# How do you write the slope-intercept form of the equation of the line passing through the points (2, 7) and (-3, -4)?

Dec 13, 2016

$y = \frac{11}{5} x + \frac{13}{5}$

#### Explanation:

First, we need to determine the slope. The formula for slope is:

$\textcolor{red}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}}$

Where $m$ is the slope and $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are the points given. Substituting the points we are given for the problem we get the slope as:

$m = \frac{- 4 - 7}{- 3 - 2}$

$m = \frac{- 11}{- 5}$

$m = \frac{11}{5}$

Now that we have the slope we can use the point-slope formula to get the equation for the line. This formula is:

$\textcolor{red}{\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)}$

Where $m$ is the slope and $\left({x}_{1} , {y}_{1}\right)$ are a given point. Substituting the slope we calculated and one of the points gives:

$y - - 4 = \frac{11}{5} \left(x - - 3\right)$

$y + 4 = \frac{11}{5} \left(x + 3\right)$

We can now solve for $y$ to get the slope-intercept form while keeping the equation balanced:

$y + 4 = \frac{11}{5} x + \frac{33}{5}$

$y + 4 - 4 = \frac{11}{5} x + \frac{33}{5} - 4$

$y + 0 = \frac{11}{5} x + \frac{33}{5} - \left(\frac{5}{5}\right) \cdot 4$

$y = \frac{11}{5} x + \frac{33}{5} - \frac{20}{5}$

$y = \frac{11}{5} x + \frac{13}{5}$