Area of Triangle Formed by Three Lines

Area of the triangle formed by three lines a₁x+b₁y+c₁=0, a₂x+b₂y+c₂=0, a₃x+b₃y+c₃=0.

Given three lines $L_1: a_1x+b_1y+c_1=0$, $L_2: a_2x+b_2y+c_2=0$, $L_3: a_3x+b_3y+c_3=0$.

Derivation

Find the three vertices by solving each pair of lines:

• $A = L_2 \cap L_3$
• $B = L_1 \cap L_3$
• $C = L_1 \cap L_2$

Each vertex is found by Cramer's rule. For example:

$$A: \quad x = \frac{b_2c_3-b_3c_2}{a_2b_3-a_3b_2}, \quad y = \frac{a_3c_2-a_2c_3}{a_2b_3-a_3b_2}$$

Substituting all three vertices into the area formula $\Delta = \frac{1}{2}|x_A(y_B-y_C)+\ldots|$ and simplifying gives:

$$\boxed{\Delta = \frac{1}{2}\left|\frac{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}^2}{(a_1b_2-a_2b_1)(a_2b_3-a_3b_2)(a_3b_1-a_1b_3)}\right|}$$

The numerator is the square of the $3\times 3$ determinant of the coefficients. The denominator is the product of the three $2\times 2$ determinants of each pair of lines.

Note: When $\Delta = 0$ in the numerator, the three lines are concurrent (no triangle formed).