A) \[2xy+x+y=0\]
B) \[x+y-2xy=0\]
C) \[x+y+2=0\]
D) \[x+y-2=0\]
Correct Answer: B
Solution :
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, \[x+y-2xy=0\].
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