The locus of the midpoint of a sliding ladder placed against a wall is a circle, whose centre is at the point of intersection of the wall and the floor.
Proof:
Proof:
The end points of the ladder are (x, 0) and (0, y).
If the length of the ladder is 1 unit, then using Theorem of Pythagoras,
x^2 + y^2 = 1
Let the midpoint of the ladder is (x', y') or (x/2, y/2) and
r is the distance between the midpoint and the centre (where the wall and floor meets).
Using distance formula:
(x/2 - 0)^2 + (y/2 - 0)^2 = r^2
(x^2 + y^2)/4 = r^2
r^2 = 1/4
Therefore, (x' - 0)^2 + (y' - 0)^2 = 1/4
x'^2 + y'^2 = 1/4
which is the equation of a circle.
