# How do you write an equation in point slope and slope intercept form given (-3, -3) and (1, -5)?

Jun 9, 2017

See a solution process below:

#### Explanation:

First, we must 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:

$m = \frac{\textcolor{red}{- 5} - \textcolor{b l u e}{- 3}}{\textcolor{red}{1} - \textcolor{b l u e}{- 3}} = \frac{\textcolor{red}{- 5} + \textcolor{b l u e}{3}}{\textcolor{red}{1} + \textcolor{b l u e}{3}} = - \frac{2}{4} = - \frac{1}{2}$

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 slope we calculated and the values from the first point in the problem gives:

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

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

We can also substitute the slope we calculated and the values from the second point in the problem giving:

$\left(y - \textcolor{red}{- 5}\right) = \textcolor{b l u e}{- \frac{1}{2}} \left(x - \textcolor{red}{1}\right)$

$\left(y + \textcolor{red}{5}\right) = \textcolor{b l u e}{- \frac{1}{2}} \left(x - \textcolor{red}{1}\right)$

We can solve for $y$ to transform this equation 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.

$y + \textcolor{red}{5} = \left(\textcolor{b l u e}{- \frac{1}{2}} \times x\right) - \left(\textcolor{b l u e}{- \frac{1}{2}} \times \textcolor{red}{1}\right)$

$y + \textcolor{red}{5} = - \frac{1}{2} x + \frac{1}{2}$

$y + \textcolor{red}{5} - 5 = - \frac{1}{2} x + \frac{1}{2} - 5$

$y + 0 = - \frac{1}{2} x + \frac{1}{2} - \left(\frac{2}{2} \times 5\right)$

$y = - \frac{1}{2} x + \frac{1}{2} - \frac{10}{2}$

$y = \textcolor{red}{- \frac{1}{2}} x - \textcolor{b l u e}{\frac{9}{2}}$