The correct option is
C Paralleogram
Given pairs
x2−3xy+2y2=0
(x−2y)(x−y)=0
x−2y=0 and x−y=0 are two sides
x2−3xy+2y2+x−2=0
x=−(1−3y)±√1+9y2−6y−4(2y2−2)2
x=−(1−3y)±√1+9y2−6y−8y2+8)2
2x=−1+3y±√y2−6y+9)
2x=−1+3y±√(y−3)2)
2x=−1+3y±(y−3)
2x=−1+3y+y−3 and 2x=−1+3y−y+3
2x−4y=−4 and 2x−2y=2
x−2y=−2 and x−y=1 are other two sides
x−2y=0 and x−y=0 and x−2y=−2 and x−y=1 are four sides
Here two opposite sides are parallel
Angle between x−2y=0 and x−y=0 by formula is
tanθ=∣∣∣m1−m21+m1m2∣∣∣
tanθ=∣∣
∣
∣∣1−121+12∣∣
∣
∣∣
tanθ=13
Here the sides are not perpendicular
Hence it is parallelogram