Given line, 2x2−xy−y2=0 ...(i) ⇒ 2x2−2xy+xy−y2=0 2x(x−y)+y(x−y)=0 (2x+y)(x−y)=0 Since, we have given that combined equation of the pair is parallel to 2x2−xy−y2=0 So, combined equation will be (2x+y+k1)(x−y+k2)=0 where k1 and k2 ...(ii) are constants. This equation satisfies (1,0). ∴ (2×1+0+k1)(1−0+k2)=0 ⇒ k2=−1 and k1=−2 On putting the values of k1 and k2 in Eq. (ii) (2x+y−2)(x−y−1)=0 ⇒ 2x2−xy−y2−4x+y+2=0