Question 337067
{{{drawing(400,400,-5,5,-5,5, 
grid(1),
line(0,2,3,-3) )}}}

The line segment drawn is the right side of the parallelogram.
Since the diagonals must intersect at the origin, and since
the diagonals of a parallelogram bisect each other, the
green line below is one-half of one of the diagonals:

{{{drawing(400,400,-5,5,-5,5, 
grid(1), green(line(0,0,0,2)),
line(0,2,3,-3) )}}}

So we draw in the other half of that diagonal which extends
down to (0,-2):

{{{drawing(400,400,-5,5,-5,5, 
grid(1), green(line(0,-2,0,2)),
line(0,2,3,-3) )}}}

Therefore (0,-2) is the lower left vertex of the
parallelogram.  So we can draw in the bottom side of the
parallelogram, as well as one-half of the other diagonal:

{{{drawing(400,400,-5,5,-5,5, line(0,-2,3,-3),
grid(1), green(line(0,-2,0,2),line(0,0,3,-3)), 
line(0,2,3,-3) )}}}

Notice that the lower left vertex is 3 units left of
and 1 unit above the lower right vertex.  Therefore
the upper left vertex must be 3 units left of and 1 unit
above the upper right vertex, which puts it at (-3,3).
So now we can finish drawing the parallelogram, and the
rest of the other diagonal:


{{{drawing(400,400,-5,5,-5,5, line(0,-2,3,-3),
grid(1), green(line(0,-2,0,2),line(-3,3,3,-3)), 
line(0,2,3,-3), line(0,2,-3,3),line(0,-2,-3,3) )}}}