1. **Problem:** Determine the domain of the function $f(x,y) = \sqrt{x + y - 1}$. 2. **Formula and rules:** The square root function $\sqrt{z}$ is defined only for $z \geq 0$. Therefore, the expression inside the square root must be non-negative: $$x + y - 1 \geq 0$$ 3. **Solve the inequality:** $$x + y - 1 \geq 0 \implies y \geq 1 - x$$ This inequality describes the domain of $f$ as all points $(x,y)$ in the plane such that $y$ is greater than or equal to $1 - x$. 4. **Interpretation:** The domain is the half-plane above (and including) the line $y = 1 - x$. 5. **Final answer:** $$\boxed{\text{Domain} = \{(x,y) \in \mathbb{R}^2 : y \geq 1 - x\}}$$