# How do you find an equation of the line that contains the following pair of points. (-4,-5) and (-8, -10)?

Jan 3, 2017

Use the point-slope formula to solve this problem. See full explanation below.

#### Explanation:

We can use the point-slope formula to solve this problem.

We first use the slope formula which requires two points.

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 points given in the problem produces:

$m = \frac{\textcolor{red}{- 10} - \textcolor{b l u e}{- 5}}{\textcolor{red}{- 8} - \textcolor{b l u e}{- 4}}$

$m = \frac{\textcolor{red}{- 10} + \textcolor{b l u e}{5}}{\textcolor{red}{- 8} + \textcolor{b l u e}{4}}$

$m = \frac{- 5}{-} 4$

$m = \frac{5}{4}$

Now with the slope and using either point from the problem we can use the point-slope formula to find the equation of the line.

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 one of the points gives:

$\left(y - \textcolor{red}{- 5}\right) = \textcolor{b l u e}{\frac{5}{4}} \left(x - \textcolor{red}{- 4}\right)$

$\left(y + \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{5}{4}} \left(x + \textcolor{red}{4}\right)$

Solving for $y$ will put this into the more familiar slope-intercept form:

$y + \textcolor{red}{5} = \textcolor{b l u e}{\frac{5}{4}} x + \left(\textcolor{b l u e}{\frac{5}{4}} \times \textcolor{red}{4}\right)$

$y + \textcolor{red}{5} = \textcolor{b l u e}{\frac{5}{4}} x + 5$

$y + \textcolor{red}{5} - 5 = \textcolor{b l u e}{\frac{5}{4}} x + 5 - 5$

$y + 0 = \textcolor{b l u e}{\frac{5}{4}} x + 0$

$y = \textcolor{b l u e}{\frac{5}{4}} x$