Area of a triangle given the vertices.

**rtblue** · March 25th 2010, 06:04 PM

What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?

I realize that you could use Distance formula, followed by Hero's Formula to obtain the answer, but i'm sure there is a faster way to go about this (that way has too many radicals and fractions, and is easy to mess up on).

Suggestions to get me started?

**Soroban** · March 25th 2010, 06:59 PM

Hello, rtblue!

As a matter of fact, there is a neat formula for this problem.

What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?

If you understand determinants, you'll like this . . .

The area of the triangle with vertices: . $(x_1,y_1),\:(x_2,y_2),\:(x_3,y_3)$ is:

. . . . . $A \;=\;\frac{1}{2}\begin{vmatrix}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}$

For your three vertices:

. . $A \;\;=\;\;\frac{1}{2}\begin{vmatrix}1&0&0 \\ 1 & 14 & 6 \\ 1 & 12 & 9 \end{vmatrix} \;\;=\;\;\frac{1}{2}\left[1\cdot\begin{vmatrix}14 & 6 \\ 12 & 9\end{vmatrix} \:-\: 0\cdot\begin{vmatrix}1&6 \\ 1&9\end{vmatrix} \:+\: 0\begin{vmatrix}1&14 \\ 1&12\end{vmatrix}\,\right]$

. . . . $=\;\;\frac{1}{2}\bigg[1(14\cdot9 - 12\cdot6) - 0 + 0\bigg] \;\;=\;\;\frac{1}{2}(126-72) \;\;=\;\;\boxed{27}$

Note: The "determinant bars" also represent absolute value.
. . . . .The determinent can turn out to be negative.

Edit: correct a silly oversight . . .
.

**rtblue** · March 25th 2010, 07:13 PM

Thanks, I understand now. (although you did forget to divide by 2 at the end

)

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