Show that the area of a triangle is given by this determinant I'm not sure how to solve this problem. Can you guys provide some input/hints? 

Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ be three points in $\mathbb{R}^{2}$.
Show that the area of $\Delta ABC$ is given by
  $\frac{1}{2}\left| \det(M) \right|$, where
  $$M = \begin{pmatrix} 1 & 1
 & 1\\  x_1 & x_2 & x_3\\  y_1 & y_2 & y_3 \end{pmatrix}$$

 A: Hint: It may be easier to show that $\frac 12\det(M)$ (i.e. without taking absolute value) computes the oriented area of the triangle.
One possible way is to show that $\det M$ is invariant under translation and shearing and gives the correct answer for the special case $A=(0,0)$, $B=(1,0)$, $C=(0,1)$. 
Note that translations correspond to adding multiples of the first row of $M$ to the second or third row, shearing corresponds to adding a multiple of the second row to the third or vice versa.
A: Here is a multi-variable calculus approach.  Suppose $S$ is the triangle spanned by vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$.  We intend to compute the double integral $$\iint_{S} 1 dydx$$ which is the double integral that gives us the area of $S$.  Now, we try to come up with a suitable change of variables, so that the legs of this triangle are on the coordinate axes.  Suppose $$u=(y_1-y_2)x+(x_2-x_1)y,v=(y_1-y_3)x+(x_3-x_1)y$$.
The Jacobian Determinant of this transformation $$\frac{\partial(u,v)}{\partial(x,y)}=\det \begin{bmatrix}\frac{\partial u}{\partial x} && \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} && \frac{\partial v}{\partial y}\end{bmatrix}=\begin{vmatrix} y_1-y_2 && x_2-x_1 \\y_1-y_3 && x_3-x_1 \\ \end{vmatrix}=(y_1-y_2)(x_3-x_1)-(y_1-y_3)(x_2-x_1)$$
But the need to take the reciprocal $\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{(y_1-y_2)(x_3-x_1)-(y_1-y_3)(x_2-x_1)}$
$S$ transforms into a right triangle $T$ whose vertices are $(0,0)$,
$(y_1-y_2)(x_3-x_1)+(x_2-x_1)(y_3-y_1),0)$,
$(0,(y_1-y_3)(x_2-x_1)+(y_2-y_1)(x_3-x_1))$.
You can verify this by plugging in the vectors $(x_2-x_1,y_2-y_1),(x_3-x_1,y_3-y_1)$ into the transformation.
So the double integral becomes under the transformation:
$$\iint_{S}1dydx=\iint_{T}\left | \frac{\partial(x,y)}{\partial(u,v)}\right | dudv=\left |\frac{1}{(y_1-y_2)(x_3-x_1)-(y_1-y_3)(x_2-x_1)} \right |\iint_{T}1dudv$$
Now we can use geometry to compute $\iint_{T}1dudv$, which is the area of $T$. Since $T$ is a right triangle, we just need to multiply its base and height and divide by $2$. Looking at the vertices, the base (measured on the $u$ axis) is $$\left|(y_1-y_2)(x_3-x_1)+(x_2-x_1)(y_3-y_1)\right|$$ and the height (measured on the $v$ axis) is $$\left|(y_1-y_3)(x_2-x_1)+(y_2-y_1)(x_3-x_1)\right|$$
Notice that the base cancels out with the the Jacobian determinant.
So the area of $S$ is: $$\iint_{S}1dydx=\frac{\left|(y_1-y_3)(x_2-x_1)+(y_2-y_1)(x_3-x_1)\right|}{2}$$
Now expand out $$\frac{\left|(y_1-y_3)(x_2-x_1)+(y_2-y_1)(x_3-x_1)\right|}{2}$$
to get:
 $$\frac{\left|y_1x_2-y_3x_2+y_3x_1+x_3y_2-x_3y_1-x_1y_2\right|}{2}$$
Rearrange the terms around:
 $$\frac{\left|-x_2y_3+y_2x_3+x_1y_3-y_1x_3-x_1y_2+x_2y_1\right|}{2}$$
And Use the fact $$x_iy_j-x_jy_i= \begin {vmatrix} x_i && x_j \\ y_i && y_j \\ \end{vmatrix}$$
to see:
$$\frac{\left|-x_2y_3+y_3x_3+x_1y_3-y_1x_3-x_1y_2+x_2y_1\right|}{2}=\frac{\left|-\begin {vmatrix} x_2 && x_3 \\ y_2 && y_3 \\ \end{vmatrix}+\begin {vmatrix} x_1 && x_3 \\ y_1 && y_3 \\ \end{vmatrix}-\begin {vmatrix} x_1 && x_2 \\ y_1 && y_2 \\ \end{vmatrix}\right|}{2}$$
Factoring out an $\left |-1\right |$ in the numerator, we see this is just $$\frac{\left|\begin {vmatrix} x_2 && x_3 \\ y_2 && y_3 \\ \end{vmatrix}-\begin {vmatrix} x_1 && x_3 \\ y_1 && y_3 \\ \end{vmatrix}+\begin {vmatrix} x_1 && x_2 \\ y_1 && y_2 \\ \end{vmatrix}\right|}{2}$$.
This is just (by the reverse of cofactor expansion):
$$\frac{1}{2}\left |\begin {vmatrix} 1 && 1 && 1 \\ x_1 && x_2 && x_3 \\ y_1 &&y_2&&y_3 \\ \end{vmatrix}\right |,$$ which verifies the determinant formula. 
A: First, note that $\begin{vmatrix} 1&1 &1 \\ x_1& x_2 &x \\ y_1& y_2& y \end{vmatrix}=0$ is an equation for the line $AB$. Clearly, because when we fill in for $A$ and $B$ the equation holds, and when working out the determinant, the equation is linear.
Also, the coefficients of $x$ and $y$, say $c_x$ and $c_y$, can be found by expanding the determinant using minors and cofactors.
Let $F(x,y)=\begin{vmatrix} 1&1 &1 \\ x_1& x_2 &x \\ y_1& y_2& y \end{vmatrix}$, and let $d$ be the distance from $C$ to $AB$.
Then the area equals $$\frac{1}{2}|AB|\cdot d=\frac{1}{2}|AB|\cdot\frac{|F(x_3,y_3)|}{\sqrt{{c_x}^2+{c_y}^2}}$$
Now, try to simplify this and you'll get the desired result.
A: Assuming the OP means the area of a triangle $ABC$:
Consider the triangle $\Delta OAB$ formed by the points $(0,0)$, $(x_1,y_1)$, and $(x_2,y_2)$.  The area of this triangle is given by 
$$A_{\Delta OAB} = \frac{1}{2} (x_1 y_2-x_2 y_1)$$
Why?  Because the area of a triangle formed by vectors $\vec{a}$ and $\vec{b}$ from the origin, having side lengths $a$ and $b$, respectively, and included angle $\theta$ is $(1/2) a b \sin{\theta} = (1/2) (\vec{a} \times \vec{b}) \cdot \vec{k}$, the length of the cross-product of $a$ and $b$, where $\vec{k}$ is a vector normal to the plane containing $\Delta OAB$, direction determined by the right-hand rule.  In terms of the coordinates of $\vec{a}$ and $\vec{b}$, $(x_1,y_1)$, and $(x_2,y_2)$, the cross product is $(x_1 y_2-x_2 y_1)$.
For a triangle $ABC$ at 3 arbitrary points in the plane, where $\vec{c} = (x_3,y_3)$, form the sum of the areas of $\Delta OAB$, $\Delta OBC$, and $\Delta OCA$/:
$$A_{\Delta ABC} = \frac{1}{2} (x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3) $$ 
Note the order of the points in the triangle is important: the points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ are in counter-clockwise order.
A: Hint: Try it and see.
For simplicity, draw the triangle in the first quadrant and drop perpendiculars from $A$, $B$, $C$ onto $A^\prime$, $B^\prime$, $C^\prime$ on the $x$-axis. The area of the triangle will be some combination (sum and difference) of the areas of the trapezoids $\square ABB^\prime A^\prime$, $\square BCC^\prime B^\prime$, $\square CAA^\prime C^\prime$. Write the areas of the trapezoids in terms of the coordinates, combine appropriately, and note that the answer agrees (in absolute value) with the half-determinant formula; then convince yourself that the same process works no matter how or where the triangle is positioned.
In the process of this elementary approach, you may come to appreciate the wisdom (and convenience) of @Hagen's answer.
A: I saw you guys' proofs. 
To be honest, it is a very simple linear algebra problem, and you only need to know very shallow knowledge of the determinant properties to solve it, which determinants work as area or volume in many calculations. And some theorem of the determinants.
First, you need to know $\frac{1}{2}\det[\mathbf{v_1},\mathbf{v_2}]$ is the area formula of the triangle whose vertices are $\mathbf{v_1},\mathbf{v_2},\mathbf{0}$ (in $R^2$), which is very easy to proof. If you still do not know it yet, remember it.
Next, translating the vertices points by the same vector at the same time does not change the area of the triangle (if the world is sane), which means if you have a triangle whose vertices are $\mathbf{a_1}, \mathbf{a_2},\mathbf{a_3}$, you can translate $\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}$ by $\mathbf{-a_1}$ and you will get new vertice vectors, a zero vector and $\mathbf{a_2-a_1}, \mathbf{a_3-a_1}$. $\frac{1}{2}\det[\mathbf{a_2-a_1}, \mathbf{a_3-a_1}]$ is the area of the triangle.
Finally, By transpose theorem, if A is an $n\times n$ matrix, $\det A=\det A^T$, and row operation theorem, If a multiple of one row of A is added to another row to produce a matrix B,
then $\det B = \det A$.
Transpose the given matrix, and then substract the first row from the rest, expand the cofactors at column 1. Then transpose back, you will get it. (I do not really want to write the details down because it is so simple)
A: Here's a proof I came up with a few years back $\to$ Let $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ be the coordinates for a triangle in the $xy$-plane with area $A$. 
$xy$-plane" />

Lift this triangle into the plane described by $z = 1$, by appending a $1$ to each of the coordinate pairs.



Then the signed volume of the tetrahedra spanned by the vectors $(x_1, y_1, 1), (x_2, y_2, 1)$ and $(x_3, y_3, 1)$ is given by:
$$V = \frac{1}{6}\det\begin{bmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{bmatrix}$$
As shown in this answer, but the absolute volume is usually calculated with the formula $|V| = \frac{1}{3}Ah$ where $A$ is the absolute area of the triangle and $h$ is the height of the tetrahedron $h = 1$. Therefore:
$$|V| = \frac{1}{6}\left|\det\begin{bmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{bmatrix}\right| = \frac{1}{3}A \;\;\; \to \;\;\; A = \frac{1}{2}\left|\det\begin{bmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{bmatrix}\right| = \frac{1}{2}\left|\det\begin{bmatrix}1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{bmatrix}\right|$$
Where the last equality comes from the fact that the absolute value of the determinant is invariant to row substitutions - see this answer.
A: Another possibility is to use the formal properties of the determinant and see how they correspond to the properties of the area. This seems lengthy, but it explains also why you have such a relation between determinant and area.
You start with the determinant:
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 
\end{array} \right|
$$
then you can subtract from the second row $x_1$ times the first row and from the third row you can subtract $y_1$ times the first row to get:
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & x_2-x_1 & x_3-x_1 \\
0 & y_2-y_1 & y_3-y_1 
\end{array} \right|.
$$
This operation does not change the determinant (which is multilinear). And it does not change the area of the triangle, because it corresponds to a translation, namely the translation of vector $-(x_1,y_1)$ which sends the first vertex of the triangle to the origin. 
Now suppose to fix ideas that $0<x_3<x_2$ and consider the triangle (which now has vertices: $p_1= (0,0)$, $p_2=(x_2-x_1,y_2-y_1)$, $p_3=(x_3-x_1, y_3-y_1)$) divided in two by the line $x=x_3-x_1$, and let $p$ be the point with $x$ coordinate equal to $p_3$ which lies on the edge $p_1 p_2$. If we consider $pp_3$ as the base of the two smaller triangles, you see that you can move the vertices $p_2$ and $p_1$ vertically (which means parallel to the common base $pp_3$ without changing the area of the resulting quadrilateral (which is the union of the two smaller triangles). Or, which is the same, you can keep $p_1$ fixed and move the base $pp_3$ vertially. If you want to move the triangle so that both the point $p$ and $p_2$ go to the $x$-axis, you find the mapping 
$$
\begin{cases}
  x' = x\\ 
  y' = y - \frac{y_2-y_1}{x_2-x_1}x
\end{cases}
$$ 
which correspond to subtract from the third line of the matrix ($y$ coordinates) a multiple of the second line ($x$ coordinate). So, again, this transformation maintains the area of the triangle, and the determinant of the matrix.
You get:
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & x_2-x_1 & x_3-x_1 \\
0 & 0       & (y_3-y_1)-\frac{y_2-y_1}{x_2-x_1}(x_3-x_1)
\end{array} \right|.
$$
So now you have a triangle (with same area of the original one) whose basis is on the $x$ axis has length $x_2-x_1$ and whose height is given by the $y$ coordinate of the third point which is the entry in the lower-right corner of the matrix. So the area is half the product of the entries on the diagonal, and in fact in this case (triangular matrix) this is indeed the determinant of the matrix.
You could also complete the diagonalization process, by moving the third vertex to the $y$ axis. In that case you would get a right triangle and a diagonal matrix.
