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