Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane. I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equations :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
$$a_3x+b_3y+c_3=0$$
Then the area of the triangle that these lines will enclose is given by the magnitude of :
$$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$
Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.
What I'm wondering is, where did this come from? And why isn't it famous? Earlier we had to calculate areas by finding the vertices and all but this does it in a minute or so and thus deserves more familiarity.

equations in the reduced form
$y=ax+b$
$y=cx+d$
$y=ex+f$
solving by the Cramer rule we get the A, B and C points
calculating the area of the ABC triangle
$ΔABC=\frac{1}{2}\left| \begin{array}{} \frac{b-d}{c-a} & \frac{bc-ad}{c-a} & 1 \\ \frac{d-f}{e-c} & \frac{de-cf}{e-c} & 1 \\ \frac{f-b}{a-e} & \frac{af-be}{a-e} & 1 \\ \end{array} \right| $
$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left| \begin{array}{} b-d & bc-ad & c-a \\ d-f & de-ef & e-c \\ f-b & af-be & a-e \\ \end{array} \right| $
$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left [ a(d-f)+c(f-b)+e(b-d) \right ] ^{2}$
$ΔABC=\frac{1}{2}\frac{\left|\begin{array}{} a & b & 1 \\ c & d & 1\\ e & f & 1 \\ \end{array}  \right| ^{2}}{\left| \begin{array}{} a & a^{2} & 1 \\ c & c^2 & 1 \\ e & e^2 & 1 \\ \end{array} \right| }$
$a=\frac{-a_1}{b_1}$
$b=\frac{-c_1}{b_1}$
$c=\frac{-a_2}{b_2}$
$d=\frac{-c_2}{b_2}$
$e=\frac{-a_3}{b_3}$
$f=\frac{-c_3}{b_3}$
$ΔABC=|\frac{1}{2}\frac{\left| \begin{array}{} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \\ \end{array} \right| ^2}{\left| \begin{array}{} a_1^2 & b_1^2 & a_1b_1 \\ a_2^2 & b_2^2 & a_2b_2 \\ a_3^2 & b_3^2 & a_3b_3 \\ \end{array} \right| }|$
$ΔABC=|\frac{1}{2}\frac{Δ^2}{δ}|$

