Question

Pls help meeeeeeeee I really need to turn in this packet

Answer 1

to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below

`(\stackrel{x_1}{300}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{600}~,~\stackrel{y_2}{8}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{600}-\underset{x_1}{300}}} \implies \cfrac{ 4 }{ 300 } \implies \cfrac{ 1 }{ 75 }`

`\begin{array} \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}{4}=\stackrel{m}{ \cfrac{ 1 }{ 75 }}(x-\stackrel{x_1}{300}) \\\\\\ y-4 = \cfrac{ 1 }{ 75 }x-4\implies {\Large \begin{array}{llll} y=\cfrac{ 1 }{ 75 }x \end{array}} \\\\[-0.35em] ~\dotfill`

`(\stackrel{x_1}{180}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{450}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{\textit{\large run}} {\underset{x_2}{450}-\underset{x_1}{180}}} \implies \cfrac{ 3 }{ 270 } \implies \cfrac{ 1 }{ 90 }`

`\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}{2}=\stackrel{m}{ \cfrac{ 1 }{ 90 }}(x-\stackrel{x_1}{180}) \\\\\\ y-2=\cfrac{ 1 }{ 90 }x-2\implies {\Large \begin{array}{llll} y=\cfrac{ 1 }{ 90 }x \end{array}}`