There is a line $e$ and two points $A$ and $B$ on it, as well as a segment $CD$ parallel to $e$. We measure on $e$ from $A$ and $B$ in the same direction the segments $AP$ and $BQ=2AP$. Determine the geometric locus of the intersection point $M$ of the lines $PC$ and $QD$, as $P$ runs along the line $e$!
I. Solution: Place the given four points in a rectangular coordinate system as follows (Figure 1):

![](https://cdn.mathpix.com/cropped/2024_05_02_48f5d4adf9cb0302c62cg-1.jpg?height=604&width=808&top_left_y=262&top_left_x=648)

$$
A(0,0), B(b, 0), C(c, k), D(d, k)
$$

Let $M\left(x_{1}, y_{1}\right)$ be a point on the geometric locus. The lines $M C$ and $M D$ intersect the line $A B$ - in this case, the $x$-axis - at points $P$ and $Q$.

The equation of the line $M C$ is:

$$
y-y_{1}=\frac{k-y_{1}}{c-x_{1}}\left(x-x_{1}\right)
$$

The equation of the line $M D$ is:

$$
y-y_{1}=\frac{k-y_{1}}{d-x_{1}}\left(x-x_{1}\right)
$$

Determine the intersection points of these lines with the $x$-axis.

If $y=0$, then

From (1), the abscissa of $P$ is

$$
x=\frac{k x_{1}-c y_{1}}{k-y_{1}}
$$

From (2), the abscissa of $Q$ is

$$
x=\frac{k x_{1}-d y_{1}}{k-y_{1}}
$$

According to the problem,

$$
2 \frac{k x_{1}-c y_{1}}{k-y_{1}}=\frac{k x_{1}-d y_{1}}{k-y_{1}}-b
$$

Thus, after rearranging, the relationship between $x_{1}$ and $y_{1}$ is

$$
2 c y_{1}-d y_{1}+b y_{1}-k x_{1}-b k=0
$$

This means that the equation of the geometric locus is:

$$
y=\frac{k}{b+2 c-d} x+\frac{b k}{b+2 c-d}
$$

This is the equation of a line. If $b+2 c-d=0$, then the above equation becomes $k\left(x_{1}+b\right)=0$, which means the equation of the geometric locus is $x=-b$, representing a line perpendicular to the $x$-axis (or the line $e=A B$).

The geometric locus can be constructed most simply by finding two of its points.

II. Solution: Measure the distance $A E=A B$ in the direction opposite to $B$ from $A$ (Figure 2), then clearly $P E=P Q$ and $B E=2 A E$, and thus $E$ is a point on the geometric locus.

![](https://cdn.mathpix.com/cropped/2024_05_02_48f5d4adf9cb0302c62cg-2.jpg?height=404&width=789&top_left_y=134&top_left_x=687)

Figure 2

Connect $E$ and $M$, and let the extension of $C D$ intersect the line $E M$ at $F$. Since $F D \| E Q$, $C F=C D$. Therefore, $F$ is a fixed point on the line $C D$ regardless of the position of $P$. In other words: any point $M$ on the geometric locus, when connected to $E$, always passes through the point $F$, meaning that the point $M$ moves along the line $E F$. It is easy to show that if $P$ runs along the line $e$, $M$ runs along the line $E F$, so the line $E F$ is the desired geometric locus. If $P C \| E F$, then $E P=F C=\frac{F D}{2}$, and thus

$$
E Q=E B-B Q=2 E A-2\left(E A-\frac{F D}{2}\right)=F D
$$

Therefore, $Q D$ is also parallel to $E F$, meaning that the point $M$ does not exist.