Let a relation R in the set N of natural numbers be defined as (x, y)$\in$R if and only if ${\mathrm{x}}^2-4\mathrm{xy}+3{\mathrm{y}}^2=0$ for all x, y$\in$N. The relation R is 

We have $\mathrm{R}=\left\{\right(\mathrm{x},\mathrm{y}):\mathrm{x}2-4\mathrm{xy}+3\mathrm{y}2=0,\mathrm{x},\mathrm{y}\in \mathrm{N}\}$
Let $\mathrm{x}\in \mathrm{N}\cdot {\mathrm{x}}^2-4\mathrm{xy}+3{\mathrm{x}}^2=4{\mathrm{x}}^2-4{\mathrm{x}}^2=0$
$\therefore (\mathrm{x},\mathrm{y})\in \mathrm{R}\therefore \mathrm{R} \mathrm{is} \mathrm{reflexive}$
We have $(\mathrm{C}{)}^2-4(\mathrm{C}\left)\right(1{)}^2=9-12+3=0\therefore (3,1)\in \mathrm{R}$
Also, $(1{)}^2-4(1\left)\right(\mathrm{C})+3(\mathrm{C}{)}^2=81-108+27=0$
$(1,3)\notin \mathrm{R}\therefore \mathrm{R} \mathrm{is} \mathrm{not} \mathrm{symmetric}$
$(9,3)\in \mathrm{R} \mathrm{because}$
$(9{)}^2-4(9\left)\right(\mathrm{C})+3(1{)}^2=9-12+27=0$
Now, $(9,1)\in \mathrm{R}\text{ if }(9{)}^2-4(9\left)\right(1)+3(1{)}^2=0\text{ if }81-36+3=48\ne 0$
Which is not so $\therefore (9,3),(3,1)\in \mathrm{R}\text{ and }(9,1)\notin \mathrm{R}\therefore \mathrm{R}$ is not transitive.