The ends of a segment of fixed length move along two intersecting perpendicular lines. What is the trajectory of the midpoint of this segment?

#
Enter a suitable coordinate system and apply the formula for the distance between two points.

## Solution

We introduce a rectangular coordinate system such that the first line is defined by the system of equations $\mathrm{z}=1, \mathrm{y}=0$, and the second line is defined by the system of equations $\mathrm{z}=-1, \mathrm{x}=0$. Thus, the first endpoint of the segment has coordinates $(\mathrm{a}, 0,1)$, and the second endpoint has coordinates $(0, \mathrm{~b},-1)$, where a and b are some numbers. The square of the distance between the endpoints of the segment is calculated by the formula $\mathrm{a}^{2}+\mathrm{b}^{2}+(1-(-1))^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}+4$. The distance between the endpoints of the segment must be equal to a fixed number $\mathrm{d}$ (the segment is rigid). Therefore, the numbers a and b are related by the equation $\mathrm{a}^{2}+\mathrm{b}^{2}+4=\mathrm{d}^{2}$. The midpoint M of the segment has coordinates $(\mathrm{a}/2, \mathrm{b}/2, (1+(-1))/2)=(\mathrm{a}/2, \mathrm{b}/2, 0)$. This means that M moves in the plane $z=0$. The equation written above transforms into the equation of a circle for the point $(\mathrm{a}/2, \mathrm{b}/2)$ in the plane

Oxy: $(\mathrm{a} / 2)^{2}+(\mathrm{b} / 2)^{2}=(\mathrm{d} / 2)^{2}-1$. Thus, the point M moves along a circle of radius $((\mathrm{d}/2)^{2}-1)^{1 / 2}$ centered at the origin, lying in the plane Oxy.

## Answer

along a circle.