Is the set of points or the set of lines on a plane "larger"? Is the set of points or the set of lines on a plane "larger",or there is a 1-1 correspondence between lines and points?
I assume the question asked by Marios was about the real plane, but planes over finite fields might also lead to interesting statements. First, let's take an example. Let $P$ be the affine plane over the two-element field $\mathbb{F}_2$. There are 4 points in this plane, but there are 6 lines (i.e. one-dimensional affine spaces) in this plane. 

The proof for $\mathbb{R}$

We have a one-to-one mapping from $\mathbb{R}^2 \setminus\{(0,0)\}$ to $\mathcal{L}$ (I couldn't find simpler) : take any element $(a,b)$ and form the line $d_{a,b}$ which is tangent at the circle of center $0$ and radius $\sqrt{a^2 + b^2}$ at the point $(a,b)$. This is injective. It is not surjective, as the lines going through $(0,0)$ (let's denote it by $\mathbb{P}_1(\mathbb{R})$ ) do not have the form $d_{a,b}$.

Therefore, the set $\mathcal{L}$ have the same cardinality as $\mathbb{P}_1(\mathbb{R}) \cup \mathbb{R}^2 \setminus\{(0,0)\}$. But it is known that there is a bijection $\phi$ between $\mathbb{P}_1(\mathbb{R})$ and $\mathbb{R} \cup \infty$ : if $d \in \mathbb{P}_1(\mathbb{R})$ , let $\phi(d)$ be $x$, where $(x,1)$ is the intersection between $d$ and the horizontal line $y=1$. If $d$ is the line $y=0$, then $\phi(d) = \infty$. This is a bijection (it is clear on a drawing).

To conclude, we have a bijection of $\mathcal{L}$ onto $\mathbb{R}^3$. It can be shown that there is a bijection of $\mathbb{R}^3$ to $\mathbb{R}$.

The general case

In the general case, we might expect the number of lines to be "greater" than the plane. 

First, note that there is a bijection of the projective space $\mathbb{P}_1(\Bbbk)$ onto $\Bbbk \cup \infty$, where $\infty$ is another point added to $\Bbbk$. When the field $\Bbbk$ is finite, it means that there is exactly as much one-dimensional vector spaces than elements in $\Bbbk$, plus one. When $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ they have the same cardinality.

Now note $\mathcal{L}_A$ the set of lines going through some point $x$ belonging to the set $A$. For instance, $\mathcal{L}_x$ is the set of lines going through $x$ and $\mathcal{L}_P$ is the set of all lines. 

Note that $\mathcal{L}_x$ has the same cardinality as $\mathbb{P}_1(\Bbbk)$. Moreover, if $x \neq y$, then $\mathcal{L}_x \cap \mathcal{L}_y$ is a singleton (there is only one line going through $x$ and $y$ when $x \neq y$). 

Using this, we show that the set of lines $\mathcal{L}$ is strictly greater than $P$, whatever be $\Bbbk$. In fact, take two points $x,y$ in $P$, we  have a bijection of $\mathcal{L}_{\{x,y\}}$ onto $(\Bbbk)^2$ (we have removed an element counted twice).
This question has several answer. Strictly speaking, there exists a bijection but it is somehow a very complicated one and not very useful either. It is not continuous, meaning that two points that are close to each other can correspond to lines that are very far. In fact, both sets, lines in the plane and points in the plane, are also in bijection with... points in a line ! Even if these sets feel very different from each other.

The second answer is that there are more lines in the plane than there are points. But in that case, the correspondence is very easy to write :

For instance, you can take the following : $(a,b)\mapsto L_{a,b}$ where $L_{a,b}$ is the line whose equation is $y=ax+b$. It is obviously injective, but not onto, because the vertical maps are not in the image. Hence, we can say that there are more lines that points.