Let's write the equation of the line that passes through the point $(5 ; 3)$ and the projection of the segment between the lines $3 x+4 y-16=0$ and $3 x+4 y-1=0$ on the $X$-axis is 1 unit long.
I. solution: Denoting the slope of the sought line by $m$, the equation is

$$
y=3+m(x-5)
$$

Thus, the abscissa of the intersection point of our line and the line given by the equation

$$
3 x+4 y-p=0
$$

is

$$
x=\frac{p+20 m-12}{4 m+3}
$$

Therefore, the projections of the endpoints of the segment in question on the $X$-axis (for $p=16$ and $p=1$) are the points

$$
x_{1}=\frac{20 m+4}{4 m+3}, \quad x_{2}=\frac{20 m-11}{4 m+3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a73aaf90a924223e44d9g-1.jpg?height=767&width=658&top_left_y=842&top_left_x=728)

The length of the segment is

$$
\left|x_{2}-x_{1}\right|=\left|\frac{15}{4 m+3}\right|=1
$$

This can be satisfied in two ways: from $x_{2}-x_{1}=-1$, we get $m=3$, and from $x_{2}-x_{1}=1$, we get $m=-4.5$. With these, the desired equations are

$$
y=3 x-12
$$

or

$$
y=-4.5 x+25.5
$$

Garai Géza (Kaposvár, Táncsics M. g. II. o. t.)

II. solution: The fact that the length of the projection is 1 holds for any line parallel to the sought line, because the two given lines are parallel. Therefore, we can choose the endpoint of the segment on one of the lines arbitrarily. Let this be the intersection point of the first line with the $Y$-axis: $(0 ; 4)$. Then the abscissa of the other endpoint is +1 or -1, and its ordinate is

$$
\begin{gathered}
y=\frac{1-3 x}{4}=-0.5, \quad \text { or } \quad 1, \quad \text { so the slopes are } \\
\frac{4-(-0.5)}{0-1}=-4.5, \quad \text { or } \quad \frac{4-1}{0-(-1)}=3 .
\end{gathered}
$$

This way, we also arrive at the lines of the previous solution.