Thread: Finding the are of a Parallelogram with coordinates...? Pre-Calculus

Hello. Pre-cal practice work for test tomorrow.

Problem gives us these coordinates: A(1,2), B(5,2), C(3,6), D(5, -3)

I graphed it. How are we supposed to find the area with these coordinates?

I know the answer is 16, because I checked the back of the book. I would like to know how to get that answer.

Thank you

Oh yes, I apologize for the misconveniance.. I will correct this when I get back to my room to access my pre-cal book.

A simple plot of the points shows that this is not a parallelogram...

It is not a parallelogram. Join any diagonal of the quadrilateral. and use the formula for finding the area of a triangle with given vertices.

Hello, theadage!

Please check for typos.
The problem doesn't make sense.

Problem gives us these coordinates: A(1,2), B(5,2), C(3,6), D(5, -3) .[1]

I graphed it. How are we supposed to find the area with these coordinates?

I know the answer is 16, because I checked the back of the book. .[2]

I would like to know how to get that answer. . So would I!

[1] The vertices are usually labeled in order around the parallelogram.
. . .The given coordinates are not.
. . .Besides, they do not form a parallelogram.

Code:

      |
      |     C
      |     o
      |    *:*
      |   * :4*
      |  *  :  *
      |Ao * * * oB
      |   * 4   *
    - + - - * - *5 - -
      |       * *
      |         o
      |         D
      |

[2] The area is not 16.

Triangle has base 4 and height 4.
. . Its area is

Triangle has base 4 and height 5.
. . Its area is

The total area is

A(1,3) B(5,2) C(3,6) D(7,6)

I know that BXH= Area.
But it's too easy... there must be a trap.
;0