Wha is the area of the quadrilateral with vertices at 
(1,1), (5,2), (4,4) and (2,3)? Put answer in square units.

First we draw the quadrilateral:

(1,1), (5,2), (4,4) and (2,3)



So we must divide the quadrilateral into two triangles, for
there is a determinant formula for finding the area of a
triangle when the coordinates of the three vertices are
given:

The area of a triangle with the vertices

(x1,y1),(x2,y2), (x3,y3)

is equal to one half the absolute 
value of this 3×3 determinant

|x1 y1  1|          
|x2 y2  1|               
|x3 y3  1|                    

Now we draw in either diagonal we choose, either of
these will do, either this

 or this 

If we use the left graph, the determinant for the
triangle with vertices (1,1), (2,3), and (4,4) is  

| 1 1 1 |
| 2 3 1 | = -3
| 4 4 1 |

I am assuming you already know how to find the
value of a determinant. If you don't. then post
again asking how.

So the area of that triangle is one half of the 
absolute value of -3.  The absolute value of -3 
is 3, and half of 3 is 1.5.

Again, using the left graph, the determinant for the
triangle with vertices (1,1), (4,4), and (5,2) is  

| 1 1 1 |
| 4 4 1 | = -9
| 5 2 1 |

So the area of that triangle is one half of the 
absolute value of -9.  The absolute value of -9 
is 9, and half of 9 is 4.5.

So the sum of the areas of those two triangles
is 

1.5 + 4.5 or 6 square units of area for the
quadrilateral.

-----------------

Now let's check it by using the graph on the right
above.

If we use the graph on the right above, the 
determinant for the triangle with vertices
 (2,3), (4,4), and (5,2) is  

| 2 3 1 |
| 4 4 1 | = -5
| 5 2 1 |


So the area of that triangle is one half of the 
absolute value of -5  The absolute value of -5
is 5, and half of 5 is 2.5.

Again, using the graph on the right, the 
determinant for the triangle with vertices 
(2,3), (5,2), and (1,1) is  

| 2 3 1 |
| 5 2 1 | = -7
| 1 1 1 |

So the area of that triangle is one half of the 
absolute value of -7. The absolute value of -7
is 7 and half of 7 is 3.5.

So the sum of the areas of those two triangles
is 

2.5 + 3.5 or, again, we get 6 square units of 
area for the quadrilateral.

Edwin