# How do you write an equation in slope intercept form given (1,-3) and (-2,-4)?

May 31, 2017

See a solution process below:

#### Explanation:

First, we need to determine the slope of the line. 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:

$m = \frac{\textcolor{red}{- 4} - \textcolor{b l u e}{- 3}}{\textcolor{red}{- 2} - \textcolor{b l u e}{1}} = \frac{\textcolor{red}{- 4} + \textcolor{b l u e}{3}}{\textcolor{red}{- 2} - \textcolor{b l u e}{1}} = \frac{- 1}{-} 3 = \frac{1}{3}$

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 substitute the slope and the values for one of the points and solve for $b$:

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

$- 3 = \frac{1}{3} + \textcolor{b l u e}{b}$

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

$- \textcolor{red}{\frac{1}{3}} - \left(\frac{3}{3} \times 3\right) = 0 + \textcolor{b l u e}{b}$

$- \textcolor{red}{\frac{1}{3}} - \frac{9}{3} = \textcolor{b l u e}{b}$

$- \frac{10}{3} = \textcolor{b l u e}{b}$

We can now substitute the slope and $b$ value we calculated into the formula giving:

$y = \textcolor{red}{\frac{1}{3}} x + \textcolor{b l u e}{- \frac{10}{3}}$

$y = \textcolor{red}{\frac{1}{3}} x - \textcolor{b l u e}{\frac{10}{3}}$