Common line of two pairs is 3x+4y=0 ∴ Equation of L is (y−1)=
−3
4
(x+1) ⇒L is 3x+4y−1=0⇒3x+4y=1 Homoginising given curve with 3x+4y−1=0, we get 2x2−xy−y2+(3x+4y)x−y(3x+4y)=0 ⇒5x2−5y2=0⇒x2−y2=0 ∴ Required equation of pair of straight lines is x2−y2=0