# How do you write an equation of a line given (5, 6) and (-7, 3)?

##### 1 Answer
Jul 9, 2017

See a solution process below:

#### Explanation:

First, we need to determine the slop 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:

$m = \frac{\textcolor{red}{3} - \textcolor{b l u e}{6}}{\textcolor{red}{- 7} - \textcolor{b l u e}{5}} = \frac{- 3}{-} 12 = \frac{1}{4}$

Next we can use the point-slope formula to write and equation for 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 $\left(\textcolor{red}{{x}_{1} , {y}_{1}}\right)$ is a point the line passes through.

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

$\left(y - \textcolor{red}{6}\right) = \textcolor{b l u e}{\frac{1}{4}} \left(x - \textcolor{red}{5}\right)$

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

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{\frac{1}{4}} \left(x - \textcolor{red}{- 7}\right)$

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{\frac{1}{4}} \left(x + \textcolor{red}{7}\right)$

We can also solve this equation for $y$ to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: v

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y - \textcolor{red}{3} = \left(\textcolor{b l u e}{\frac{1}{4}} \times x\right) + \left(\textcolor{b l u e}{\frac{1}{4}} \times \textcolor{red}{7}\right)$

$y - \textcolor{red}{3} = \frac{1}{4} x + \frac{7}{4}$

$y - \textcolor{red}{3} + 3 = \frac{1}{4} x + \frac{7}{4} + 3$

$y = \frac{1}{4} x + \frac{7}{4} + \left(\frac{4}{4} \times 3\right)$

$y = \frac{1}{4} x + \frac{7}{4} + \frac{12}{4}$

$y = \textcolor{red}{\frac{1}{4}} x + \textcolor{b l u e}{\frac{19}{4}}$