which set of ordered pairs (x,y) could represent a linear function? a = {(-1,4), (1,2), (3,-1), (5,-3)} b =…

Explanation:

Step1: Recall slope - formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For a linear function, the slope between any two points is constant.

Step2: Calculate slopes for set A

For points $(-1,4)$ and $(1,2)$: $m_1=\frac{2 - 4}{1-(-1)}=\frac{-2}{2}=-1$. For points $(1,2)$ and $(3,-1)$: $m_2=\frac{-1 - 2}{3 - 1}=\frac{-3}{2}$. Since $m_1\neq m_2$, set A is not a linear function.

Step3: Calculate slopes for set B

For points $(-5,6)$ and $(-1,1)$: $m_1=\frac{1 - 6}{-1-(-5)}=\frac{-5}{4}$. For points $(-1,1)$ and $(1,-2)$: $m_2=\frac{-2 - 1}{1-(-1)}=\frac{-3}{2}$. Since $m_1\neq m_2$, set B is not a linear function.

Step4: Calculate slopes for set C

For points $(-4,-7)$ and $(-2,-4)$: $m_1=\frac{-4-(-7)}{-2-(-4)}=\frac{3}{2}$. For points $(-2,-4)$ and $(0,-1)$: $m_2=\frac{-1-(-4)}{0 - (-2)}=\frac{3}{2}$. For points $(0,-1)$ and $(2,2)$: $m_3=\frac{2-(-1)}{2 - 0}=\frac{3}{2}$. Since the slope is constant ($m = \frac{3}{2}$), set C represents a linear function.

Step5: Calculate slopes for set D

For points $(2,8)$ and $(3,5)$: $m_1=\frac{5 - 8}{3 - 2}=-3$. For points $(3,5)$ and $(4,3)$: $m_2=\frac{3 - 5}{4 - 3}=-2$. Since $m_1\neq m_2$, set D is not a linear function.

Answer:

C. ${(-4,-7),(-2,-4),(0,-1),(2,2)}$