Intersection of lines with submanifold

I come across an interesting problem below which confuses me a lot.

Let $$L$$ be a submanifold of $$\mathbb{R}^n$$ with codimension > 1 . Prove that: (1) if $$x$$ $$\notin L$$, for almost every line $$l$$ passing through $$x$$, we have $$l$$ $$\cap$$ $$L$$ = $$\emptyset$$. (2) if $$x$$ $$\in L$$, for almost every line $$l$$ passing through $$x$$, we have $$l$$ $$\cap$$ $$L$$ = {$$x$$}.

The definition of "almost every" is not given clearly. I guess it means that when we equate lines passing through $$x$$ with $$\mathbb{R}^n$$\{$$x$$} the points can't satisfy the results above are measured zero.

Edit: It is more proper to consider measure zero in $$\mathbb{RP}^{n-1}$$

Considering conclusion containing "almost every", I attempted to use parametric transversality theorem to solve it. If we can prove that almost every line intersects transversally with $$L$$ while it is impossible for $$l$$ $$\cap$$ $$L$$ $$\neq$$ $$\emptyset$$ due to the restriction on the codimension. However, I found this is ridiculous because when $$n=3$$ , $$L$$ also a line whose codimension is 2 and $$x \in L$$ , every line passing through $$x$$ does not intersect transversally with $$L$$ . So we can't solve both situations at the same time as preconceived.

Is my idea correct? If it is, how should I construct the function satisfying the conditon for parametric transversality theorem? Appreciate for any help!

Parametric Transversality Theorem: Suppose that $$F:X\times S\to Y$$ is a smooth map of manifolds and $$Z$$ is a submanifold of $$Y$$, all manifolds without boundary. If $$F$$ is transverse to $$Z$$ then for almost every $$s\in S$$ the map $$f_s : x\mapsto F(x,s)$$ is transverse to $$Z$$.

Let $$x\in \mathbb R^n$$. Define $$F: (L\setminus \{x\}) \times \mathbb R\to \mathbb R^n$$, $$F(y, t) = tx + (1-t)y.$$
Clearly this map is smooth. Since $$(L\setminus \{x\} \times \mathbb R)$$ has dimension strictly less then $$n= \dim \mathbb R^n$$, the image of $$F$$ is of measure zero.
That means (i) when $$x\notin L$$, the union of the sets of lines in $$\mathbb R^n$$ passing through $$x$$ and $$L$$ is of measure zero in $$\mathbb R^n$$, and when (ii) $$x\in L$$, the union of sets of lines joining $$x$$ to some $$y\in L$$ in $$\mathbb R^n$$ is of measure zero.
In general, there is a more natural way to talk about measure zero in this setting: for each $$x\in \mathbb R^n$$, let $$\mathbb{RP}^{n-1}_x$$ be the set of lines in $$\mathbb R^n$$ passing through $$x$$. Then $$\mathbb {RP}^{n-1}_x$$ is a smooth manifold of dimension $$n-1$$ and "sets of measure zero" is well defined. Define $$G : L\setminus\{x\} \to \mathbb{RP}^{n-1}_x,$$ where for each $$y\in L$$, $$G(y)$$ is the line in $$\mathbb R^n$$ passing though $$x, y$$. Again one can check that $$G$$ is smooth. Since $$L$$ has codimension $$>1$$, $$\dim L < n-1=\dim \mathbb{RP}^{n-1}_x$$. Thus the image is of measure zero.
Note that we use just a weak form of Sard theorem (about smooth maps $$f : M \to N$$ with $$\dim M < \dim N$$).