Area of a triangle by determinant (linear algebra) A triangle has corners at $(0,0), (3,4)$, and $(2,-1)$. Compute its area using a determinant.
I'm not sure how to set up the matrix to use a determinant to find the area. Thanks for any help!!
 A: Suppose a triangle has three corners: $(x_1,y_1),\ (x_2,y_2),\ (x_3,y_3)$.
Then area of that triangle is defined by:
$$
\Delta =\dfrac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1 \\ 
x_2 & y_2 & 1\\
x_3 & y_3 & 1
\end{vmatrix} \text{ sq. unit}
$$
A: You would find the distance between the points $(3,4)$ and $(2,-1)$, then the distance from the origin of the line through these two points (base and height).
Do it in abstract. The line through $(x_1,y_1)$ and $(x_2,y_2)$ has equation
$$
(y_1-y_2)(x-x_2)-(x_1-x_2)(y-y_2)=0
$$
that can be rewritten as
$$
x(y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1)=0
$$
so the distance from it to the origin is
$$
h=\frac{|0(y_1-y_2)-0(x_1-x_2)+(x_1y_2-x_2y_1)|}
       {\sqrt{(y_1-y_2)^2+(x_1-x_2)^2\,}}
$$
and the base is
$$
b=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
$$
Therefore the area is
$$
A=\frac{1}{2}bh=
\frac{1}{2}|x_1y_2-x_2y_1|=
\frac{1}{2}\left|\det\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}\right|
$$
In your case
$$
A=\frac{1}{2}\left|\det\begin{bmatrix}3&2\\4&-1\end{bmatrix}\right|
$$
A: Are you aware that the image of the unit square $[0,1]\times[0,1]$ under a linear transformation with matrix $M$ has area $|\text{det}(M)|$? So if we take half the square - i.e., the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$ - the image of that under the same linear transformation has area $\frac{1}{2}|\text{det}(M)|$.
You want your triangle to be the image of the triangle described above under some linear transformation. Of course that transformation must send $(0,0)$ to $(0,0)$. You also want it to send $(1,0)$ and $(0,1)$ to the other two vertices of the triangle. Now all you have to do is write down the matrix for such a linear transformation.
