Question

Write the equation of the line passing through (-1,0) and (0, -3).

ulas
3x - y = 3
x - 3y = 3
Light
6:32
3x + y = -3
Eliminate
Tools
x + 3y = -3

Answer 1

`\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} \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`

Answer 2

3x + y = -3

Explanation:

The slope-intercept form of an equation of a line:

`y=mx+b`

m - slope

b - y-intercept → (0, b)

The formula of a slope:

`m=(y_2-y_1)/(x_2-x_1)`

We have two points (-1, 0) and (0, -3) → b = -3.

Calculate the slope:

`m=(-3-0)/(0-(-1))=(-3)/(1)=-3`

Finally we have the equation:

`y=-3x-3`

Convert it to the standard form (Ax + By = C):

`y=-3x-3` add 3x to both sides

`3x+y=-3`