135x2−136xy+33y2=0. Its pair of angular bisectors, is h(x2−y2)=(a−b)xy ⇒−68(x2−y2)=102xy −2(x2−y2)=3xy ⇒2x2+3xy−2y2=0 ∴2x2+4xy−xy−2y2=0 ⇒(2x−y)(x+2y)=0 Given lines makes equal angles with 2x−y+c=0, x+2y+d=0 for any c, d values.
