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A straight line through the point $(1,1)$ meets the $x$-axis at ' $A$ ' and the $y$-axis at ' $B$ '. The locus of the mid-point of $A B$ is
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The correct answer is:
$x+y-2 x y=0$
Equation of line passing through point $(1,1)$ is,
$y-1=m(x-1)$
Line (i) meets $x$-axis, so $y=0$
$\therefore \frac{-1}{m}=x-1 \Rightarrow x=1-\frac{1}{m}$
Line (i) meets $y$-axis, so $x=0$
$\therefore \quad y-1=-m \Rightarrow y=1-m$
Let mid point of $A B$ be $(h, k)$,
Then $h=\frac{0+(1-(1 / m))}{2} ; k=\frac{0+(1-m)}{2}$
$\begin{aligned}
& m=\frac{1}{1-2 h} ; m=1-2 k \\
& \quad 1-2 k=\frac{1}{1-2 h} \\
& \Rightarrow 1-2 k-2 h+4 h k=1 \Rightarrow-2 h-2 k+4 h k=0
\end{aligned}$
Hence the Locus of mid point is, $x+y-2 x y=0$
$y-1=m(x-1)$
Line (i) meets $x$-axis, so $y=0$
$\therefore \frac{-1}{m}=x-1 \Rightarrow x=1-\frac{1}{m}$
Line (i) meets $y$-axis, so $x=0$
$\therefore \quad y-1=-m \Rightarrow y=1-m$
Let mid point of $A B$ be $(h, k)$,
Then $h=\frac{0+(1-(1 / m))}{2} ; k=\frac{0+(1-m)}{2}$
$\begin{aligned}
& m=\frac{1}{1-2 h} ; m=1-2 k \\
& \quad 1-2 k=\frac{1}{1-2 h} \\
& \Rightarrow 1-2 k-2 h+4 h k=1 \Rightarrow-2 h-2 k+4 h k=0
\end{aligned}$
Hence the Locus of mid point is, $x+y-2 x y=0$
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