Question

which set of ordered pairs (x,y) could represent a linear function? a = {(-2,-1), (1,1), (4,2), (7,4)} b = {(-7,-7), (-3,-5), (3,-2), (7,0)} c = {(-2,-3), (0,0), (2,3), (5,6)} d = {(-6,2), (-2,1), (1,0), (4,-1)}

Explanation:

Step1: Recall linear function slope formula

For a linear function, the slope \( m \) between any two points \((x_1,y_1)\) and \((x_2,y_2)\) should be constant, where \( m=\frac{y_2 - y_1}{x_2 - x_1} \). We will calculate the slope between consecutive points for each set.

Step2: Calculate slope for set A

• Between \((-2,-1)\) and \((1,1)\): \( m_1=\frac{1 - (-1)}{1 - (-2)}=\frac{2}{3} \)
• Between \((1,1)\) and \((4,2)\): \( m_2=\frac{2 - 1}{4 - 1}=\frac{1}{3} \)

Since \( m_1
eq m_2 \), A is not linear.

Step3: Calculate slope for set B

• Between \((-7,-7)\) and \((-3,-5)\): \( m_1=\frac{-5 - (-7)}{-3 - (-7)}=\frac{2}{4}=\frac{1}{2} \)
• Between \((-3,-5)\) and \((3,-2)\): \( m_2=\frac{-2 - (-5)}{3 - (-3)}=\frac{3}{6}=\frac{1}{2} \)
• Between \((3,-2)\) and \((7,0)\): \( m_3=\frac{0 - (-2)}{7 - 3}=\frac{2}{4}=\frac{1}{2} \)

All slopes are equal (\(\frac{1}{2}\)), so B is a candidate. Let's check others to be sure.

Step4: Calculate slope for set C

• Between \((-2,-3)\) and \((0,0)\): \( m_1=\frac{0 - (-3)}{0 - (-2)}=\frac{3}{2} \)
• Between \((0,0)\) and \((2,3)\): \( m_2=\frac{3 - 0}{2 - 0}=\frac{3}{2} \)
• Between \((2,3)\) and \((5,6)\): \( m_3=\frac{6 - 3}{5 - 2}=\frac{3}{3}=1 \)

Since \( m_2
eq m_3 \), C is not linear.

Step5: Calculate slope for set D

• Between \((-6,2)\) and \((-2,1)\): \( m_1=\frac{1 - 2}{-2 - (-6)}=\frac{-1}{4}=-\frac{1}{4} \)
• Between \((-2,1)\) and \((1,0)\): \( m_2=\frac{0 - 1}{1 - (-2)}=\frac{-1}{3} \)

Since \( m_1
eq m_2 \), D is not linear.

Answer:

B. \(\{(-7, -7),\ (-3, -5),\ (3, -2),\ (7, 0)\}\)