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Which set of ordered pairs $(x, y)$ could represent a linear function?
\[
\begin{array}{l}
\mathbf{A}=\{(-6,5),(-3,2),(0,-1),(3,-4)\} \\
\mathbf{B}=\{(-3,5),(-1,4),(1,3),(4,1)\} \\
\mathbf{C}=\{(2,-3),(4,-1),(7,1),(9,3)\} \\
\mathbf{D}=\{(-6,9),(-4,3),(-1,-3),(2,-9)\}
\end{array}
\]

Answer
A
B
C
D

Step 1 ：Calculate the slope for each set of ordered pairs in option A using the formula \((y2 - y1) / (x2 - x1)\).

Step 2 ：The slope between the first two points in A is \((-3 / 3) = -1\).

Step 3 ：The slope between the second and third points in A is \((-3 / 3) = -1\).

Step 4 ：The slope between the third and fourth points in A is \((-3 / 3) = -1\).

Step 5 ：Since all the slopes in A are the same, A could represent a linear function.

Step 6 ：Calculate the slope for each set of ordered pairs in option B using the formula \((y2 - y1) / (x2 - x1)\).

Step 7 ：The slope between the first two points in B is \((-1 / 2) = -0.5\).

Step 8 ：The slope between the second and third points in B is \((-1 / 2) = -0.5\).

Step 9 ：The slope between the third and fourth points in B is \((-2 / 3) = -0.67\).

Step 10 ：Since all the slopes in B are not the same, B could not represent a linear function.

Step 11 ：Calculate the slope for each set of ordered pairs in option C using the formula \((y2 - y1) / (x2 - x1)\).

Step 12 ：The slope between the first two points in C is \((2 / 2) = 1\).

Step 13 ：The slope between the second and third points in C is \((2 / 3) = 0.67\).

Step 14 ：The slope between the third and fourth points in C is \((2 / 2) = 1\).

Step 15 ：Since all the slopes in C are not the same, C could not represent a linear function.

Step 16 ：Calculate the slope for each set of ordered pairs in option D using the formula \((y2 - y1) / (x2 - x1)\).

Step 17 ：The slope between the first two points in D is \((-6 / 2) = -3\).

Step 18 ：The slope between the second and third points in D is \((-6 / 3) = -2\).

Step 19 ：The slope between the third and fourth points in D is \((-6 / 3) = -2\).

Step 20 ：Since all the slopes in D are not the same, D could not represent a linear function.

Step 21 ：Therefore, the answer is \(\boxed{A}\).