In a Cartesian coordinate system, we know that the line $e$ passes through the point $(1 ; 1)$, and the line $f$ passes through the point $(-1 ; 1)$. We also know that the difference in their slopes is 2. Determine the geometric locus of the intersection point of lines $e$ and $f$.
Solution. Let $(x ; y)$ be the intersection point of the two lines. $(x ; y)$ cannot be $(1 ; 1)$ or $(-1 ; 1)$, because in that case the two lines would be identical.

The slope of $e$ is $m_{e}=\frac{y-1}{x-1}$, and the slope of $f$ is $m_{f}=\frac{y-1}{x+1}$.

1. Case: $m_{e}>m_{f}$, i.e., $m_{e}-m_{f}=2$.

$$
\begin{aligned}
\frac{y-1}{x-1}-\frac{y-1}{x+1}=2, & \frac{2(y-1)}{x^{2}-1}=2 \\
y-1=x^{2}-1, & y=x^{2}
\end{aligned}
$$

2. Case: $m_{f}>m_{e}$, i.e., $m_{e}-m_{f}=-2$.

$$
\begin{aligned}
& \frac{y-1}{x-1}-\frac{y-1}{x+1}=-2, \quad \frac{2(y-1)}{x^{2}-1}=-2 \\
& y-1=1-x^{2}, \quad y=2-x^{2}
\end{aligned}
$$

The intersection points of lines $e$ and $f$ thus lie on one of two parabolas. Each step in our reasoning is reversible if $x \neq \pm 1$, so the set of intersection points is the union of the parabolas defined by $y=x^{2}$ and $y=2-x^{2}$, excluding the points $(1 ; 1)$ and $(-1 ; 1)$ (which we already excluded at the beginning of the solution).

Note. Many interpreted the statement that the difference in slopes is 2 to mean that the first line is steeper (some took the second line to be steeper), and did not consider the other case.