Let R=\ (x, y): x, y ∈ N. and .x^2-4 x y+3 y^2=0\ , where N is the...

\mathrm{R}=\left\{(x, y): x, y \in \mathrm{N}\right. and \left.x^2-4 x y+3 y^2=0\right\}$Now,$x^2-4 x y+3 y^2=0 \Rightarrow(x-y)(x-3 y)=0 \therefore x=y$or$x=3 y \therefore \quad \mathrm{R}=\{(1,1),(3,1),(2,2),(6,2),(3,3), (9,3), \ldots \ldots\}$Since$(1,1),(2,2),(3,3), \ldots \ldots$are present in the relation, therefore$\mathrm{R}$is reflexive. Since$(3,1)$is an element of$R$but$(1,3)$is not the element of$\mathrm{R}$, therefore$\mathrm{R}$is not symmetric Here$(3,1) \in R$and$(1,1) \in R \Rightarrow(3,1) \in R (6,2) \in \mathrm{R} and (2,2) \in \mathrm{R} \Rightarrow(6,2) \in \mathrm{R}$For all such$(a, b) \in \mathrm{R}$and$(b, c) \in \mathrm{R} \Rightarrow(a, c) \in \mathrm{R}$Hence$\mathrm{R} is transitive.