Question

In the 4th edition. Problem 1.1-9 asks for the 3 possible 4th points in a parallelogram given (1,1), (4,2), and (1,3).
Is there an analytic way to solve that? How else could you solve it? I am missing some important piece of background math on this one and it's point baffles.

Answer 1

Here is one way to approach the problem: The first key is taking the three points and figuring out the possible sides of the possible parallelograms. The sides are formed by taking two points and determining the vector between them. So for example, one possible side is between the points (1, 1) and (4, 2); this side is given by the difference vector (4, 2) - (1, 1) = (3, 1) as shown in the following diagram:



The second key is that you can use this vector in conjunction with the third point (i.e., the one other than the two you used to find the first side) to create another side of the parallelogram and find a fourth point, as shown in the following diagram:



What you're basically doing is adding the difference vector to the vector representing the third point. Since you're adding the same vector to the third point as the one that goes between the first and second points, the sides formed by those vectors are guaranteed to be parallel.

The third key is that instead of adding the vector representing the first side to the third point, you can add its negative (or, in other words, subtract the difference vector), as shown in the following diagram:



Finally, you can repeat this process: take any two of the three points, find the vector that goes between them (first side of a parallelogram), and then add this vector or its negative to the third point (forming a parallel side of the parallelogram) to find a fourth point. Keep doing this until you find all of the possible first sides (there will be three of these vectors) and all of the possible fourth points (there will be six of these, but some of them will be the same).