Given, (a,b)R(c,d)⇒ad(b−c)=bc(a−d)
⇒(a,b)R(c,d)⇔bcb−c=ada−d
⇒c1−b1=d1−a1
⇒a1−b1=d1−c1
Reflexive:
(a,b)R(a,b)⇒a1−b1=b1−a1
It is false. So, given relation is not reflexive.
Symmetric:
(a,b)R(c,d)⇒a1−b1=d1−c1
⇒c1−d1=b1−a1
∴(c,d)R(a,b)
It is symmetric.
Transitive:
(a,b)R(c,d)⇒a1−b1=d1−c1
(c,d)R(e,f)=c1−d1=e1−f1
a1−b1=e1−f1
⇒(a,b)R(e,f)
So, not transitive.