Question

A stick oflength 10 units rests against the floor and a wall of a room. If the stick begins to slide on the floor then the locus of its middle point is:

Answer 1

x^2+y^2=25

Answer 2

Let the stick of length $L=10$ units have its ends on the x-axis and y-axis. Let these points be $(a,0)$ and $(0,b)$. By Pythagoras theorem, $a^2+b^2 = L^2 = 100$. Let the midpoint of the stick be $(x,y)$. Using the midpoint formula, $x = a/2$ and $y = b/2$. This implies $a=2x$ and $b=2y$. Substitute these into the equation $a^2+b^2=100$: $(2x)^2 + (2y)^2 = 100$ $4x^2 + 4y^2 = 100$ $x^2 + y^2 = 25$ Since the stick is in the first quadrant, $a \ge 0$ and $b \ge 0$, which means $x \ge 0$ and $y \ge 0$. The locus is a quarter circle centered at the origin with radius 5.

The locus of the middle point is a circle with equation $x^2+y^2=25$. Considering the physical constraints, it is a quarter circle in the first quadrant.