In my lecture notes we have the following:
$$P \oplus Q \oplus R =O \Leftrightarrow P, Q, R \text{ are collinear }$$
So $$P \oplus Q \oplus O =O \Leftrightarrow Q=-P$$
that means that $Q=-P$ and $P$ is at a perpendicular to the $x-$axis line.
($O$ is the point at infinity)
Can you explain to me why the following stands?
So $$P \oplus Q \oplus O =O \Leftrightarrow Q=-P$$
that means that $Q=-P$ and $P$ is at a perpendicular to the $x-$axis line.
Edit:
If $P=(x, y)$, then $-P=(x, y^{\star})$.
$y$ and $y^{\star}$ are roots of $$y^2+(a_1x+a_3)y=x^3+a_2x^2+a_4x+a_6$$
So $$y+y^{\star}=-a_1 x - a_3 \Rightarrow y^{\star}=-y-a_1x-a_3$$
We have done the first step.
If $P=(x, y)$ then $-P=(x, -y-a_1x-a_3)$.
Points of order $2$.
If $P=(x, y)$ is of order $2$ $$\Leftrightarrow P \oplus P=O \Leftrightarrow P=-P \Leftrightarrow (x, y)=(x, -y-a_1x-a_3) \\ \Leftrightarrow y=-y-a_1x-a_3 \Leftrightarrow 2y=a_1x-a_3 \overset{ ch K \neq 2}{ \Leftrightarrow } y=\frac{-a_1x-a_3}{2} $$
Can you explain to me why $P=(x, y)$ and $-P=(x, -y-a_1x-a_3)$ are points of order $2$ ???
One has that $O$ is the neutral element with respect to the group law $\oplus$.
So $P \oplus Q \oplus O = P \oplus Q$ and $P \oplus Q = O$ is essentially the definition of $Q = - P$ as $-P$ denotes the inverse element of $P$, and $Q$ apparently is the inverse element of $Q$ as $P \oplus Q = 0$.
For the second part of the question recall from the geometric construction used to add points that the point at infinity $O$ can be thought of as "infinitely far away" on the $y$-axis. Thus the line passing through $O$ and $P$ will be parallel to the $y$-axis, and its (third) intersection with the curve, so $Q$, is thus what you described.
For the edit: A point $P$, and more generally an element of a commutative group, is called of order $2$ if $2P = P + P = 0$, equivalently $P = -P$ (and $P$ is non-zero). The point is to establish when a point is of order $2$. We have for a general point $P= (x,y)$ that the $x,y$ need to fulfill the equation, else they would not be points on the curve. Now, above we saw that the inverse of the point $-P$ is the "other" point with very same $x$. So we find the second coordinate of this point. Then $P$ is of order $2$ if in fact $P = - P$ so that the two second coordinates coincide. To this end we compare the two expressions for the second coordinate of $P$ and $-P$. If they are equal $P= -P$ and $P$ is a point of order $2$, if not then not as $P \neq -P$.