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 \[\left( x-2y+2 \right)\left( x-y-1 \right)=0\] \[\Rightarrow \,\,\,x-2y=0,\text{ }x-y=0\] and \[x-2y+2=0,\text{ }x-y-1=0\] Since, the lines \[x-2y=0,\text{ }x-2y+2=0\]and\[x-y=0,\text{ }x-y-1=0\]are parallel. Also, angle between \[x-2y=0\]and \[x-y=0\]is not\[90{}^\circ \]. \[\therefore \] It is a parallelogram.
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