Question

Mathematics grade 9
answer??

Answer 1

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

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-1}-\stackrel{y1}{(-3)}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{0}}} \implies \cfrac{ -1 +3 }{ 5 } \implies \cfrac{ 2 }{ 5 }`

`\begin{array}c \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}{5}}(x-\stackrel{x_1}{0}) \implies y +3 = \cfrac{2}{5} ( x -0) \\\\\\ y + 3 =\cfrac{2}{5}x\implies y=\cfrac{2}{5}x-3 \\\\[-0.35em] ~\dotfill`

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-11}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-11}-\stackrel{y1}{(-8)}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{0}}} \implies \cfrac{ -11 +8 }{ 5 } \implies \cfrac{ -3 }{ 5 } \implies -\cfrac{3}{5}`

`\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}{(-8)}=\stackrel{m}{-\cfrac{3}{5}}(x-\stackrel{x_1}{0}) \implies y +8 = -\cfrac{3}{5} ( x -0) \\\\\\ y+8=-\cfrac{3}{5}x\implies y=-\cfrac{3}{5}x-8 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill~ \begin{cases} y=\cfrac{2}{5}x-3 \\\\\\ y=-\cfrac{3}{5}x-8 \end{cases}~\hfill~`