Question

A rod of length l slides with its ends on two perpendicular lines. Then, the locus of its mid point is

• A. $x^2+y^2=\frac{l^2}{4}$
• B. $x^2+y^2=\frac{l^2}{2}$
• C. $x^2-y^2=\frac{l^2}{4}$
• D. None of these

Answer 1

Answer: (A)

Let both of the ends of the rod are on $x$-axis and $y$-axis. Let $AB$ be rod of length $l$ and coordinates of $A$ and $B$ be $(a,0)$ and $(0,b)$ respectively.

Let $P(h,k)$ be the mid point of the rod $AB$.

Now, in $\Delta OAB$,
$O{A}^2+O{B}^2=A{B}^2$
$a^2+b^2=l^2$
$\Rightarrow (2h{)}^2+(2k{)}^2=l^2$ [using Eq. (i)]
$\Rightarrow h^2+k^2=\frac{l^2}{4}$
$\therefore$ The equation of locus is
$x^2+y^2=\frac{l^2}{4}$