Cardinality of set of all rectangles on a plane

54 Views Asked by At

The set of rectangles in $\mathbb{R}^2$ with sides parallel to axes and bottom-left corner in origin can be described as:$$\mathcal{R}=\{[0,a]\times[0,b]:a,b\in\mathbb{R}\}$$ There is bijection $\mathcal{R}\to\mathbb{R}^2$ given by $[0,a]\times[0,b]\mapsto(a,b)$
Hence I get: $|\mathbb{R}^2|=|\mathbb{R}| = \mathfrak{c}$

How would you expand this on all rectangles including translated and rotated? My idea is to extend cartesian product with coordinates of bottom-left corner - $[x,y]$ and rotation - $\phi$: $$\mathcal{R}=\{[0,a]\times[0,b] \times [x, y] \times \phi:a,b,x,y,\phi\in\mathbb{R}\}$$ With bijection: $$\mathcal{R}\to\mathbb{R}^5 \quad \text{due to:} \quad[0,a]\times[0,b]\times[x,y]\times\phi\mapsto(a,b,x,y,\phi)$$ Of cardinality: $$|\mathbb{R}^5|=|\mathbb{R}| = \mathfrak{c}$$ I was also thinking about directly defining unique rectangle on a plane with three points, but the above, if correct, requires less data - $(a,b,x,y,\phi)$ vs $(x_1,y_1,x_2,y_2,x_3,y_3)$.
Anyway, would love to hear your opinions or interesting ways of defining unique rotated/translated rectangle on a plane with minimum variables.