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 AB is [UPSEAT 2004]

Equation of LINE passing through point (1, 1) is,                    \[y-1=m(x-1)\]                                    ......(i)                    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 \[x=0\]                    \[\therefore \]  \[y-1=-m\Rightarrow y=1-m\]                    LET mid point of AB be (h, k),                    Then \[h=\frac{0+(1-(1/m))}{2}\];\[k=\frac{0+(1-m)}{2}\]                    \[m=\frac{1}{1-2h}\] ; \[m=1-2k\]                    \ \[1-2k=\frac{1}{1-2h}\]                    Þ \[1-2k-2h+4hk=1\] Þ \[-2h-2k+4hk=0\]                    Hence the Locus of mid point is, \[x+y-2xy=0\].