# How do you write an equation of a line given (9,3) and m=-2/3?

Jan 16, 2017

Use the point-slope formula. See 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 point and slope from the problem gives the equation:

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{2}{3}} \left(x - \textcolor{red}{9}\right)$

We can also convert to the more familiar slope-intercept form if we solve 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 color(blue)(b is the y-intercept value.

$y - \textcolor{red}{3} = \left(\textcolor{b l u e}{- \frac{2}{3}} \times x\right) - \left(\textcolor{b l u e}{- \frac{2}{3}} \times \textcolor{red}{9}\right)$

$y - \textcolor{red}{3} = - \frac{2}{3} x - \left(- 6\right)$

$y - \textcolor{red}{3} = - \frac{2}{3} x + 6$

$y - \textcolor{red}{3} + 3 = - \frac{2}{3} x + 6 + 3$

$y - 0 = - \frac{2}{3} x + 9$

$y = - \frac{2}{3} x + 9$