 gode 4. In the Cartesian coordinate system, the curve represented by
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$
$(a, b$ are two distinct positive numbers) is ( ).
(A) Triangle
(B) Square
(C) Non-square Rectangle
(D) Non-square Rhombus
If we remove the absolute value in equation (1), we need to consider four cases, which is rather complicated. By using symmetry, we can avoid the hassle of classification.
$\because$ In (1), interchanging $x$ and $y$ does not change the equation,
$\therefore$ it is symmetric about the line $y=2$;
$\therefore$ it is symmetric about the line $y=-x$.

From this, we can rule out (A).
Substituting $y=x$ into the original equation, we can easily find the two intersection points with the axis of symmetry to be $A(a, a), A^{\prime}(-a,-a)$.

Similarly, we can find the two intersection points with the axis of symmetry $y=-x$ to be $B(b, -b), B^{\prime}(-b, b)$, thus we can also rule out (B).
If it is a rectangle, then the point $(2 a, 0)$ on the $x$-axis should satisfy the original equation (as shown in the figure), but
$$
\begin{array}{l}
\frac{2 a}{2 a}+\frac{2 a}{2 b}=1+\frac{a}{b} \\
\neq 0 .
\end{array}
$$

Based on this, we can also rule out (C). Therefore, the only option left is (D).