1. **Problem statement:** The random vector $(X,Y)$ is uniformly distributed over region $A$ defined by vertices $(1,1)$, $(1,0)$, $(2,0)$, and $(2,2)$. 2. **Describe region $A$ algebraically:** - The $x$-coordinate ranges between 1 and 2: $1 \leq x \leq 2$. - The lower boundary is the $x$-axis segment $y=0$, for $1 \leq x \leq 2$. - The upper boundary is piecewise: - For $1 \leq x < 2$: $y \leq 1$ - At $x=2$: $y$ goes from 0 up to 2, so $0 \leq y \leq 2$ To combine smoothly, since the polygon is bounded by points $(1,1)$ to $(2,1)$ horizontally, and then from $(2,1)$ to $(2,2)$ vertically, the region can be described as: $$A = \{(x,y) : 1 \leq x < 2, 0 \leq y \leq 1 \} \cup \{(x,y) : x = 2, 0 \leq y \leq 2 \}$$ For the continuous region (neglecting single line $x=2$), the uniform distribution mainly covers $\{(x,y): 1 \leq x \leq 2, 0 \leq y \leq 1\}$ plus the vertical line at $x=2$ up to $y=2$ (which has zero area), so effectively: $$A = \{(x,y) : 1 \leq x \leq 2, 0 \leq y \leq 1\}$$ 3. **Joint density function:** - Since $(X,Y)$ is uniform on $A$, - The area of $A$ is base $= 2 - 1 = 1$ and height $=1$ so area $=1 \times 1 =1$. Hence, $$f_{X,Y}(x,y) = \begin{cases} 1, & 1 \leq x \leq 2, 0 \leq y \leq 1 \\ 0, & \text{otherwise} \end{cases}$$ This is a uniform distribution over the rectangle from $x=1$ to 2 and $y=0$ to 1. 4. **Conclude:** Region $A$ algebraically is $1 \leq x \leq 2$, $0 \leq y \leq 1$, and the joint density is constant $1$ over this region.