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(B) The given lines are $4x + 3y = 1$ and $3x + 4y = 1$. Solving these simultaneously,we get $x = 1/7$ and $y = 1/7$. So,the intersection point is $A(

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$x + y - 14xy = 0$

Vedclass · Answer

(B) The given lines are $4x + 3y = 1$ and $3x + 4y = 1$. Solving these simultaneously,we get $x = 1/7$ and $y = 1/7$. So,the intersection point is $A(1/7, 1/7)$.Let the equation of the line passing through $A(1/7, 1/7)$ be $y - 1/7 = m(x - 1/7)$.This line meets the $x$-axis at $P$ (where $y=0$) and the $y$-axis at $Q$ (where $x=0$).For $P$,$-1/7 = m(x_1 - 1/7) \implies x_1 = 1/7 - 1/(7m) = (m-1)/(7m)$. So $P = ((m-1)/(7m), 0)$.For $Q$,$y_1 - 1/7 = m(-1/7) \implies y_1 = (1-m)/7$. So $Q = (0, (1-m)/7)$.Let $(h, k)$ be the midpoint of $PQ$. Then $h = (m-1)/(14m)$ and $k = (1-m)/14$.From $k = (1-m)/14$,we get $14k = 1-m$,so $m = 1-14k$.Substituting $m$ into $h = (m-1)/(14m)$,we get $h = (1-14k-1)/(14(1-14k)) = -14k/(14(1-14k)) = -k/(1-14k)$.Thus,$h(1-14k) = -k \implies h - 14hk = -k \implies h + k = 14hk$.Replacing $(h, k)$ with $(x, y)$,the locus is $x + y = 14xy$ or $x + y - 14xy = 0$.

If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$ meets the coordinate axes at the points $P$ and $Q$,then the locus of the midpoint of $PQ$ is:

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