# “Conic sections” that are really just two straight lines

My teacher was teaching co-ordinate geometry and today he said that the following equation will always represent a conic section:$$ax^2+by^2+2hxy+2gx+2fy+c=0$$ Then he said that if the determinant of the following matrix is zero then the equation can be split into two linear factors, and thus will represent a couple of straight lines.$$\begin{bmatrix}a & h & g\\h & b & f\\ g & f & c\end{bmatrix}$$ What is the proof of this theorem?

• A very useful observation is that if that matrix is $A$ and we set $v = (\begin{matrix}x & y & 1 \end{matrix})^T$, then the conic is given by the equation $$v^T A v = 0$$ – Hurkyl Aug 11 '14 at 16:53

When a conic consists of two lines, moreover called a degenerate conic, it can be written as follows:

$$(ax+by+c)(dx+ey+f)=0$$

Its determinant is (MathGem's hint 2):

$$\frac{1}{8}\begin{vmatrix} 2ad & ae+bd & af+cd \\ ae+bd & 2be & bf+ce \\ af+cd & bf+ce & 2cf\end{vmatrix}$$

which is $0$.

Hint:

1. Ask yourself when a conic section becomes degenerate?

2. Calculate the determinant of that matrix.

Can you see a relation between 1 and 2?