# How do you write an equation in slope intercept form using the points (4,-3) , (2,3)?

Jan 18, 2017

$y = - 3 x + 9$

#### Explanation:

First, we will write and equation in point-slope form and then convert to slope-intercept form.

To use the point-slope form we must first determine the slope.

The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$

Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.

Substituting the values from the points in the problem gives:

$m = \frac{\textcolor{red}{3} - \textcolor{b l u e}{- 3}}{\textcolor{red}{2} - \textcolor{b l u e}{4}}$

$m = \frac{\textcolor{red}{3} + \textcolor{b l u e}{3}}{\textcolor{red}{2} - \textcolor{b l u e}{4}}$

$m = \frac{6}{-} 2 = - 3$

We can now use this calculated slope and either point to write the equation in point-slope form.

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.

Again, substituting gives:

$\left(y - \textcolor{red}{- 3}\right) = \textcolor{b l u e}{- 3} \left(x - \textcolor{red}{4}\right)$

$\left(y + \textcolor{red}{3}\right) = \textcolor{b l u e}{- 3} \left(x - \textcolor{red}{4}\right)$

We can now convert this to slope-intercept form.

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.

We can solve our equation for $y$:

$\left(y + \textcolor{red}{3}\right) = \textcolor{b l u e}{- 3} \left(x - \textcolor{red}{4}\right)$

$y + \textcolor{red}{3} = \left(\textcolor{b l u e}{- 3} \times x\right) - \left(\textcolor{b l u e}{- 3} \times \textcolor{red}{4}\right)$

$y + 3 = - 3 x - \left(- 12\right)$

$y + 3 = - 3 x + 12$

$y + 3 - \textcolor{red}{3} = - 3 x + 12 - \textcolor{red}{3}$

$y + 0 = - 3 x + 9$

$y = - 3 x + 9$