To solve this problem, let's try to find a point on the line. It appears that (6, 8) is a coordinate on the line. We can substitute this x-value of 6 into the functions and question whether it equals the y-value of 8.
Equation 1:
[tex]8 + 2 \stackrel{?}{=} \dfrac{6}{5} (6 + 1)[/tex]
[tex]10 \neq \dfrac{42}{5}[/tex]
The first option choice is not the equation of our line.
Equation 2:
[tex](8 - 2) \stackrel{?}{=} \dfrac{6}{5} (6 - 1)[/tex]
[tex]6 = 6 \,\,\checkmark[/tex]
The second choice is the equation of our line. For more verification, we will check the other equations as well.
Equation 3:
[tex]8 + 1 \stackrel{?}{=} \dfrac{6}{5} (6 + 2)[/tex]
[tex]9 \neq \dfrac{48}{5}[/tex]
The third choice is not the equation of our line.
Equation 4:
[tex]8 - 1 \stackrel{?}{=} \dfrac{6}{5} (6 - 2)[/tex]
[tex]7 \neq \dfrac{24}{5}[/tex]
The fourth option is not the equation of our line.
The second choice is the equation of the line.
