[tex]\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-1)}}}\implies \cfrac{-3}{0+1}\implies -3 \\\\\\ \begin{array}{|c|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}{0}=\stackrel{m}{-3}[x-\stackrel{x_1}{(-1)}] \\\\\\ y=-3(x+1)\implies y = -3x-3[/tex]
Answer:
Step-by-step explanation:
The slope-intercept form of an equation of a line:
[tex]y=mx+b[/tex]
m - slope
b - y-intercept → (0, b)
The formula of a slope:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
We have two points (-1, 0) and (0, -3) → b = -3.
Calculate the slope:
[tex]m=\dfrac{-3-0}{0-(-1)}=\dfrac{-3}{1}=-3[/tex]
Finally we have the equation:
[tex]y=-3x-3[/tex]
Convert it to the standard form (Ax + By = C):
[tex]y=-3x-3[/tex] add 3x to both sides
[tex]3x+y=-3[/tex]
