Finding the area of a triangle given vertices.

[needshelp22]

I am currently trying to do a math packet for summer work. The problem states, "What is the area of the triangle that has the following vertices: (-3,4) (6,3) (2,1)." To solve this problem I have set it into a matrix.
It looks like this:
| -3 4 1 |
| 6 3 1 |
| 2 -1 1 |

I found two problems in my notes that are exactly the same format, and can find out each of them without a problem. I find the determinate of the above matrix, and take half of it. The only problem is, that whenever I find the determinate of this matrix I end up getting a negative number! -40 to be exact. Taking half of that would give me -20, and I don't think that an area can be negative...

 

[Denis]

www.mathopenref.com/coordtrianglearea.html

 

[soroban]

Hello, needshelp22!

Your determinant has an extra minus-sign . . .

What is the area of the triangle that has vertices: (-3,4), (6,3), (2,1).

\(\displaystyle \text{Area} \;=\;\frac{1}{2}\left|\;\begin{array}{|ccc|}\text{-}3 & 4 & 1 \\ 6 & 3 & 1 \\ 2 & 1 & 1 \end{array}\;\right| \quad\text{ (absolute value of the determinent)}\)

\(\displaystyle \text{We have: }\:A \;=\;\tfrac{1}{2}\bigg|\text{-}3(2-1) - 4(6-2) + 1(6-4)\,\bigg| \;=\;\tfrac{1}{2}\bigg|\text{-}3 - 16 + 2\,\bigg|\)

. . . . . \(\displaystyle A \;=\;\tfrac{1}{2}|\text{-}17| \;=\;\frac{17}{2}\)