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.

Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":

$$\begin{align} x \cos\theta + y \sin\theta - p &= 0 \\ x \cos\phi + y \sin\phi - q &= 0 \\ x \cos\psi + y \sin\psi - r &= 0 \end{align}$$

with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:

Then

$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ Likewise, $$C_2 = \sin B \qquad C_3 = \sin C$$

Moreover, $$D := \left|\begin{array}{ccc} \cos\theta & \sin\theta & - p \\ \cos\phi & \sin\phi & - q \\ \cos\psi & \sin\psi & - r \end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$

Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$

Therefore,

$$\begin{align} D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt] &= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt] &= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt] &= -\frac{2\;|\triangle ABC|}{d} \end{align}$$

Also, $$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$

Finally: $$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$

Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ be the vertices of the triangle. Then the three non-concurrent, non parallel lines can be represented by: $$(y_2-y_1)(x-x_1)-(x_2-x_1)(y-y_1)=0$$ $$(y_3-y_2)(x-x_2)-(x_3-x_2)(y-y_2)=0$$ $$(y_1-y_3)(x-x_3)-(x_1-x_3)(y-y_3)=0$$ The coefficients in the OP's question are thus represented by: $$ a_1 = (y_2-y_1) \quad ; \quad b_1 = -(x_2-x_1) \quad ; \quad c_1 = -(y_2-y_1)x_1+(x_2-x_1)y_1 \\ a_2 = (y_3-y_2) \quad ; \quad b_2 = -(x_3-x_2) \quad ; \quad c_2 = -(y_3-y_2)x_2+(x_3-x_2)y_2 \\ a_3 = (y_1-y_3) \quad ; \quad b_3 = -(x_1-x_3) \quad ; \quad c_3 = -(y_1-y_3)x_3+(x_1-x_3)y_3 $$ Now straightforward calculation should reveal that: $$ \frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3} =\frac{1}{2} det\begin{bmatrix}(x_2-x_1) & (y_2-y_1)\\(x_3-x_1) & (y_3-y_1)\end{bmatrix} $$ Where the latter (half) determinant certainly represents the area of the triangle:

The algebra is somewhat tedious. Therefore I've invoked MAPLE to save time and effort:

A := array([[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]]);
C1 := det(array([[a2,b2],[a3,b3]]));
C2 := -det(array([[a1,b1],[a3,b3]]));
C3 := det(array([[a1,b1],[a2,b2]]));
B := array([[x2-x1,y2-y1],[x3-x1,y3-y1]]);
verify(det(A)^2/(2*C1*C2*C3),det(B)/2,equal);
                        true

Following is a derivation of the formula using homogeneous coordinates.

For any point $p = (u,v)$ on the plane, we will use $\vec{p} = (x,y,z) = (zu,zv, z)$ to denote any choice of its homogeneous coordinates. For any line $\ell = \{ (u,v) : au+bv+c = 0 \}$, we will associate it with a vector $\vec{\ell} = (a,b,c)$ as use it as a homogeneous coordinates of $\ell$.

In terms of homogeneous coordinates, the condition that point $p$ lies on line $\ell$ can be expressed as

$$au + bv + c = 0 \quad\iff\quad ax + by + cz = 0 \quad\iff\quad \vec{\ell}\cdot \vec{p} = 0$$

Let $\ell_1, \ell_2, \ell_3$ be $3$ lines on the plane forming the sides of a triangle $T$ in counterclockwise order.

Let $p_1 = (u_1,v_1)$ be the intersection of $\ell_2$ and $\ell_3$, we have $$p_1 \in \ell_2 \cap \ell_3 \quad\implies\quad \vec{\ell}_2\cdot \vec{p}_1 = \vec{\ell}_3 \cdot \vec{p}_1 = 0 \quad\implies\quad \vec{p}_1 \propto \vec{\ell}_2 \times \vec{\ell}_3$$ Let $p_2 = (u_2,v_2)$ be the intersection of $\ell_3$ and $\ell_1$, we have $$p_2 \in \ell_3 \cap \ell_1 \quad\implies\quad \vec{\ell}_3\cdot \vec{p}_2 = \vec{\ell}_1 \cdot \vec{p}_2 = 0 \quad\implies\quad \vec{p}_2 \propto \vec{\ell}_3 \times \vec{\ell}_1$$ Let $p_3 = (u_3,v_3)$ be the intersection of $\ell_1$ and $\ell_2$. we have $$p_3 \in \ell_1 \cap \ell_2 \quad\implies\quad \vec{\ell}_1\cdot \vec{p}_3 = \vec{\ell}_2 \cdot \vec{p}_3 = 0 \quad\implies\quad \vec{p}_3 \propto \vec{\ell}_1 \times \vec{\ell}_2 $$

Since $\vec{p}_1,\vec{p}_2,\vec{p}_3$ are homogeneous coordinates and rescaling them doesn't change the underlying points $p_1, p_2, p_3$, we can just pick $$ \vec{p}_1 = \vec{\ell}_2 \times \vec{\ell}_3,\quad \vec{p}_2 = \vec{\ell}_3 \times \vec{\ell}_1\quad\text{ and }\quad \vec{p}_3 = \vec{\ell}_1 \times \vec{\ell}_2 $$

In terms of them, the area of triangle $T$ is given by $$\verb/Area/(T) = \frac12 \left| \begin{matrix} u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ 1 & 1 & 1\\ \end{matrix}\right| = \frac{1}{2z_1z_2z_3} \left|\begin{matrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\\ z_1 & z_2 & z_3\\ \end{matrix}\right| = \frac{ \vec{p}_1\cdot (\vec{p_2} \times \vec{p}_3) }{ 2 (\vec{p}_1\cdot \hat{z})(\vec{p}_2\cdot \hat{z})(\vec{p}_3\cdot \hat{z}) } $$

Notice $$\require{cancel} \vec{p}_2 \times \vec{p}_3 = (\vec{\ell}_3 \times \vec{\ell}_1) \times (\vec{\ell}_1 \times \vec{\ell}_2) = ((\vec{\ell}_3 \times \vec{\ell}_1) \cdot \vec{\ell}_2) \vec{\ell}_1 - \color{red}{\cancelto{0}{\color{gray}{((\vec{\ell}_3 \times \vec{\ell}_1) \cdot \vec{\ell}_1)}}} \vec{\ell}_2 = \Delta \vec{\ell}_1 $$ where $\Delta = \vec{\ell}_1\cdot (\vec{\ell}_2 \times \vec{\ell}_3) = \left|\begin{matrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{matrix}\right|$, we find $$\vec{p}_1\cdot(\vec{p}_2\times\vec{p}_3) = \Delta( \vec{\ell}_2 \times \vec{\ell}_3 ) \cdot \vec{\ell}_1 = \Delta^2 $$ Together with $$\vec{p}_1 \cdot \hat{z} = (\vec{\ell}_2 \times \vec{\ell}_3 ) \cdot \hat{z} = \left|\begin{matrix} a_2 & a_3 & 0\\ b_2 & b_3 & 0\\ c_2 & c_3 & 1\\ \end{matrix}\right| = C_1 \stackrel{def}{=} \left|\begin{matrix} a_2 & a_3\\ b_2 & b_3\\ \end{matrix}\right| $$ and similar relations $$\vec{p}_2\cdot\hat{z} = C_2 \stackrel{def}{=} \left|\begin{matrix} a_3 & a_1\\ b_3 & b_1\\ \end{matrix}\right| \quad\text{ and }\quad \vec{p}_3\cdot\hat{z} = C_3 \stackrel{def}{=} \left|\begin{matrix} a_1 & a_2\\ b_1 & b_2\\ \end{matrix}\right|$$ we obtain $$\verb/Area/(T) = \frac{\Delta^2}{2C_1C_2C_3} = \frac{\left|\begin{matrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{matrix}\right|^2 }{2C_1C_2C_3} $$