# How do you write the point slope form of the equation given (-5,-4) and (4,2)?

Feb 8, 2017

$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x + \textcolor{red}{5}\right)$

Or

$\left(y - \textcolor{red}{2}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{4}\right)$

#### Explanation:

First, we must 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:

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

$m = \frac{\textcolor{red}{2} + \textcolor{b l u e}{4}}{\textcolor{red}{4} + \textcolor{b l u e}{5}} = \frac{6}{9} = \frac{3 \times 2}{3 \times 3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 3} = \frac{2}{3}$

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.

We can substitute the slope we calculated and the first point giving:

$\left(y - \textcolor{red}{- 4}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{- 5}\right)$

$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x + \textcolor{red}{5}\right)$

We can also substitute the slope we calculated and the second point giving:

$\left(y - \textcolor{red}{2}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{4}\right)$