Expert: Josh - 11/5/2003

Question
can you help me with one question:

if the points A, B, C and D form a parallelogram and A is (-1,8) , B is (3, - 2) C(-5,0) What are the possible points for D?
How do you graph and show all possible parallelograms?
Do all the possible points D: D1, D2 and D3 form an equilateral triangle?
What is special about the points A, B and C?  

Answer
Hi Karen,

Before we discuss this problem, please draw a scaled diagram showing the points A=(-1,8),B=(3,-2),C=(-5,0) on the 2D plane.
Draw straight lines joining A and C, C and B.

Then ask yourself, what are the properties of a parallelogram. Well, by definition, the opposite sides in this quadrilateral must be parallel. How can we phrase this differently. This is the same as saying that opposite sides have the same gradient or slope. ...[#A]

It is necessary to calculate the slopes of the straight lines AC, CB and AB. We write mAC, mCB, mAB for these.

mAC=(yA-yC)/(xA-xC)=(8-0)/(-1--5)=+2
mCB=(yC-yB)/(xC-xB)=(0--2)/(-5-3)=-1/4
mAB=(yA-yB)/(xA-xB)=(8--2)/(-1-3)=+2.5

Using argument [#A], we draw a straight line (L1) passing through point A with slope mCB and another line (L2) passing through point B with slope mAC.
Exercise: You can work out L3 yourself, which passes through point C, with slope mAB. (Do this now)

L1: (y-yA)=mCB*(x-xA)
L2: (y-yB)=mAC*(x-xB)
L3:

Distance constraint:
by definition, the opposite sides in a parallelogram must have equal length...[#B]

So, work out distance d(A,C) and d(C,B)
Exercise: You can work out d(A,B) yourself.

d(A,C)=sqrt[(yA-yC)^2+(xA-xC)^2]=sqrt(80)
d(C,B)=sqrt[(yA-yC)^2+(xA-xC)^2]=sqrt(68)
d(A,B)=

There are 3 cases to consider.
[I] D is a distance of d(C,B) from point A on L1, in the positive x direction. D is also a distance of d(A,C) from point B on L2, in the positive y direction.

[II] D is a distance of d(C,B) from point A on L1, in the negative x direction. Hence, D is also a distance of d(A,B) from point C on L3, in the positive y direction.

[III] D is a distance of d(A,C) from point B on L2, in the negative y direction. Hence, D is also a distance of d(A,B) from point C on L3, in the negative y direction.

-Now, all your problems are solved, you can graph all possible parallelograms on the Cartesian (xy) plane.

-To check if D1,D2,D3 (from case I,II,III above) form an equilateral triangle, check their distance of separation and see whether d(D1,D2)=d(D1,D3)=d(D2,D3) is true or false.

-Special feature: you can see quite easily (can prove this yourself) that A,B,C are the midpoints of the sides of the triangle subtended by D1,D2,D3.

Cheers.