• Nov 2nd 2008, 08:45 PM
Hellreaver
I need to find the area of of a triangle with the given vertices:

A(1,1) B(2,2) C(3,-3)

I found the lengths of these, and used 1/2llallllbllllallllbll.
Which I got to result in the area equals 12. Is this correct?

If I do it using area= P1P2P3= 1/2 $\begin{bmatrix}x1&y1&1\\x2&y2&1\\x3&y3&1 \end{bmatrix}$
I end up with:

-1/2 $\begin{bmatrix}1&1&1\\2&2&1\\3&-3&1 \end{bmatrix}$
Which can be row reduced to:
$\begin{bmatrix}1&1&1\\0&0&-1\\0&-6&-2 \end{bmatrix}$

I then use cofactor expansion:
(-1/2)[1 $\begin{bmatrix}0&-1\\-6&-2 \end{bmatrix}$]
When I find the determinant of the cofactor i get:
(-1/2)(-6)
This gives me 3 as an answer...
Which way is right? Or are both wrong?
• Nov 3rd 2008, 03:54 AM
mr fantastic
Quote:

Originally Posted by Hellreaver
I need to find the area of of a triangle with the given vertices:

A(1,1) B(2,2) C(3,-3)

I found the lengths of these, and used 1/2llallllbllllallllbll.
Which I got to result in the area equals 12. Is this correct?

If I do it using area= P1P2P3= 1/2 $\begin{bmatrix}x1&y1&1\\x2&y2&1\\x3&y3&1 \end{bmatrix}$
I end up with:

-1/2 $\begin{bmatrix}1&1&1\\2&2&1\\3&-3&1 \end{bmatrix}$
Which can be row reduced to:
$\begin{bmatrix}1&1&1\\0&0&-1\\0&-6&-2 \end{bmatrix}$

I then use cofactor expansion:
(-1/2)[1 $\begin{bmatrix}0&-1\\-6&-2 \end{bmatrix}$]
When I find the determinant of the cofactor i get:
(-1/2)(-6)
This gives me 3 as an answer...
Which way is right? Or are both wrong?

I get 3 square units.

There are any number of area formulae from geometry you can use to get this answer.