I've stumbled upon the determinant way of testing line concurrency which says that, if three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ and $a_3x+b_3y+c_3=0$ are concurrent, then $$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$$

I want to understand this using the concept of vectors. So far I can tell that this is referring to a transformation along the line normals and showing that it will have a 1-D span.

But what does the matrix have to do with $c_1,c_2,c_3$ since the only $a,b$ define the normal of the line?

  • $\begingroup$ The concurrency of three lines should depend symmetrically on the three lines involved, however it is calculated. Also this will be more than a pure vector question as the concurrency of three vectors depends not just on their direction, but also on how they are positioned in space - take three concurrent lines and translate one, and the lines will no longer be concurrent, though the direction vectors will be the same. $\endgroup$ Commented Oct 25, 2018 at 7:51
  • $\begingroup$ Do you understand why, if you consider the matrix entries as coordinates of points, the vanishing determinant means that the points are colinear? $\endgroup$
    – amd
    Commented Oct 25, 2018 at 8:04
  • 3
    $\begingroup$ Given three lines, generally they intersect in three points. The null determinant is equivalent to the affirmation that the area of the triangle with vertex at those three points is null. $\endgroup$
    – Cesareo
    Commented Oct 25, 2018 at 8:26
  • $\begingroup$ @MarkBennet That thought did occur to me, but that still doesn't explain the presence of $c$ particularly. $\endgroup$ Commented Oct 25, 2018 at 15:29
  • $\begingroup$ @amd Yes, I understand that. But what I don't understand here, is the coordinates being the coefficients of the lines. $\endgroup$ Commented Oct 25, 2018 at 15:37

1 Answer 1


The three lines are concurrent iff the system of equations is consistent. This will be the case when the augmented matrix $$A = \left[\begin{array}{cc|c}a_1&b_1&-c_1\\a_2&b_2&-c_2\\a_3&b_3&-c_3\end{array}\right]$$ has the same rank as the $3\times2$ coefficient matrix. The latter has rank at most 2, so a necessary condition is that $\det A=0$. Multiplying a column by $-1$ changes the sign of the determinant, but we only care about its vanishing, hence we can also use the matrix in your question.

In order to understand this in terms of normals, it might be helpful to make the equations homogeneous by moving to $\mathbb R^3$. The equation $a_ix+b_iy+c_iz=0$ describes a plane through the origin with normal $(a_i,b_i,c_i)^T$. For the three planes to intersect somewhere other than the origin, the homogeneous system’s coefficient matrix must be rank-deficient, which in turn is equivalent to its determinant vanishing. We connect these planes back to lines in $\mathbb R^2$ by intersecting them with the plane $z=1$. (This is one of the standard models of the real projective plane.)

  • $\begingroup$ The projective plane interpretation also relates to the argument that if the lines coincide at $(x_0,y_0)$, then $(x_0,y_0,1)$ is in the null space of the original matrix - where $[x_0 : y_0 : 1]$ is the homogeneous coordinates of $(x_0,y_0)$ after embedding $\mathbb{R}^2 \hookrightarrow \mathbb{R} \mathbb{P}^2$. $\endgroup$ Commented Oct 25, 2018 at 21:14
  • $\begingroup$ @DanielSchepler Yep. I didn’t really want to elaborate further here, though. My first thought when seeing the question was “point-line duality.” $\endgroup$
    – amd
    Commented Oct 25, 2018 at 21:16

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