# How do you write an equation of a line given slope of m=-1/3 and passes through the point (-1,1)?

Mar 4, 2018

$y = - \frac{1}{3} x + \frac{2}{3}$

#### Explanation:

Use point slope form: $y - {y}_{1} = m \left(x - {x}_{1}\right)$ You can use this equation if you have the given point and the slope of the line. Hope this helps!

Mar 4, 2018

The equation of the line is $y = - \frac{1}{3} x + \frac{2}{3}$.

#### Explanation:

You can write the line in point-slope form, which uses a point and slope to make an equation. It looks like this:

$\textcolor{w h i t e}{\implies} y - {y}_{1} = m \left(x - {x}_{1}\right)$

where $m$ is the slope and $\left({x}_{1} , {y}_{1}\right)$ is a point on the line.

Now we can apply this to our problem. We have a slope of $- \frac{1}{3}$ and a point $\left(- 1 , 1\right)$. Plug in the values to the point-slope formula:

$\textcolor{w h i t e}{\implies} y - {y}_{1} = m \left(x - {x}_{1}\right)$

$\implies y - 1 = - \frac{1}{3} \left(x - \left(- 1\right)\right)$

$\textcolor{w h i t e}{\implies} y - 1 = - \frac{1}{3} \left(x + 1\right)$

$\textcolor{w h i t e}{\implies} y - 1 = - \frac{1}{3} x - \frac{1}{3}$

$\textcolor{w h i t e}{\implies} y = - \frac{1}{3} x + \frac{2}{3}$

That's the equation of the line. Here's what it looks like:

graph{-(1/3)x+(2/3) [-8.89, 8.89, -4.444, 4.445]}