A rod of length 2l slides with its ends on tw | vedclass

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(A) Let the two perpendicular lines be the coordinate axes $OX$ and $OY$. Let the ends of the rod be $A(0, a)$ on the $y$-axis and $B(b, 0)$ on the $x

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$x^2+y^2=l^2$

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(A) Let the two perpendicular lines be the coordinate axes $OX$ and $OY$. Let the ends of the rod be $A(0, a)$ on the $y$-axis and $B(b, 0)$ on the $x$-axis.The length of the rod is $AB = 2l$. By the distance formula,$\sqrt{a^2+b^2} = 2l$,which implies $a^2+b^2 = 4l^2$.Let $P(x, y)$ be the mid-point of the rod $AB$. Then $x = \frac{0+b}{2} = \frac{b}{2}$ and $y = \frac{a+0}{2} = \frac{a}{2}$.This gives $b = 2x$ and $a = 2y$.Substituting these values into the equation $a^2+b^2 = 4l^2$,we get $(2y)^2 + (2x)^2 = 4l^2$.$4x^2 + 4y^2 = 4l^2$,which simplifies to $x^2+y^2 = l^2$.Thus,the locus of the mid-point is $x^2+y^2 = l^2$.

$A$ rod of length $2l$ slides with its ends on two perpendicular lines. The locus of its mid-point is

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