Let R= (x, y): x, y ∈ N . and .x^2-4 x y+3 y^2=0, where N is the set of all natural numbers. Then the relation R is :

$\mathrm{R}=\left\{(x, y): x, y \in \mathrm{N}\right.$R={(x,y):x,y∈N and $\left.x^2-4 x y+3 y^2=0\right\}$x2−4xy+3y2=0} Now, $x^2-4 x y+3 y^2=0$x2−4xy+3y2=0 $\Rightarrow(x-y)(x-3 y)=0$⇒(x−y)(x−3y)=0 $\therefore x=y$∴x=y or $x=3 y$x=3y $\therefore \quad \mathrm{R}=\{(1,1),(3,1),(2,2),(6,2),(3,3)$∴R={(1,1),(3,1),(2,2),(6,2),(3,3), $(9,3), \ldots \ldots\}$(9,3),……} Since $(1,1),(2,2),(3,3), \ldots \ldots$(1,1),(2,2),(3,3),…… are present in the relation, therefore $\mathrm{R}$R is reflexive. Since $(3,1)$(3,1) is an element of $R$R but $(1,3)$(1,3) is not the element of $\mathrm{R}$R, therefore $\mathrm{R}$R is not symmetric Here $(3,1) \in R$(3,1)∈R and $(1,1) \in R \Rightarrow(3,1) \in R$(1,1)∈R⇒(3,1)∈R $(6,2) \in \mathrm{R}$(6,2)∈R and $(2,2) \in \mathrm{R} \Rightarrow(6,2) \in \mathrm{R}$(2,2)∈R⇒(6,2)∈R For all such $(a, b) \in \mathrm{R}$(a,b)∈R and $(b, c) \in \mathrm{R}$(b,c)∈R $\Rightarrow(a, c) \in \mathrm{R}$⇒(a,c)∈R Hence $\mathrm{R}$R is transitive.