If the edges of a triangle of a triangle are k, l, and m, show that $k^2-lm$ is never a degenerate

Question

If $k$, $l$, $m$ are the edges of a triangle with non-zero area, how can you show that

$$k^2-lm = 0$$

this will not reduce to a product to two lines $l_1l_2=0$?

Answer 1

Let

$$\begin{cases}a_1x+b_1y+c_1&=&0& \ (k)\\ a_2x+b_2y+c_2&=&0& \ (l)\\ a_3x+b_3y+c_3&=&0& \ (m)\\\end{cases}$$

$$k^2-lm=$$

$$=\begin{pmatrix}x \ y \ 1\end{pmatrix}\begin{pmatrix}a_1\\ b_1\\ c_1\end{pmatrix} \begin{pmatrix}a_1 b_1 c_1\end{pmatrix}\begin{pmatrix}x\\ y\\ 1\end{pmatrix}-\begin{pmatrix}x \ y \ 1\end{pmatrix}\begin{pmatrix}a_2\\ b_2\\ c_2\end{pmatrix}\begin{pmatrix}a_3 \ b_3 \ c_3\end{pmatrix}\begin{pmatrix}x\\ y\\ 1\end{pmatrix}$$

$$=\begin{pmatrix}x \ y \ 1\end{pmatrix}\left[\begin{pmatrix}a_1\\ b_1\\ c_1\end{pmatrix} \begin{pmatrix}a_1 b_1 c_1\end{pmatrix}-\begin{pmatrix}a_2\\ b_2\\ c_2\end{pmatrix}\begin{pmatrix}a_3 \ b_3 \ c_3\end{pmatrix}\right]\begin{pmatrix}x\\ y\\ 1\end{pmatrix}$$

$$=\begin{pmatrix}x \ y \ 1\end{pmatrix}\underbrace{\left[\begin{pmatrix}a_1&a_2\\ b_1&b_2\\ c_1&c_2\end{pmatrix} \begin{pmatrix}\ a_1 & \ b_1 & \ c_1\\-a_3 & -b_3 & -c_3\end{pmatrix}\right]}_M\begin{pmatrix}x\\ y\\ 1\end{pmatrix}$$

Matrix $M$ is a rank-2 matrix.

Whereas the matrix $N$ associated with $l_1l_2=(ax+by+c)(a'x+b'y+c')=0$ is

$$=\begin{pmatrix}x \ y \ 1\end{pmatrix}\underbrace{\left[\begin{pmatrix}a\\ b\\ c\end{pmatrix} \begin{pmatrix}a'& b'& c'\end{pmatrix}\right]}_N\begin{pmatrix}x\\ y\\ 1\end{pmatrix}$$

is rank-1.

Remark : This question could be casted into the framework of pencil of conics.