Question regarding analytic geometry Problem:
Given a convex polygon with $N$ sides, say $[A_1A_2...A_N]$. We know all the coordinates of the vertices of the polygon $A_1=(x_1,y_1)$, $A_2=(x_2,y_2)$, ..., $A_N=(x_N,y_N)$. We will move all sides of the polygon, for $d$ units, parallel to their respective perpendicular vector away from the centroid. What is the new area of the polygon?
Ideas:
Since we know al the points of the convex polygon, we can calculate its initial area using the Shoelace formula:
$${\displaystyle {\begin{aligned}\mathbf {A} &={1 \over 2}\left|\left(\sum _{i=1}^{n-1}x_{i}y_{i+1}\right)+x_{n}y_{1}-\left(\sum _{i=1}^{n-1}x_{i+1}y_{i}\right)-x_{1}y_{n}\right|\\[4pt]&={1 \over 2}|x_{1}y_{2}+x_{2}y_{3}+\cdots +x_{n-1}y_{n}+x_{n}y_{1}-x_{2}y_{1}-x_{3}y_{2}-\cdots -x_{n}y_{n-1}-x_{1}y_{n}|\end{aligned}}}$$
Then, we also know that the two figures (the initial and final one) are similar, because all of their sides are parallel. Additionally,
$$\frac{A_{initial}}{A_{final}} = \bigg( \frac{(A_1A_2)_{initial}}{(A_1A_2)_{final}} \bigg)^2$$
We can calculate $(A_1A_2)_{initial}$, because we know the coordinates of all the vertices. However, how can we calculate $(A_1A_2)_{final}$? Is there a formula using similar figures?
 A: You are mistaken in your assumption that (in the general case) the initial and final polygon are similar. What's true is that the interior angles do not change, because the lines creating the respective arms are parallel.
That's enough to ensure similiarity for triangles, but not for polygons with more vertices. A simple example is if your initial polygon is a $2\times4$ rectangle and you have $d=1$, then your final polygon will be a $4\times6$ rectangle, which is not similar to the initial one!
IMHO the way to go here (and make it a real analytic geometry problem) is to actually calculate the vertex coordinates $A'_i=(x'_i, y'_i)$ of the final polygon and apply the shoelace formula to that polygon!
How to do that? The side $A_{i-1}A_i$ of the initial polygon gets moved "outwards" by length $d$. That means the line containing the moved side has the same ascension as the original line containing $A_{i-1}A_i$, which is
$$m_{i-1} = \frac{y_i - y_{i-1}}{x_i - x_{i-1}}. \label{eq1}\tag{1}$$
All you need to find the exact line equation is one point on it. That means you need to quantify the "move by length $d$ perpendicular outwards" part.
First, you need to find one perpendicular vector to the original side. The vector from $A_{i-1}$ to $A_i$ is $(x_i-x_{i-1}, y_i-y_{i-1})$. If you turn that vector clockwise by 90°, you get $v_{i-1} = (y_i-y_{i-1}, x_{i-1} - x_i)$.
Second, you need to scale that vector such that it has length $d$, which makes it
$$v'_{i-1} =\frac{d}{\sqrt{(y_i-y_{i-1})^2+(x_{i-1}-x_i)^2}}(y_i-y_{i-1}, x_{i-1} - x_i).$$
Third, you need to decide if moving "parallel to their respective perpendicular vector away from the centroid" for the side $A_{i-1}A_i$ is represented by that vector $v'_{i-1}$ or it's negative. We'll deal with this later, but introduce a sign value $s_{i-1}$ here that is either $+1$ or $-1$ and get the final movement vector as
$$v''_{i-1} =s_{i-1}\frac{d}{\sqrt{(y_i-y_{i-1})^2+(x_{i-1}-x_i)^2}}(y_i-y_{i-1}, x_{i-1} - x_i).$$
You apply that movement vector to any point on the line $A_{i-1}A_i$ (say $A_{i-1}$) and have a point of the moved line. Together with \eqref{eq1} you now can build the line equation of the line that containes the moved $A_{i-1}A_i$ line!
OK, now you can also build the line equation for the  moved side $A_iA_{i+1}$ in exactly the same way. And if you calculate the point where those 2 lines intersect, you find the point $A'_i$ and it's coordinates. And you can do that for all the vertices of the final polygon, giving you the coordinates that you need for the shoelace formula.
That formula is not going to be pretty, as all the square roots and fractions from the formulas above will still be there, but it is doable.
Now we still have one problem left to decide, what should the sign factor $s_i$ for each side be? If you move along the side $A_{i-1}A_i$ from $A_{i-1}$ to $A_i$, the interior of the polygon may either be the the left or to the right.
If the interior is to the left, then you need to move the side to the right to move it away from the centroid (since it's convex, the centroid is always in the interior). That corresponds to turining the vector from $A_{i-1}$ to $A_i$ clockwise 90°, so in that case we have $s_{i-1}=+1$.
If the interior is to the right of the side, however, then we'de have to turn the vector from $A_{i-1}$ to $A_i$ counterclockwise by 90° to move away from the controid, in which case we have $s_{i-1}=-1$.
Now note that the direction where the interior is left or right can't change from one side to the next (you'd have the interior suddenly end without a border and appear "from nowhere" on the other side), so it is the same direction for all sides.
So our all our $s_{i}, i=1,\ldots, n$ are actually the same $s$, but which? The wikipedia page for the shoelace formula says

If the points are labeled sequentially in the counterclockwise
direction, then the sum of the above determinants is positive and the
absolute value signs can be omitted; if they are labeled in the
clockwise direction, the sum of the determinants will be negative.
This is because the formula can be viewed as a special case of Green's
theorem.

The formula is written as determinants, but leads to the same as yours inside the absolute value bars on the outside. Notice that being "labeled sequentially in the counterclockwise direction" is the same as saying the interior is to the left, so this is a way to find out the sign factor $s$. If
$$x_{1}y_{2}+x_{2}y_{3}+\cdots +x_{n-1}y_{n}+x_{n}y_{1}-x_{2}y_{1}-x_{3}y_{2}-\cdots -x_{n}y_{n-1}-x_{1}y_{n}$$
is positive, then our $s$ equals 1, otherwise it equals $-1$.
