Question

12. The points (0.5,) and (7, 13) are on the graph of a proportional relationship.

a. What is the constant of proportionality?
b. Name one more point on the graph.
c. Write an equation that represents the proportional relationship.

Answer 1

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

let's do a), c) and last b).

a)

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{13}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{13}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{7}-\underset{x_1}{0}}} \implies \cfrac{ 8 }{ 7 }`

c)

well, we know it passes through (7 , 13) and we know its slope, so let's use that

`(\stackrel{x_1}{7}~,~\stackrel{y_1}{13})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{8}{7} \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{13}=\stackrel{m}{ \cfrac{8}{7}}(x-\stackrel{x_1}{7}) \\\\\\ y-13=\cfrac{8}{7}x-8\implies y=\cfrac{8}{7}x+5`

b)

another point? well, hmmm let's pick a random "x" value hmmm say 7/8, so

`y=\cfrac{8}{7}x+5\qquad \qquad \boxed{x=\cfrac{7}{8}}\hspace{3em}y=\cfrac{8}{7}\stackrel{x}{\left( \cfrac{7}{8} \right)}+5\implies y=1+5\implies \boxed{y=6} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{another~point}{{\Large \begin{array}{llll} \left((7)/(8)~~,~~6 \right) \end{array}}}~\hfill`