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(A) First,find the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$. Multiplying the second equation by $2$,we get $4x - 2y = 2$. Addi

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$x + 3y - 10xy = 0$

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(A) First,find the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$. Multiplying the second equation by $2$,we get $4x - 2y = 2$. Adding this to the first equation: $(x + 2y) + (4x - 2y) = 1 + 2$,which gives $5x = 3$,so $x = 3/5$. Substituting $x = 3/5$ into $x + 2y = 1$,we get $3/5 + 2y = 1$,so $2y = 2/5$,which gives $y = 1/5$. The point of intersection is $P(3/5, 1/5)$. Let the variable line be $\frac{x}{a} + \frac{y}{b} = 1$. Since it passes through $P(3/5, 1/5)$,we have $\frac{3}{5a} + \frac{1}{5b} = 1$. The midpoint $M(h, k)$ of $AB$ is given by $h = a/2$ and $k = b/2$,so $a = 2h$ and $b = 2k$. Substituting these into the equation: $\frac{3}{5(2h)} + \frac{1}{5(2k)} = 1$. This simplifies to $\frac{3}{10h} + \frac{1}{10k} = 1$,which is $3k + h = 10hk$. Replacing $(h, k)$ with $(x, y)$,the locus is $x + 3y = 10xy$,or $x + 3y - 10xy = 0$.

$A$ variable straight line passes through the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$ and meets the coordinate axes at $A$ and $B$. The locus of the midpoint of $AB$ is:

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