SOLUTION: Find the area of a parallelogram with vertices (0, 0), (3, 3), (4, 6), and (7, 9) using the formula which relates the area to the absolute value of the determinant of the matrix.

To find out what order to taken the vertices, we plot
the parallelogram:




To get the positive value of the area, take the coordinates 
going counter-clockwise around the polygon. Put the first
pair of coordinate (0,0) again at the bottom:



To evaluate that determinant, add all the NW to SE diagonal
products and subtract all the NE to SW diagonal products.

(0×3)+(3×9)+(7×6)+(4×0)-(0×3)-(3×7)-(9×4)-(6×0) =

  0  +  27 +  42 +   0 -   0 -  21 -  36 -   0  =

 12.  

Multiply by :

So 

Edwin

Let V1 = (a,b) and V2 = (c,d) be two vectors that are two adjacent sides of the parallelogram released from one common vertex. 


Then the area of the parallelogram is the absolute value of the determinant


S = abs (det  ) = | ad - bc |.


In your case, it is convenient to take point (0,0) as the common vertex.

Then two vectors/adjacent sides are 

V1 = (3-0,3-0) = (3,3),     (1)    and

V2 = (4-0,6-0) = (4,6).     (2)

Do not mix these vectors (1) and (2) with the point coordinates.  
Again, (1) and (2) represent the vectors that are two adjacent sides of the parallelogram with the common vertex (0,0).

Next, form this matrix   = .


Calculate its determinant  det = 3*6 - 4*3 = 18 - 12 = 6.


Take the absolute value of the determinant:  |det| = |6| = 6.


In this case the absolute value is the same as the determinant, but in other cases it may not be so.


Your area is 6:  S = 6 square units.


So, I answered your question and described the procedure.