To solve this, we are going to use the slope formula, and the point slope formula.
Slope formula: [tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]
where
[tex]m[/tex] is the slope.
[tex](x_{1},y_{1})[/tex] are the coordinates of the first point.
[tex](x_{2},y_{2})[/tex] are the coordinates of the first point.
Point-slope formula: [tex]y-y_{1}=m(x-x_{1})[/tex]
Plan A. Coordinates of the first point: (1,14) (number of discs, cost). Coordinates of the second point: (2,17). Lets replace those values in our slope formula:
[tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]
[tex]m= \frac{17-14}{2-1} [/tex]
[tex]m=3[/tex]
Now that we have the slope, we can use the point slope formula:
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-14=3(x-1)[/tex]
[tex]y-14=3x-3[/tex]
[tex]y=3x+11[/tex]
Remember that in a linear equation of the form [tex]y=mx+b[/tex], [tex]m[/tex] is the slope and [tex]b[/tex] is the y-intercept. In our model the y-intercept is the membership fee.
Since [tex]b=11[/tex] in our linear equation, we can conclude that the membership fee is $11
Plan B. Coordinates of the first point (1,12). Coordinates of the second point (2,16).
[tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]
[tex]m= \frac{16-12}{2-1} [/tex]
[tex]m=4[/tex]
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-12=4(x-1)[/tex]
[tex]y-12=4x-4[/tex]
[tex]y=4x+8[/tex]
Membership fee: $8
Plan C. Coordinates of the first point: (1,10). Coordinates of the second point: (2,15).
[tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]
[tex]m= \frac{15-10}{2-1} [/tex]
[tex]m=5[/tex]
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-10=5(x-1)[/tex]
[tex]y-10=5x-5[/tex]
[tex]y=5x+5[/tex]
Membership fee: $5
Part D. Coordinates of the first point: (1,12). Coordinates of the second point (2,21).
[tex]m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]
[tex]m= \frac{21-12}{2-1} [/tex]
[tex]m=9[/tex]
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-12=9(x-1)[/tex]
[tex]y-12=9x-9[/tex]
[tex]y=9x+3[/tex]
Membership fee: $3
We can conclude that the plan that has the smallest one-time membership is Plan D.
