If 2x2−3xy+y2=0  represents two sides of a triangle and lx + my + n = 0 is the third side then the locus of in centre of the triangle is

Given 2x2−3xy+y2=0 2x2−2xy−xy+y2=02x(x−y)−y(x−y)=0x−y=0             2x−y=0Let (α,β) be in-centre⊥ Distance from centre to both the lines are equal (radius of the incircle)α−β2=2α−β55(α−β)2=2(2α−β)2∴ Locus ⇒3x2+2xy−3y2=0