A variable straight line passes through the points of the intersection of the lines x + 2y = 1 and 2x - y = 1 and meets the co-ordinates axes in axes in Anad B. Prove that the locus of the midpoint of AB is 10xy = x + 3y.
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$$\begin{gathered} \mathrm{given}, \\ \mathrm{x}+2\mathrm{y}=1 \\ 2\mathrm{x}-\mathrm{y}=1 \\ \mathrm{multiplying} 1\mathrm{st} \mathrm{equation} \mathrm{by} 2 \mathrm{and} \mathrm{subtracting} \mathrm{from} \mathrm{second}: \\ 2\mathrm{x}+4\mathrm{y}=2 \\ 2\mathrm{x}-\mathrm{y}=1 \\ 2\mathrm{x}+4\mathrm{y}-2\mathrm{x}+\mathrm{y}=2-1 \\ 5\mathrm{y}=1 \\ \mathrm{y}=\frac{1}{5} \\ \mathrm{x}=1-\frac{2}{5}=\frac{3}{5} \\ \mathrm{point} \mathrm{of} \mathrm{intersection}\left(\frac{3}{5},\frac{1}{5}\right) \\ \mathrm{let} \mathrm{line} \mathrm{be}: \\ \left(\mathrm{y}-\frac{1}{5}\right)=\mathrm{m}\left(\mathrm{x}-\frac{3}{5}\right) \\ \mathrm{on} \mathrm{x}-\mathrm{axis}, \mathrm{y}=0 \\ -\frac{1}{5}=\mathrm{mx}-\frac{3\mathrm{m}}{5} \\ \frac{3\mathrm{m}}{5}-\frac{1}{5}=\mathrm{mx} \\ \mathrm{x}=\frac{3}{5}-\frac{1}{5\mathrm{m}} \\ \mathrm{on} \mathrm{y}-\mathrm{axis}: \\ \mathrm{x}=0 \\ \mathrm{y}-\frac{1}{5}=\frac{-3\mathrm{m}}{5} \\ \mathrm{y}=\frac{1}{5}-\frac{3\mathrm{m}}{5} \\ \mathrm{A}\left(\frac{3}{5}-\frac{1}{5\mathrm{m}},0\right) \\ \mathrm{B}\left(0,\frac{1}{5}-\frac{3\mathrm{m}}{5}\right) \\ \mathrm{let} \mathrm{mid}-\mathrm{point} \mathrm{be} \left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{h}=\frac{0+\frac{3}{5}-\frac{1}{5\mathrm{m}}}{2} \\ 2\mathrm{h}=\frac{3}{5}-\frac{1}{5\mathrm{m}} \\ 10\mathrm{h}=3-\frac{1}{\mathrm{m}} \\ \frac{1}{\mathrm{m}}=3-10\mathrm{h} \\ \mathrm{m}=\frac{1}{3-10\mathrm{h}} \\ \mathrm{k}=\frac{0+\frac{1}{5}-\frac{3\mathrm{m}}{5}}{2} \\ 2\mathrm{k}=\frac{1}{5}-\frac{3\mathrm{m}}{5} \\ 10\mathrm{k}=1-3\mathrm{m} \\ 3\mathrm{m}=1-10\mathrm{k} \\ \mathrm{m}=\frac{1-10\mathrm{k}}{3} \\ \mathrm{so} \mathrm{since} \mathrm{m}=\mathrm{m} \\ \mathrm{we} \mathrm{have}, \\ \frac{1}{3-10\mathrm{h}}=\frac{1-10\mathrm{k}}{3} \\ 3=\left(1-10\mathrm{k}\right)\left(3-10\mathrm{h}\right) \\ 3=3-10\mathrm{h}-30\mathrm{k}+100\mathrm{kh} \\ 100\mathrm{kh}=10\mathrm{h}+30\mathrm{k} \\ \mathrm{dividing} \mathrm{by} 10 \mathrm{whole} : \\ 10\mathrm{kh}=\mathrm{h}+3\mathrm{k} \\ \mathrm{replacing} \mathrm{k} \mathrm{by} \mathrm{h} \mathrm{by} \mathrm{x} \\ \mathbf{10}\mathbf{xy}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{proved}\mathbf{!} \end{gathered}$$

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