[proofplan] The statement is ultimately the uniqueness of inverses in the group law on $E(k)$, but we also record how the geometric chord-and-tangent construction realizes that inverse. First we handle the cases involving the identity point $O$. For affine points, the vertical line through $P$ meets the cubic at $P$, at its reflected point across the $x$-axis, and at $O$; the chord-and-tangent rule then says that this reflected point is exactly the group inverse. Conversely, if $P+Q=O$, the third-intersection rule forces the line through $P$ and $Q$ to pass through $O$, hence to be vertical in the affine chart, so $Q$ is the reflected point. [/proofplan]

[step:Handle the cases where one point is the identity] Since $O$ is the identity element of the group $E(k)$, for every $R \in E(k)$ one has \begin{align*} O+R = R \end{align*} and the inverse of $O$ is $-O=O$. If $P=O$, then $P+Q=O$ is equivalent to $Q=O$, which is equivalent to $Q=-O=-P$. If $Q=O$, then $P+Q=O$ is equivalent to $P=O$, which is equivalent to $Q=O=-P$. Thus it remains to consider the case where both $P$ and $Q$ are affine points. [/step]

[step:Identify the inverse of an affine point by the vertical line]Let $P \in E(k)\setminus\{O\}$ be an affine point, and write \begin{align*} P = (x_P,y_P) \in k^2. \end{align*} Define the affine point \begin{align*} P' := (x_P,-y_P) \in k^2. \end{align*} Since $P \in E(k)$, we have \begin{align*} y_P^2 = x_P^3 + ax_P + b. \end{align*} Because $(-y_P)^2 = y_P^2$, the point $P'$ also satisfies the affine equation of $E$, so $P' \in E(k)$. Let $L \subset \mathbb{P}^2_k$ be the projective line with equation $X-x_PZ=0$. This line contains $P$, $P'$, and $O=[0:1:0]$. In the chord-and-tangent group law, when a line meets $E$ in three points $A,B,C$, counted with intersection multiplicity, the sum satisfies \begin{align*} A+B+C=O. \end{align*} Applying this to the line $L$ gives \begin{align*} P+P'+O=O. \end{align*} Since $O$ is the identity element, this reduces to \begin{align*} P+P'=O. \end{align*} Therefore $P'$ is the inverse of $P$, so \begin{align*} -P = (x_P,-y_P). \end{align*}[/step]

[guided]Let us spell out why the reflected point is the inverse. Start with an affine point \begin{align*} P = (x_P,y_P) \in E(k)\setminus\{O\}. \end{align*} The Weierstrass equation in the affine chart $Z=1$ is \begin{align*} y^2 = x^3+ax+b. \end{align*} Define \begin{align*} P' := (x_P,-y_P). \end{align*} This point lies on $E$ because replacing $y_P$ by $-y_P$ does not change the square: \begin{align*} (-y_P)^2 = y_P^2 = x_P^3+ax_P+b. \end{align*} Now consider the projective line \begin{align*} L := \{[X:Y:Z]\in \mathbb{P}^2_k : X-x_PZ=0\}. \end{align*} In the affine chart $Z=1$, this is the vertical line $x=x_P$. It contains both $P$ and $P'$. It also contains the point at infinity $O=[0:1:0]$, since substituting $X=0$ and $Z=0$ into $X-x_PZ=0$ gives $0=0$. The chord-and-tangent group law says that if a projective line meets the cubic in three points $A,B,C$, with intersection multiplicity counted, then \begin{align*} A+B+C=O. \end{align*} For the vertical line $L$, the three intersection points are $P$, $P'$, and $O$, again counted with multiplicity in the tangent case $y_P=0$. Hence \begin{align*} P+P'+O=O. \end{align*} Because $O$ is the identity element of the group law, adding $O$ does not change the sum, so \begin{align*} P+P'=O. \end{align*} Thus $P'$ is the group inverse of $P$. Therefore \begin{align*} -P = P' = (x_P,-y_P). \end{align*}[/guided]

[step:Prove that the reflected point sums with $P$ to $O$] Assume $Q=-P$. If $P=O$, the result was handled above. Otherwise $P=(x_P,y_P)$ is affine, and the previous step gives \begin{align*} Q=-P=(x_P,-y_P). \end{align*} The same vertical-line computation gives \begin{align*} P+Q=O. \end{align*} Thus $Q=-P$ implies $P+Q=O$. [/step]

[step:Use uniqueness of inverses to prove the converse] Assume \begin{align*} P+Q=O. \end{align*} By the definition of $-P$ as the group inverse of $P$, one also has \begin{align*} P+(-P)=O. \end{align*} In a group, inverses are unique: if $P+R=O$ and $P+S=O$, then adding $-P$ on the left gives \begin{align*} R = O+R = (-P+P)+R = -P+(P+R) = -P+O = -P, \end{align*} and the same calculation with $S$ gives $S=-P$. Applying this with $R=Q$ gives \begin{align*} Q=-P. \end{align*} Therefore $P+Q=O$ implies $Q=-P$. [/step]

[step:Conclude the equivalence] The forward implication shows that any point $Q$ satisfying $P+Q=O$ must be the inverse $-P$. The reverse implication shows that the inverse $-P$ always satisfies $P+(-P)=O$. Hence, for all $P,Q \in E(k)$, \begin{align*} P+Q=O \end{align*} if and only if \begin{align*} Q=-P. \end{align*} This proves the vertical line criterion for inverses on the Weierstrass cubic. [/step]