A) square but not rhombus
B) rhombus
C) parallelogram
D) rectangle but not a square
Correct Answer: C
Solution :
Given pair of lines are\[{{x}^{2}}-3xy+2{{y}^{2}}=0\]and \[{{x}^{2}}-3xy+2{{y}^{2}}+x-2=0\]. \[\therefore \] \[(x-2y)(x-y)=0\] and \[(x-2y+2)(x-y-1)=0\] \[\Rightarrow \]\[x-2y=0\],\[x-y=0\]and\[x-2y+2=0\], \[x-y-1=0\] Since, the lines\[x-2y=0\],\[x-2y+2=0\]and \[x-y=0\],\[x-y-1=0\]are parallel. Also, angle between\[x-2y=0\]and\[x-y=0\]is not\[{{90}^{o}}\]. \[\therefore \]It is a parallelogram.
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