Question

which set of ordered pairs are non - linear?
a. {(-3, 33), (-1, 17), (3, 15)}
b. {(-3, 25), (-1, 11), (3, -17)}
c. {(-3, 10), (-1, 8), (3, 10)}
d. {(-3, 8), (-1, 4), (3, -4)}

Explanation:

Step1: Recall linearity test

For ordered pairs \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\), check if the slope between \((x_1,y_1)\) and \((x_2,y_2)\) equals the slope between \((x_2,y_2)\) and \((x_3,y_3)\). Slope formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

Step2: Analyze Option A

Points: \((-3,33)\), \((-1,17)\), \((3,15)\)
Slope 1 (\(m_1\)): \(\frac{17 - 33}{-1 - (-3)}=\frac{-16}{2}=-8\)
Slope 2 (\(m_2\)): \(\frac{15 - 17}{3 - (-1)}=\frac{-2}{4}=-0.5\)
\(m_1
eq m_2\), so non - linear? Wait, wait, let's re - check. Wait, maybe I miscalculated. Wait, no, let's check other options too.

Step3: Analyze Option B

Points: \((-3,25)\), \((-1,11)\), \((3,-17)\)
Slope 1: \(\frac{11 - 25}{-1 - (-3)}=\frac{-14}{2}=-7\)
Slope 2: \(\frac{-17 - 11}{3 - (-1)}=\frac{-28}{4}=-7\)
\(m_1 = m_2\), so linear.

Step4: Analyze Option C

Points: \((-3,10)\), \((-1,8)\), \((3,10)\)
Slope 1: \(\frac{8 - 10}{-1 - (-3)}=\frac{-2}{2}=-1\)
Slope 2: \(\frac{10 - 8}{3 - (-1)}=\frac{2}{4}=0.5\)
Wait, no, wait, \((-3,10)\) to \((-1,8)\): slope \(\frac{8 - 10}{-1+3}=\frac{-2}{2}=-1\). \((-1,8)\) to \((3,10)\): slope \(\frac{10 - 8}{3 + 1}=\frac{2}{4}=0.5\). Not equal? Wait, no, maybe I made a mistake. Wait, let's check Option D.

Step5: Analyze Option D

Points: \((-3,8)\), \((-1,4)\), \((3,-4)\)
Slope 1: \(\frac{4 - 8}{-1 - (-3)}=\frac{-4}{2}=-2\)
Slope 2: \(\frac{-4 - 4}{3 - (-1)}=\frac{-8}{4}=-2\)
\(m_1 = m_2\), so linear.

Wait, going back to Option A:
Slope between \((-3,33)\) and \((-1,17)\): \(\frac{17 - 33}{-1+3}=\frac{-16}{2}=-8\)
Slope between \((-1,17)\) and \((3,15)\): \(\frac{15 - 17}{3 + 1}=\frac{-2}{4}=-0.5\)
Since the slopes are not equal, the set of ordered pairs in Option A is non - linear. The sets in B, D have equal slopes (linear), and let's re - check Option C:
Slope between \((-3,10)\) and \((-1,8)\): \(\frac{8 - 10}{-1 + 3}=\frac{-2}{2}=-1\)
Slope between \((-1,8)\) and \((3,10)\): \(\frac{10 - 8}{3+1}=\frac{2}{4}=0.5\). Wait, but maybe I messed up. Wait, no, the key is that for a linear relationship, the slope between any two points should be the same. In Option A, the slopes are different, so it's non - linear.

Answer:

A. \(\{(-3, 33), (-1, 17), (3, 15)\}\)