-insert the variables *[tex \Large a] and *[tex \Large b] at the very end.  That will simplify the calculations considerably.  So, set *[tex \Large a = 0] and *[tex \Large b = 0] so that point *[tex \Large P] is *[tex \Large P(0,\,0)]. 


The coordinates of *[tex \Large S] are the solution set of the system formed by the equations of the lines containing the segments *[tex \Large RS] and *[tex \Large PS].  In order to solve such a system, we have to develop the two equations.  So far, we have one point in each of the equations, namely *[tex \Large P(0,\,0)] and *[tex \Large R(-1,\,7)].  However, in order to derive the two equations we either need the slope of these lines or another point on the lines.  Another point is not directly available, but fortunately we are dealing with a parallelogram, and we know that the slopes of parallel lines are equal.


So, we first need to derive the slope of the lines containing the segments *[tex \Large PQ] and *[tex \Large QR].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{PQ}\ =\ \frac{0 - 4}{0 - 3} =\ \frac{4}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{QR}\ =\ \frac{4 - 7}{3 - (-1)} =\ -\frac{3}{4}]


Having the two slopes, we can write an equation for the two missing lines.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{RS}:\ y - 7 =\ \frac{4}{3}(x - (-1)) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{RS}:\ 3y - 21 = 4x + 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{RS}:\ 4x - 3y = -25]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{SP}:\ y - 0 =\ -\frac{3}{4}(x - 0)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{SP}:\ 4y - 0 = -3x + 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_{SP}:\ 3x + 4y = 0]


Now solve the system:


Eq 1:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x - 3y = -25]


Eq 2:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x + 4y = 0]


Multiply Eq 1 by 4 and Eq 2 by 3:


Eq 3:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16x - 12y = -100]


Eq 4:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x + 12y = 0]


Add Eq 3 to Eq 4:


Eq 5:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25x + 0y = -100 \ \ \Rightarrow\ \ x = -4]


Multiply Eq 1 by 3 and Eq 2 by -4:


Eq 3:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x - 9y = -75]


Eq 4:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -12x - 16y = 0]


Add Eq 3 to Eq 4:


Eq 5:*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0x - 25y = -75 \ \ \Rightarrow\ \ y = 3]


Hence, the solution set of the system, and therefore the point of intersection of the two missing sides of the normalized parallelogram is *[tex \Large \left(-4,3\right)].


Re-inserting the *[tex \Large a] and *[tex \Large b] offsets, you get *[tex \Large S(a - 4,\,b + 3)]


It may be instructive to note that *[tex \Large PQRS] is not just a parallelogram.  Noting that the two slopes that we calculated are negative reciprocals tells us that the adjacent sides are perpendicular, so *[tex \Large PQRS] is at least a rectangle.  Also, a little work with the distance formula (left as an exercise for the student) shows that the two given adjacent sides, and therefore all four sides, are equal in measure.  Therefore *[tex \Large PQRS] is a square.  

 
John
*[tex \LARGE e^{i\pi} + 1 = 0]
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