Conics consisting of two points/lines makes them rank 2 While studying conics, I came across this concept and example:

Degenerate conics. If the matrix $C$ is not of full rank, then the conic is termed degenerate. Degenerate point conics include two lines (rank 2), and a repeated line (rank 1).
Example. The conic
$$C = \mathbf{l}\mathbf{m}^T + \mathbf{m} \mathbf{l}^T$$
is composed of two lines $\mathbf{l}$ and $\mathbf{m}$. Points on $\mathbf{l}$ satisfy $\mathbf{l}^T \mathbf{x} = 0$, and are on the conic since $\mathbf{x}^T C \mathbf{x} = (\mathbf{x}^T \mathbf{l})(\mathbf{m}^T \mathbf{x}) + (\mathbf{x}^T \mathbf{m})(\mathbf{l}^T \mathbf{x}) = 0$. Similarly, points satisfying $\mathbf{m}^T \mathbf{x} = 0$ also satisfy $\mathbf{x}^T C \mathbf{x} = 0$. The matrix $C$ is symmetric and has rank 2. The null vector is $\mathbf{x} = \mathbf{l} \times \mathbf{m}$ which is the intersection point of $\mathbf{l}$ and $\mathbf{m}$.
Degenerate line conics include two points (rank 2), and a repeated point (rank 1). For example, the line conic $C^* = \mathbf{x} \mathbf{y}^T + \mathbf{y} \mathbf{x}^T$ has rank 2 and consists of lines passing through either of the two points $\mathbf{x}$ and $\mathbf{y}$. Note that for matrices that are not invertible $(C^*)^* \not= C$.

I'm wondering why these conics consisting of two points/lines makes them rank 2 (and why is the repeated point for the latter rank 1)? I'd really appreciate clarification of this example. Thank you.

For the two-point/line degenerate conics, the explanation is already there in the text: “The null vector is $\mathbf x=\mathbf l\times\mathbf m$” [emphasis mine]. We can drill down into this statement a bit, though.  
What is the dimension of the null space of $\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$? Well, $$(\mathbf l\mathbf m^T+\mathbf m\mathbf l^T)\mathbf x = (\mathbf m^T\mathbf x)\mathbf l+(\mathbf l^T\mathbf x)\mathbf m = 0.\tag{*}$$ If $\mathbf l$ and $\mathbf m$ are linearly independent, in which case they represent distinct lines, (*) implies that $\mathbf l^T\mathbf x = \mathbf m^T\mathbf x = 0$, in other words, that $\mathbf x$ is orthogonal to both $\mathbf l$ and $\mathbf m$. These vectors are all elements of $\mathbb R^3$, so $\dim\operatorname{span}\{\mathbf l,\mathbf m\} = 2$, and the dimension of its orthogonal complement and therefore also the nullity of $\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$ is $1$. Indeed, the orthogonal complement of the span of $\mathbf l$ and $\mathbf m$ is spanned by $\mathbf l\times\mathbf m$.  
On the other hand, if $\mathbf l$ and $\mathbf m$ are linearly dependent, so that both represent the same line, then $\mathbf l = c\mathbf m$ for some $c\ne0$, and $\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$ is a scalar multiple of $\mathbf m\mathbf m^T$. If $\mathbf m\mathbf m^T\mathbf x=0$, then we must have $\mathbf m^T\mathbf x=0$, so the null space of the matrix consists of all vectors orthogonal to $\mathbf m$. This is a two-dimensional space, making the rank of the matrix $1$. One can also see this directly: the columns of $\mathbf m\mathbf m^T$ are all scalar multiples of $\mathbf m$, so its column space is spanned by $\mathbf m$—its rank is $1$.
