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

Hint:

If rod ends slide on perpendicular axes, midpoint always lies on a circle with radius \(l/2\) and centre at origin.

Answer 1

The Correct Option is A

Step 1: Assume endpoints of rod.
Let the ends of rod be at \(A(a,0)\) on \(x\)-axis and \(B(0,b)\) on \(y\)-axis.
Step 2: Use length condition.
Distance \(AB=l\):
\[ AB^2=a^2+b^2=l^2 \] Step 3: Midpoint coordinates.
Midpoint \(M(x,y)\) is:
\[ x=\frac{a}{2},\quad y=\frac{b}{2} \] So:
\[ a=2x,\quad b=2y \] Step 4: Substitute into length equation.
\[ (2x)^2+(2y)^2=l^2 \Rightarrow 4x^2+4y^2=l^2 \Rightarrow x^2+y^2=\frac{l^2}{4} \] Final Answer: \[ \boxed{x^2+y^2=\frac{l^2}{4}} \]