# How do you write an equation of a line with point (3,-2), slope 1/3?

Mar 22, 2017

See the entire solution process below:

#### Explanation:

The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the values from the point and the slope given in the problems allows us to write this equation:

$\left(y - \textcolor{red}{- 2}\right) = \textcolor{b l u e}{\frac{1}{3}} \left(x - \textcolor{red}{3}\right)$

$\left(y + \textcolor{red}{2}\right) = \textcolor{b l u e}{\frac{1}{3}} \left(x - \textcolor{red}{3}\right)$

We can also transform this to the slope-intercept form by solving for $y$. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y + \textcolor{red}{2} = \left(\textcolor{b l u e}{\frac{1}{3}} \times x\right) - \left(\textcolor{b l u e}{\frac{1}{3}} \times \textcolor{red}{3}\right)$

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

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

$y + 0 = \frac{1}{3} x - 3$

$y = \textcolor{red}{\frac{1}{3}} x - \textcolor{b l u e}{3}$