Given the parallel lines $a$ and $b$, a point $P$, and a segment $d$. We need to construct a line passing through point $P$ such that the absolute value of the difference between the segments $PA$ and $PB$ is equal to $d$, where $A$ and $B$ are the intersection points of the desired line with the given lines. What are the conditions for solvability?
Let the line farther from $P$ be denoted by $a$ and let the distance of $P$ from $a$ be $a_{0}$, and from $b$ be $b_{0}$. Then the distance between the two parallel lines is $d=a_{0}-b_{0}$, or $a_{0}+b_{0}$, depending on whether the point $P$ is outside (1st figure) or inside (2nd figure) the plane defined by $a$ and $b$. (The latter case can otherwise be reduced to the former by reflecting the line $b$ over the point $P$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_e416e15b7c1135655144g-1.jpg?height=322&width=501&top_left_y=343&top_left_x=801)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_e416e15b7c1135655144g-1.jpg?height=303&width=425&top_left_y=778&top_left_x=839)

2nd figure

Draw any line through $P$, which intersects $a$ and $b$ at points $A^{*}$ and $B^{*}$, respectively. Measure the distance $PB^{*}$ from point $A$ and denote the resulting point by $Q$. Let the orthogonal projection of $Q$ onto $a$ be $Q^{\prime}$ and the orthogonal projection of $P$ onto $b$ be $P^{\prime}$. The right triangles $PP^{\prime}B^{*}$ and $QQ^{\prime}A^{*}$ are congruent because their hypotenuses are equal by the given condition and $\angle B^{*} = \angle A^{*}$, as corresponding (1st figure) or alternate interior angles (2nd figure). Therefore,

$$
b_{0}=PP^{\prime}=QQ^{\prime}
$$

Thus, the distance of point $Q$ from $a$ is always $b_{0}$, regardless of the direction of the arbitrary line. The geometric locus of points $Q$ for which $PQ=PA-BB$ is a line $q$ that is $b_{0}$ units away from $a$ and $a_{0}-b_{0}$ units away from $P$. The desired line $PQ=d$ intersects the line $q$ at points $Q_{1}$ and $Q_{2}$, which, when connected to $P$, provide the solution. There are 2, 1, or 0 solutions depending on whether $d \gtreqless a_{0}-b_{0}>0$.

$$
a_{0}-b_{0}=0 \text{ in which case the line } q \text{ degenerates to the point } Q \equiv P.
$$

In this case, there is no solution unless $d$ is equal to zero, in which case any line passing through $P$ is a solution.