# How do you write equation of a line in slope-intercept form that has a slope of -1/4 and passes through the point (8, -1)?

##### 1 Answer
Mar 5, 2017

$y = \textcolor{red}{- \frac{1}{4}} x + \textcolor{b l u e}{1}$

#### Explanation:

First, we can write an equation in point-slope form. 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 and point from the problem gives:

$\left(y - \textcolor{red}{- 1}\right) = \textcolor{b l u e}{- \frac{1}{4}} \left(x - \textcolor{red}{8}\right)$

$\left(y + \textcolor{red}{1}\right) = \textcolor{b l u e}{- \frac{1}{4}} \left(x - \textcolor{red}{8}\right)$

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 now solve the point-slope form of the equation for $y$ to put it in slope-intercept form.

$y + \textcolor{red}{1} = \left(\textcolor{b l u e}{- \frac{1}{4}} \times x\right) - \left(\textcolor{b l u e}{- \frac{1}{4}} \times \textcolor{red}{8}\right)$

$y + \textcolor{red}{1} = - \frac{1}{4} x - \left(- \frac{8}{4}\right)$

$y + \textcolor{red}{1} = - \frac{1}{4} x + 2$

$y + \textcolor{red}{1} - 1 = - \frac{1}{4} x + 2 - 1$

$y + 0 = - \frac{1}{4} x + 1$

$y = \textcolor{red}{- \frac{1}{4}} x + \textcolor{b l u e}{1}$