2. Define " $\triangle$ " as an operation on ordered pairs of real numbers, as shown below, $(a, b) \triangle(c, d)=(a c+b d, a d+b c)$. If for any real numbers $a, b$ we have $(a, b) \triangle(x, y)=(a, b)$, then $(x, y)$ is ( ).
(A) $(0,1)$
(B) $(1,0)$
(C) $(-1,0)$
(D) $(0,-1)$
2. (B).

By definition, we have
$$
(a, b)=(a, b) \doteq(x, y)=(a x+b y, a y+b x) \text {. }
$$

Then, \(a x+b y=a, a y+b x=b\).
Adding the two equations, we get
$$
(a+b) x+(a+b) y=a+b \text {. }
$$

By the arbitrariness of \(a, b\), we know \(x+y=1\).
Subtracting the two equations, we get \((a-b) x-(a-b) y=a-b\).
Similarly, we get \(x-y=1\).
Therefore, \(x=1, y=0\).