Given pairs of lines 2x2−5xy+2y2+6x−3y=0 ⇒(2y−x−3)(y−2x)=0 ∴ Pairs of line are 2y−x−3=0 and y−2x=0 Solving these equation we get, x=1,y=2 ∴α=1 Equation of line passing through origin is y−2x=0 ∵ Slope of line m=2 ∴β=2 Equation of angular bisector of line 2y−x−3=0 and y−2x=0 is
2y−x−3
√5
=±
y−2x
√5
2y−x−3=y−2x or 2y−x−3=−y+2x ∵x+y−3=0 or 3x−3y+3=0 x+y−3=0 or x−y+1=0 γ=−3 ∴γ<α<β i.e., −3<1<2