Condition for Concurrency of Three Lines

Three lines a₁x+b₁y+c₁=0, a₂x+b₂y+c₂=0, a₃x+b₃y+c₃=0 are concurrent if and only if this determinant vanishes.

Three lines $L_1: a_1x+b_1y+c_1=0$, $L_2: a_2x+b_2y+c_2=0$, $L_3: a_3x+b_3y+c_3=0$ are concurrent if they all pass through a common point.

Derivation

Find the intersection of $L_1$ and $L_2$ by solving simultaneously. By Cramer's rule:

$$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \qquad y = \frac{a_2c_1 - a_1c_2}{a_1b_2 - a_2b_1}$$

For $L_3$ to pass through this point, substitute into $L_3 = 0$:

$$a_3(b_1c_2-b_2c_1) + b_3(a_2c_1-a_1c_2) + c_3(a_1b_2-a_2b_1) = 0$$

This is exactly the expansion of:

$$\boxed{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0}$$

Alternative via family of lines

$L_3$ is concurrent with $L_1$ and $L_2$ if and only if $L_3$ is a member of the family $L_1 + \lambda L_2 = 0$ for some $\lambda$ — i.e., $L_3 = L_1 + \lambda L_2$ for some $\lambda$. This means the rows of the determinant are linearly dependent, which is exactly the $\Delta = 0$ condition.