Answer :
Solution
For this case we just need to use the concept of slope given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Case a
[tex]m=\frac{-2+2}{1+1}=\frac{-2+2}{2-1}=\frac{-2+2}{3-2}=0[/tex]Then is linear
Case b
[tex]m=\frac{-4+7}{0+2}\ne\frac{-3+4}{1-0}[/tex]Case c
[tex]m=\frac{1-0}{0+1}\ne\frac{4-1}{1-0}[/tex]Case d
[tex]m=\frac{-2+4}{0+4}=\frac{-1+2}{2-0}=\frac{0+1}{4-2}=\frac{1}{2}[/tex]is linear
Then the best options are b and c