6. In the Cartesian coordinate system, the curve represented by the equation $\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1$ (where $a, b$ are two distinct positive numbers) is
(A) Triangle
(B) Square
(C) Non-square rectangle
(D) Non-square rhombus
(D)
6. 【Analysis and Solution】Rotate the Cartesian coordinate system $X O Y$ counterclockwise by $45^{\circ}$ to obtain the new coordinate system $X^{\prime} O Y^{\prime}$. Point $P$ has coordinates $\left(x^{\prime}, y^{\prime}\right)$ in the coordinate system $X^{\prime} O Y^{\prime}$ and coordinates $(x, y)$ in the coordinate system $X O Y$. Then,
$$
\left\{\begin{array}{l}
x^{\prime}=\frac{1}{\sqrt{2}}(x+y), \\
y^{\prime}=\frac{1}{\sqrt{2}}(-x+y) .
\end{array}\right.
$$

The equation given in the problem is $\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1$.
Transforming it, we get $\frac{\left|x^{\prime}\right|}{a}+\frac{\left|y^{\prime}\right|}{b}=\sqrt{2}$.
Clearly, (2) represents a curve that is symmetric about the $x^{\prime}$-axis and the $y^{\prime}$-axis. In the first quadrant of $X^{\prime} O Y^{\prime}$, (2) becomes $\frac{x^{\prime}}{a}+\frac{y^{\prime}}{b}=\sqrt{2}$, which is the line segment $A B$, where $A(\sqrt{2} a, 0)$ and $B(0, \sqrt{2} b)$.

By symmetry, in the second quadrant, equation (2) is the line segment $B C$, where $C(-\sqrt{2} a, 0)$; in the third quadrant, equation (2) is the line segment $C D$, where $D(0,-\sqrt{2} b)$; and in the fourth quadrant, equation (2) is the line segment $A D$.

By symmetry, $A B=B C=C D=A D$. Since $a \neq b$, $A C \neq B D$, so $A B C D$ is a rhombus that is not a square.