Given pair of lines are x2−3xy+2y2=0 and ‌x2−3xy+2y2+x−2=0. ∴‌‌(x−2y)(x−y)=0 and ‌‌(x−2y+2)(x−y−1)=0 ⇒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∘. ∴ It is a parallelogram.