A

Step 1: Calculate the downstream and upstream speeds.
The effective speed of the boat going downstream is given by \(v_b + v_r\), and for upstream, it is \(v_b - v_r\).
Step 2: Use the formula \(distance = speed \times time\). For downstream, the formula becomes \(x = (v_b + v_r) t_1\) and for upstream, it becomes \(x = (v_b - v_r) t_2\).
Step 3: Rearranging the equations gives us two relationships: \(t_1 = \frac{x}{(v_b + v_r)}\) and \(t_2 = \frac{x}{(v_b - v_r)}\).
Step 4: From the ratios, we have \(\frac{t_1}{t_2} = \frac{(v_b - v_r)}{(v_b + v_r)}\).
Step 5: Therefore, simplifying gives: \(t_1 = (v_b - v_r)t_2\) which implies that \( (v_b - v_r) t_2 = vt_1\) or \( (v_b - v_r) t_2 = vt_1\) leads us to option [A].
Hence, the answer is A.