Given: A wire of total resistance $9\,\Omega$ is bent into an equilateral triangle.

Since the triangle is equilateral, the wire is divided equally into 3 equal sides.

Therefore, resistance of each side is $R_{\text{side}}=\frac{9}{3}=3\,\Omega$

We need the equivalent resistance between any two vertices, say $A$ and $B$.

Between $A$ and $B$, there are two possible paths:

• Direct side $AB$: resistance $=3\,\Omega$
• Path through the third vertex $C$: resistance $=3+3=6\,\Omega$

These two paths are in parallel between $A$ and $B$.

So, $R_{\text{eq}}=\frac{3\times 6}{3+6}$

$R_{\text{eq}}=\frac{18}{9}=2\,\Omega$

Hence, the equivalent resistance across any two vertices is $\boxed{2\,\Omega}$