As shown in the figure, $AC$ is a ladder that slides downwards against the wall $AB$.
- Distance from $A=x$
- Distance from $C=y$
Options:
- $\text{(A)} \quad x > y$
- $\text{(B)} \quad y > x$
- $\text{(C)} \quad x = y$
- $\text{(D)} \quad \text{Relationship cannot be determined.}$
From my understanding, If a ladder slides down the wall, the length of the ladder remains the same, hence, $x$ (which gets reduced in number) should be equal to $y$ (which gets increased in number) in order to maintain the length of the ladder.
However the answer is D. i.e. Relationship cannot be determined.
We have that
$$(AB-x)^2+(BC+y)^2=AB^2+BC^2$$
Which in the plane $x-y$ is a circle passing through the origin, for example for $AB=2$ and $BC=1$ we have
thus for some value $x=x_0$ we have
$y=x$ for $x=x_0$
$y>x$ for $x<x_0$
$y<x$ for $x>x_0$