Thanks!!
$\displaystyle
proj_y(x) = \frac{xy}{|y|^2}y = \frac{17}{37}(6,1) = (\frac{102}{37},\frac{17}{37})
$
This defines a point on the vector y (notice the slope of the new vector is 1/6). Now observe that the vector z from this point to x is perpendicular to the vector y. If we connect x,y tip to tail and complete the parallelogram the vector y-x (the line from the point y to the point x) is the diagonal and defines 2 identical triangles which compose the parallelogram.
What is neat about the projection we found above is that the length of z is exactly the height of that triangle with base length |y|. This is all we need to find the area of the triangle. Twice this area is the area of the parallelogram.