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A rod of length $l$ slides with its ends on two perpendicular lines. Then, the locus of its mid point is
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Verified Answer
The correct answer is:
$x^{2}+y^{2}=\frac{1^{2}}{4}$
Let both of the ends of the rod are on $x$-axis and $y$-axis. Let $A B$ 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 $A B$.
$$
\begin{array}{l}
\text { Then, } h=\frac{0+a}{2}=\frac{a}{2}...(i) \\
\qquad \begin{array}{l}
k=\frac{b+0}{2}=\frac{b}{2} \\
\ln \Delta O A B \\
O A^{2}+O B^{2}=A B^{2}
\end{array} \\
\Rightarrow a^{2}+b^{2}=l^{2} \\
\Rightarrow(2 h)^{2}+(2 k)^{2}=l^{2} \quad[\text { using eq. (i) }] \\
\Rightarrow h^{2}+k^{2}=\frac{l^{2}}{4}
\end{array}
$$
$\therefore$ The equation of locus is
$$
x^{2}+y^{2}=\frac{l^{2}}{4}
$$
Let $P(h, k)$ be the mid point of the rod $A B$.
$$
\begin{array}{l}
\text { Then, } h=\frac{0+a}{2}=\frac{a}{2}...(i) \\
\qquad \begin{array}{l}
k=\frac{b+0}{2}=\frac{b}{2} \\
\ln \Delta O A B \\
O A^{2}+O B^{2}=A B^{2}
\end{array} \\
\Rightarrow a^{2}+b^{2}=l^{2} \\
\Rightarrow(2 h)^{2}+(2 k)^{2}=l^{2} \quad[\text { using eq. (i) }] \\
\Rightarrow h^{2}+k^{2}=\frac{l^{2}}{4}
\end{array}
$$
$\therefore$ The equation of locus is
$$
x^{2}+y^{2}=\frac{l^{2}}{4}
$$
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