Question 186973
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Since you are given 3 points, you can select 3 pairs of points.  Using the two-point form of the equation of a line, you can create three equations representing three lines.

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

For *[tex \Large P_1(-2,-4)] and *[tex \Large P_2(-5,-1)]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_1: y + 4 = \left(\frac{-4 + 1}{-2 + 5}\right)(x + 2) \ \ \Rightarrow\ \ y = -x -6]

For *[tex \Large P_2(-5,-1)] and *[tex \Large P_3(0,-4)]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_2: y + 1 = \left(\frac{-1 + 4}{-5 - 0}\right)(x + 5) \ \ \Rightarrow\ \ y = -\frac{3}{5}x -4]

For *[tex \Large P_1(-2,-4)] and *[tex \Large P_3(0,-4)]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_3: y + 4 = \left(\frac{-4 + 4}{-2 - 0}\right)(x + 2) \ \ \Rightarrow\ \ y = - 4]

Now, using:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_\alpha \parallel L_\beta \ \ \Leftrightarrow\ \ m_\alpha = m_\beta]

Create the equation of the parallel to each of the above lines through the given point <i><b>not</b></i> on that line using the point-slope form of the equation of a line:

*[tex \Large m_1 = -1] through *[tex \Large P_3(0,-4):]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_4: y + 4 = -1(x - 0) \ \ \Rightarrow\ \ y = -x -4]

*[tex \Large m_2 = -\frac{3}{5}] through *[tex \Large P_2(-5,-1):]

(Left as an exercise for the student)

*[tex \Large m_3 = 0] through *[tex \Large P_2(-5,-1):]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L_6: y + 1 = 0(x + 5) \ \ \Rightarrow\ \ y = -1]

Now that we have *[tex \Large L_4 \parallel L_1] and *[tex \Large L_6 \parallel L_3] we only need to find the point of intersection between *[tex \Large L_4] and *[tex \Large L_6]

Since we know that

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y=-1]

from the equation for *[tex \Large L_6], we can just substitute into the equation for *[tex \Large L_4]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \   -1 = -x -4 \ \ \Rightarrow\ \ x = -3]

Hence, Point D for the first parallelogram is (-3,-1).

Now all you have to do is derive the equation for *[tex \Large L_5] and find the points of intersection with the other two lines to get the other two coordinate pairs for the other two possibilities for Point D.

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