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(B) Given lines are $x+2y=4$ $(i)$ and $2x+y=4$ (ii).Solving equations $(i)$ and (ii), we get the point of intersection as $(\frac{4}{3}, \frac{4}{3})

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$2(x+y)=3xy$

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(B) Given lines are $x+2y=4$ $(i)$ and $2x+y=4$ (ii).Solving equations $(i)$ and (ii), we get the point of intersection as $(\frac{4}{3}, \frac{4}{3})$.Let the equation of the line passing through this point be $\frac{x}{a} + \frac{y}{b} = 1$.Since it passes through $(\frac{4}{3}, \frac{4}{3})$, we have $\frac{4}{3a} + \frac{4}{3b} = 1$, which simplifies to $\frac{1}{a} + \frac{1}{b} = \frac{3}{4}$ (iii).Let the mid-point of $AB$ be $(h, k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$, so $a = 2h$ and $b = 2k$.Substituting these into equation (iii), we get $\frac{1}{2h} + \frac{1}{2k} = \frac{3}{4}$.Multiplying by $2hk$, we get $k + h = \frac{3}{2}hk$, or $2(h+k) = 3hk$.Replacing $(h, k)$ with $(x, y)$, the locus is $2(x+y) = 3xy$.

$A$ straight line through the point of intersection of the lines $x+2y=4$ and $2x+y=4$ meets the coordinate axes at $A$ and $B$. The locus of the mid-point of $AB$ is

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