I am trying to solve the following exercise, in the context of basic affine geometry (I saw it in a exercise sheet from a first course on affine and lineal geometry):
Consider the affine plane $ \mathbb{R}^2 $ and the curve $$ L = \lbrace (x, y) \in \mathbb{R}^2 : y = 2x^3 - 3x^2 + x - 2 \rbrace $$ Show that all the lines crossing $ L $ on three points such that one is the middle point of the other two have a point in common. Find the coordinates of this point.
At first I tried to get a description of what points $ P = (p, q), Q = (a, b) $ in the curve have the property of $ (P + Q) / 2 $ being also in the curve.
The equation $$ \frac{q + b}{2} = 2 \left( \frac{p + a}{2} \right)^3 - 3\left( \frac{p + a}{2} \right)^2 + \left( \frac{p + a}{2} \right) -2 $$ becomes the following expression after replacing $ q = 2p^3 - 3p^2 + p - 2 $ and $ b = 2a^3 - 3a^2 + a - 2 $, \begin{equation} 3p^3 + p^2(-3a - 3) + p(-3a^2 + 6a) + (3a^3 - 3a^2 + 4) = 0 \ \ \ \ (1) \end{equation}
Now I take a line $ y = m x + n $ that crosses $ L $ in two points $ P, Q $ like before, $$ \begin{cases} q = m p + n \\ b = m a + n \end{cases} \Longrightarrow q - b = m(p - a) = (2p^3 - 3p^2 + p - 2) - (2a^3 - 3a^2 + a - 2) $$ $$ = 2(p^3 - a^3) - 3(p^2 - a^2) + (p - a) = (p - a) \cdot (2(p^2 - pa + a^2) - 3(p + a) + 1) $$ We suppose $ P, Q $ to be different points, so $ p \neq a $, and then $$ m = 2(p^2 - pa + a^2) - 3(p + a) + 1 $$ And now we can compute $ n $, $$ n = q - mp = (2p^3 - 3p^2 + p - 2) - (2(p^2 - pa + a^2) - 3(p + a) + 1) p $$ $$ = (2p^3 - 3p^2 + p - 2) + (- 2p^3 + 2 p^2a - 2pa^2 + 3p^2 + 3pa - p) = -2 + 2 p^2a - 2pa^2 + 3pa - p $$ But now I have a lot of ugly expressions involving $ p $, $ a $... I am not sure on how to combine (1) with this.
Any hints?
HINT.-The curve has two extremuns at the roots of the derivative $y'=6x^2-6x+1$ and the corresponding points are $$P_1=\left(\frac{3-\sqrt3}{6},-1.904\right),P_2=\left(\frac{3+\sqrt3}{6},-2.096\right)$$ The midpoint of these two points is $M=\left(\dfrac 12,-2\right)$ and all line passing by this point $M$ and cutting the curve at two other points, these points will have $M$ as midpoint (it is easy to prove by some straightforward calculation, the case of vertical line also included vía the point at infinity).
NOTE.-If you make a rectangular change of coordinates by axis $x=\dfrac12$ and $y=-2$ with origin $M$ it can help you to see very clear somme kind of symmetry of the curve aroud the point $M$.