[proofplan]
We write the unique nonvertical line through $P$ and $Q$, substitute its equation into the Weierstrass equation, and obtain a monic cubic polynomial in the $x$-coordinate. Since $x_1$ and $x_2$ are roots, polynomial division identifies the third intersection abscissa $x_R$ and comparison of the $x^2$ coefficient gives $x_R=m^2-x_1-x_2$. The nonsingularity hypothesis ensures that $E$ is an elliptic curve with the standard chord-and-tangent group law. That group law defines $P+Q$ as the reflection of the third intersection point across the $x$-axis, which gives the stated formula for the ordinate.
[/proofplan]
[step:Write the line through the two given affine points]
Because $x_1 \neq x_2$, the slope
\begin{align*}
m=\frac{y_2-y_1}{x_2-x_1}
\end{align*}
is defined. Define the affine [linear map](/page/Linear%20Map) $\ell:\mathbb{R}\to \mathbb{R}$ by
\begin{align*}
\ell(x)=m(x-x_1)+y_1.
\end{align*}
Then $\ell(x_1)=y_1$. Also,
\begin{align*}
\ell(x_2)=m(x_2-x_1)+y_1=\frac{y_2-y_1}{x_2-x_1}(x_2-x_1)+y_1=y_2.
\end{align*}
Thus the graph $y=\ell(x)$ is the line through $P$ and $Q$.
[/step]
[step:Substitute the line into the cubic and identify the third intersection abscissa]
Define the polynomial map $G:\mathbb{R}\to \mathbb{R}$ by
\begin{align*}
G(x)=x^3+ax+b-\ell(x)^2.
\end{align*}
A real number $x \in \mathbb{R}$ satisfies $G(x)=0$ exactly when the point $(x,\ell(x))$ lies on $E$, because
\begin{align*}
G(x)=0
\iff
\ell(x)^2=x^3+ax+b.
\end{align*}
Since $P,Q \in E(\mathbb{R})$ and $\ell(x_1)=y_1$, $\ell(x_2)=y_2$, we have $G(x_1)=0$ and $G(x_2)=0$.
The roots $x_1$ and $x_2$ are distinct. Hence $(x-x_1)(x-x_2)$ divides $G(x)$. Since $G$ is monic of degree $3$, there is a unique real number $x_R$ such that
\begin{align*}
G(x)=(x-x_1)(x-x_2)(x-x_R)
\end{align*}
for every $x \in \mathbb{R}$. Define $R=(x_R,y_R)\in \mathbb{R}^2$, where $y_R=\ell(x_R)$. Then $G(x_R)=0$, so $R \in E(\mathbb{R})$.
[guided]
The point of substituting the line into the elliptic curve equation is that intersections of a line with the cubic curve become roots of a single-variable cubic polynomial. Define the polynomial map $G:\mathbb{R}\to \mathbb{R}$ by
\begin{align*}
G(x)=x^3+ax+b-\ell(x)^2.
\end{align*}
For any real number $x$, the point on the line with abscissa $x$ is $(x,\ell(x))$. This point lies on $E$ exactly when its coordinates satisfy the Weierstrass equation:
\begin{align*}
(x,\ell(x))\in E(\mathbb{R}) \iff \ell(x)^2=x^3+ax+b \iff x^3+ax+b-\ell(x)^2=0 \iff G(x)=0.
\end{align*}
Now $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ lie on $E(\mathbb{R})$. The line function satisfies $\ell(x_1)=y_1$ directly from its definition. For $x_2$, the definition of $m$ gives
\begin{align*}
\ell(x_2)=m(x_2-x_1)+y_1=\frac{y_2-y_1}{x_2-x_1}(x_2-x_1)+y_1=y_2.
\end{align*}
Therefore $G(x_1)=x_1^3+ax_1+b-y_1^2=0$ and $G(x_2)=x_2^3+ax_2+b-y_2^2=0$.
The hypothesis $x_1\neq x_2$ matters here: it means the two linear factors $x-x_1$ and $x-x_2$ are distinct. Thus $(x-x_1)(x-x_2)$ divides $G(x)$.
Because $\ell(x)=m(x-x_1)+y_1$ is affine linear, the term $\ell(x)^2$ has degree at most $2$. Hence $G(x)=x^3+ax+b-\ell(x)^2$ is a monic cubic polynomial. Dividing this monic cubic by the monic quadratic $(x-x_1)(x-x_2)$ leaves a monic linear factor, so there is a unique real number $x_R$ such that
\begin{align*}
G(x)=(x-x_1)(x-x_2)(x-x_R)
\end{align*}
for every $x \in \mathbb{R}$. Define the corresponding point on the line by $R=(x_R,y_R)$, where $y_R=\ell(x_R)$. Since $G(x_R)=0$, the equivalence above gives $R\in E(\mathbb{R})$. This is the third intersection point of the line with the cubic, counted with multiplicity.
[/guided]
[/step]
[step:Compare the quadratic coefficients to compute the third abscissa]
First expand $\ell(x)^2$:
\begin{align*}
\ell(x)^2=\bigl(m(x-x_1)+y_1\bigr)^2=m^2x^2+2m(y_1-mx_1)x+(y_1-mx_1)^2.
\end{align*}
Therefore the coefficient of $x^2$ in
\begin{align*}
G(x)=x^3+ax+b-\ell(x)^2
\end{align*}
is $-m^2$.
On the other hand,
\begin{align*}
(x-x_1)(x-x_2)(x-x_R)
=x^3-(x_1+x_2+x_R)x^2+\text{terms of degree at most }1.
\end{align*}
Comparing the coefficients of $x^2$ in the identity
\begin{align*}
G(x)=(x-x_1)(x-x_2)(x-x_R)
\end{align*}
gives
\begin{align*}
-(x_1+x_2+x_R)=-m^2.
\end{align*}
Hence
\begin{align*}
x_R=m^2-x_1-x_2.
\end{align*}
[/step]
[step:Reflect the third intersection point to obtain the group-law sum]
By the chord-and-tangent definition of the group law on $E$, the sum $P+Q$ is the inverse of the third intersection point $R$. For the Weierstrass equation $y^2=x^3+ax+b$, the inverse of an affine point $(x,y)\in E(\mathbb{R})$ is its reflection across the $x$-axis, namely $(x,-y)$. Therefore
\begin{align*}
P+Q=-R=(x_R,-y_R).
\end{align*}
Thus $x_3=x_R$, and the preceding step gives
\begin{align*}
x_3=m^2-x_1-x_2.
\end{align*}
Since $y_R=\ell(x_R)=m(x_R-x_1)+y_1$, we obtain
\begin{align*}
y_3=-y_R=-\bigl(m(x_R-x_1)+y_1\bigr)=m(x_1-x_R)-y_1=m(x_1-x_3)-y_1.
\end{align*}
This proves the stated chord formula for $P+Q$.
[/step]
