In Jean Gallier, Basics of Affine Geometry (Ch. 2) shows an example of curved affine plan. He says: "this should dispell any idea that affine spaces are dull".
Let's consider the set ( circular paraboloid) $$\mathfrak{P}=\{P=(x,y,z)\in \mathbb{R}\times \mathbb{R}\times \mathbb{R}\mid x^2+y^2=z\}.$$ Its elements are the points $P=(x_1,x_2,x_1^2+x_2^2).$ We will organize $\mathfrak{P}$ as a two dimensional affine space.
The vector space associated to the affine space $\mathfrak{P}$ is $\mathbb{R}^2$. Tha action is the mapping
$$\mathfrak{P}\times \mathbb{R} \mapsto \mathfrak{P}$$
$$((x,y,x^2+y^2)\;,\;\begin{bmatrix} u\\v\end{bmatrix})\;\mapsto\;(x+u,y+v,(x+u)^2+(y+v)^2).$$
We also use the structural function of the affine space, $$\phi \;:\;\mathfrak{P}\times \mathfrak{P} \rightarrow \mathbb{R}^2,\;\;\phi(A,B)=\begin{bmatrix}b_1-a_1\\b_2-a_2 \end{bmatrix},$$ for $A=(a_1,a_2,a_1^2+a_2^2)\;,\;B=(b_1,b_2,b_1^2+b_2^2).$
In fact, $\mathfrak{P}$ is isomorphic to $\mathbb{A}_2$ (materialized by the plan $xOy$) by projection $pr_3$.
Let's take the example $A=(1,0,1) \in \mathfrak{P}\;,\;B=(0,1,1)\in \mathfrak{P}\;,\;\phi(A,B)=\begin{bmatrix}-1\\1 \end{bmatrix}.$ We choose the affine frame with the origin in point $A$ and a linear basis $\overrightarrow{e_1}, \overrightarrow{e_2}$:
$$\mathfrak{R}=(A=(1,0,1);\overrightarrow{e_1}=\begin{bmatrix}1\\0 \end{bmatrix},\;\overrightarrow{e_2}=\begin{bmatrix}0\\1 \end{bmatrix}).$$
For an arbitrary point $P=(x,y,x^2+y^2)\in \mathfrak{P}$, the vector $\phi(A,P)=\begin{bmatrix}x-1\\y\end{bmatrix}=(x-1)\overrightarrow{e_1}+y\overrightarrow{e_2}\;\;\Rightarrow\;\;P(x-1,y)$ are the affine Cartesian coordinates of $P$ in the frame $\mathfrak{R}$. In particular in the frame $\mathfrak{R}$ we have $A=(1,0,1)=A(1-1,0)=A(0,0)\;,\;B=(0,1,1)=B(0-1,1)=B(-1,1).$
Conversely, the point $C(c_1,c_2)$ that has the $(c_1,c_2)$ coordinates in the frame $\mathfrak{R}$ corresponds to the point $C=(c_1+1,c_2, c_1^2+c_2^2+2c_1+1) \in \mathfrak{P}.$
Let us now investigate the colinear points with A and B. Points $A,B,C=(c_1,c_2,c_1^2+c_2^2)$ are affinely dependent iff the vectors $\overrightarrow{AC}\;,\;\overrightarrow{BC}$ are linerarly dependent. So
$$\overrightarrow{AC}=\phi(A,C)=\begin{bmatrix}c_1-a_1\\c_2-a_c \end{bmatrix}=\begin{bmatrix}c_1-1\\c_c \end{bmatrix}$$
$$\overrightarrow{BC}=\phi(B,C)=\begin{bmatrix}c_1-b_1\\c_2-b_2) \end{bmatrix}=\begin{bmatrix}c_1\\c_2-1 \end{bmatrix}$$
are linearly dependent. This means $\exists\;(\lambda,\mu)\neq (0,0)$ so that
$\lambda \cdot \overrightarrow{AC}\;=\;\mu \cdot \overrightarrow{BC}$ $\Leftrightarrow \lambda \begin{bmatrix}c_1-1\\c_2 \end{bmatrix}=\mu \begin{bmatrix}c_1\\c_2-1 \end{bmatrix}\Leftrightarrow$
$\begin{cases}-\lambda+(\lambda-\mu)c_1=0\\\mu+(\lambda-\mu)c_2=0 \end{cases} \tag{1}$
If $\lambda=\mu$, then we get from (1), $-\lambda=\mu=0$, case excluded. If $\lambda \neq \mu$, then
$$c_1=\frac{\lambda}{\lambda-\mu}\;,\;c_2=\frac{-\mu}{\lambda-\mu}.$$
Designate $\alpha :=\frac{\lambda}{\lambda-\mu}\;\Rightarrow\;c_2=1-\alpha$, so points "colinear" with $A,\;B$ are
$$C=(\alpha,1-\alpha,1-2\alpha+2\alpha ^2) \tag{2}$$
which is expressed in the frame $\mathfrak{R}$: $C(\alpha -1, 1-\alpha).$
In a particular case, we get the "middle" of the bipoint $\{A,B\}$ which is in the frame $\mathfrak{R}$:
$$\frac{1}{2}A+\frac{1}{2}B=\left (-\frac{1}{2},\frac{1}{2}\right )$$
to which corresponds thepoint $$C=\left (-\frac{1}{2}+1, \frac{1}{2},(-\frac{1}{2})^2+(\frac{1}{2})^2+2(-\frac{1}{2})+1\right )=C(\frac{1}{2},\frac{1}{2},\frac{1}{2}) \in \mathfrak{P}.$$
We can obtain from the points (2) the equation of the "line" $AB$.
In the frame $\mathfrak{R}$ it is "x+y=0".(Attention, here x and y are the coordinates in $\mathfrak{R}$ of a generic point M(x,y).)
In the frame $Oxyz$, we need to remove the parameter $\alpha$ from (2)
$$\begin{cases}x=\alpha\\y=1-\alpha\\z=1-2\alpha+2\alpha^2 .\end{cases}$$
We obtain the equivalent equations
$$\begin{cases}x+y=1\\x^2+y^2=z\end{cases}$$
that is, the intersection of paraboloid $\mathfrak{P}$ with the vertical plane $x+y=1$.
If we choose in the $xOy$ plan the frame with the origin in $C'(\frac{1}{2},\frac{1}{2})$(projection of the midle of bipunct {A,B} above) and the "X" axis along the line $x+y=1$ we obtain(avoiding certain not too heavy calculations) the change of coordinates
$$\begin{cases}Y=z\\X=\frac{1}{\sqrt{2}}(1-2x) \end{cases}$$
in which the equation of the "line AB" is written $Y=\frac{1}{2}+X^2$ that is a parabola.
$$\blacksquare$$
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