jmc

We use the distance formula: $\sqrt{((-1)-2{)}^2+((-1)-2{)}^2}=\sqrt{9+9}=\boxed{3\sqrt{2}}$.

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We note that the points $(2,2)$, $(-1,-1)$, and $(2,-1)$ form an isosceles right triangle (a 45-45-90 triangle) with legs of length 3. Therefore, the hypotenuse has length $\boxed{3\sqrt{2}}$.