Question

Pls help me, it is due today! Thank You very much to whoever helps me!​

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}{3}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{1}-\stackrel{y1}{3}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{3}}} \implies \cfrac{ -2 }{ 3 } \implies - \cfrac{2}{3}`

`\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}{3}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y-3=-\cfrac{ 2 }{ 3 }x+2\implies {\Large \begin{array}{llll} y=-\cfrac{ 2 }{ 3 }x+5 \end{array}}`