Question

A Stick Of Length $10Units$ 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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Solution

Let the midpoint of Rod be h and kThe coordinates of A and B will be$A=\left(0,2k\right)AndB=\left(2h,0\right)$Distance from A to BAB = fixed (i.e length of rad is fixed)$\sqrt{{\left(2k\right)}^{2}+{\left(2h\right)}^{2}}=10$$4\left({k}^{2}+{h}^{2}\right)=100\phantom{\rule{0ex}{0ex}}{k}^{2}+{h}^{2}=25$Now by expressing in a form of locus, we get:$\mathbf{\therefore }\mathbf{}{\mathbit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbit{y}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{25}$Therefore, the locus of its middle point is ${x}^{2}+{y}^{2}=25$

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