If three corners of a parallelogram are known solve for the 3 possible 4th corners. An example would be three corners being the points: (1,1), (4,2) and (1,3). I understand the specific solution for this example: (4,4), (4,0) or (-2,2). Which I reasoned when i drew it out. The example came from a linear algebra textbook and i'm curious if it can be done some other way.
 A: The three possibilities come from adding the coordinates of two points and subtracting the third.  So $(4,0)=(4,2)+(1,1)-(1,3)$.  There are three choices of which point to subtract.  You are translating the point you subtract to the origin, then adding the vectors to the other two points to find the opposite one, then translating back.  So you could look at it  as $(4,0)=[(4,2)-(1,3)]+[(1,1)-(1,3)]+(1,3)$
A: Given 3 vertices, these vertices define 2 vectors; the unknown endpoint corresponds to the sum of these 2 vectors.
Example: $(1,1)$ and $(1,3)$ form the vector $(0,2)$.  $(1,1)$ and $(4,2)$ form the vector $(3,1)$.  Sum these vectors to get $(3,3)$.  Then add the vertex $(1,1)$ back to get $(4,4)$ as the formerly unknown corner.
A: let the vertices of parallelogram be vectors a,b,c,d where a = (1,1);
b = (4,2), c = (1,3), and let d = (x,y).
Now for parallelogram there can be three possibility i.e.
a - b || c - d or
a - c || d - b or
a - d || b - c (as opposite sides are parallel)
solving from above three statements and equating respective components on LHS and RHS.
we can have (x,y) = (4,4) or (4,0) or (-2,1)
