I have just begun learning about dilations. In my book (Lang's Geometry 2nd edition), dilation by $r$ relative to a point $0$ is defined to be the association which to each point $P$ of the plane associates the point $P'$ that lies on the ray $R_{op}$ at a distance from $0$ equal to $r$ times that of $P$ from $O$.
What I'm having trouble with is proving the followng thereom, which the book presents without a rigorous proof:
Let $S$ be an arbitrary region in the plane with area $A$. Let $rS$ be the image of $S$ under a dilation by a positive number $r$. Then the area of $rS$ is $r^2·A$.
I know the proof for arbitrary triangles and rectangles, but how to prove it for an arbitrary shape, be it a polygon or not? Thanks very much in advance.
let us express the relationship between $(x,y)$, the generic point of $P$, and $(X,Y)$ its image in $P'$ with vector/matrix notations:
$$\tag{1}\begin{cases}X=rx\\Y=ry \end{cases} \ \ \iff \ \ \begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}r&0\\0&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
Then it suffices, (in the spirit of Serge Lang's book), to use the geometrical interpretation of the determinant of the matrix of the linear transformation $f$, which is $r^2$, as:
$$\tag{2} \text{final area = initial area} \ \times \ r^2$$
$Edit:$ In fact formula (2) comes from what could consider as the "ultimate form" of the change of variables formula :
$$\int 1_{f(A)} = \int 1_A \det(f).$$
where $1_X$ is the characteristic function of (measurable!) set $A$ : $1$ on $A$, zero elsewhere).
Note that $\det(f)$ appears as the jacobian of transformation $f$.