(b) : We have, $(a,b)R(c,d) \Leftrightarrow ad(b - c) = bc(a - d)$\\ $\Rightarrow \frac{b - c}{bc} = \frac{a - d}{ad}$ and $\frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$\\ For reflexive : $(a,b)R(a,b) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$ which is false.\\ Hence it is not reflexive.\\ For symmetric, $(a,b)R(c,d) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$\\ $\Rightarrow \frac{1}{c} - \frac{1}{d} = \frac{1}{b} - \frac{1}{a} \Rightarrow (c,d)R(a,b)$ Hence, it is symmetric.\\ For transitive : $(a,b)R(c,d) \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$\\ and $(c,d)R(e,f) \Rightarrow \frac{1}{c} - \frac{1}{d} = \frac{1}{f} - \frac{1}{e}$\\ $\Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f} \neq (a,b)R(e,f)$ \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{13} Hence, it is not transitive.