Question

Help pls ghmthmjrmyurmuyjmuyj

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}{0}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{0}-\stackrel{y1}{3}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{0}}} \implies \cfrac{ -3 }{ 5 } \implies -\cfrac{3}{5}`

`\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}{3}=\stackrel{m}{-\cfrac{3}{5}}(x-\stackrel{x_1}{0})\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5(y-3)=5\left( -\cfrac{3}{5}(x-0) \right)} \\\\\\ 5y-15=-3(x-0)\implies 5y-15=-3x \\\\\\ 3x+5y-15=0\implies \stackrel{ \textit{\small A} }{3} x+\stackrel{ \textit{\small B} }{5} y=15`