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.