A stick of length l rests against the floor a | vedclass

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(B) Let the coordinates of the ends of the stick be $(a, 0)$ on the floor and $(0, b)$ on the wall. The length of the stick is $l$,so $a^2 + b^2 = l^2

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Circle

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(B) Let the coordinates of the ends of the stick be $(a, 0)$ on the floor and $(0, b)$ on the wall. The length of the stick is $l$,so $a^2 + b^2 = l^2$.Let the middle point of the stick be $(h, k)$.By the midpoint formula,$h = \frac{a+0}{2} = \frac{a}{2}$ and $k = \frac{0+b}{2} = \frac{b}{2}$.This gives $a = 2h$ and $b = 2k$.Substituting these into the equation $a^2 + b^2 = l^2$,we get $(2h)^2 + (2k)^2 = l^2$.$4h^2 + 4k^2 = l^2 \Rightarrow h^2 + k^2 = \frac{l^2}{4}$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = \left(\frac{l}{2}\right)^2$,which represents a circle.

$A$ stick of length $l$ rests against the floor and a wall of a room. If the stick begins to slide on the floor,then the locus of its middle point is

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