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(D) Let the two lines be $L_1: x + y = 3$ and $L_2: x - y = -3$. The intersection of these two lines gives one vertex of the parallelogram,say $A$. So

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$(3, 6)$

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(D) Let the two lines be $L_1: x + y = 3$ and $L_2: x - y = -3$. The intersection of these two lines gives one vertex of the parallelogram,say $A$. Solving $x + y = 3$ and $x - y = -3$ by adding them gives $2x = 0 \Rightarrow x = 0$,and substituting $x = 0$ into $x + y = 3$ gives $y = 3$. So,$A = (0, 3)$.Let the diagonals intersect at $P(2, 4)$. In a parallelogram,the diagonals bisect each other. If $A(0, 3)$ is a vertex,the opposite vertex $C(x_c, y_c)$ satisfies $\frac{0 + x_c}{2} = 2$ and $\frac{3 + y_c}{2} = 4$,which gives $x_c = 4$ and $y_c = 5$. Thus,$C = (4, 5)$.Since the sides are parallel to the given lines,the other two vertices $B$ and $D$ lie on lines parallel to $L_1$ and $L_2$ passing through $C(4, 5)$. The line through $C$ parallel to $x + y = 3$ is $x + y = 4 + 5 = 9$. The line through $C$ parallel to $x - y = -3$ is $x - y = 4 - 5 = -1$.Vertex $B$ is the intersection of $x + y = 3$ and $x - y = -1$. Adding gives $2x = 2 \Rightarrow x = 1$,and $y = 2$. So $B = (1, 2)$.Vertex $D$ is the intersection of $x + y = 9$ and $x - y = -3$. Adding gives $2x = 6 \Rightarrow x = 3$,and $y = 6$. So $D = (3, 6)$.Comparing with the options,$(3, 6)$ is a vertex.

Two sides of a parallelogram are along the lines $x + y = 3$ and $x - y + 3 = 0$. If its diagonals intersect at $(2, 4)$,then one of its vertices is

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