# How do you find a equation of the line containing the given pair of points (3,1) and (9,3)?

Feb 14, 2017

See the entire solution process below:

#### Explanation:

First we need to determine the slope 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:

$m = \frac{\textcolor{red}{3} - \textcolor{b l u e}{1}}{\textcolor{red}{9} - \textcolor{b l u e}{3}} = \frac{2}{6} = \frac{1}{3}$

Now, we can use the point-slope formula to find an 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 $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the slope we calculate and the first point from the problem gives:

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

We can also substitute the slope we calculate and the second point from the problem giving:

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{\frac{1}{3}} \left(x - \textcolor{red}{9}\right)$

We can also solve for the slope-intercept form. 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.

Solving our second equation for $y$ gives:

$y - \textcolor{red}{3} = \left(\textcolor{b l u e}{\frac{1}{3}} \times x\right) - \left(\textcolor{b l u e}{\frac{1}{3}} \times \textcolor{red}{9}\right)$

$y - \textcolor{red}{3} = \frac{1}{3} x - \frac{9}{3}$

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

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

$y = \textcolor{red}{\frac{1}{3}} x + \textcolor{b l u e}{0}$ or $y = \frac{1}{3} x$

Four equations which solve this problem are:

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

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{\frac{1}{3}} \left(x - \textcolor{red}{9}\right)$ or

$y = \textcolor{red}{\frac{1}{3}} x + \textcolor{b l u e}{0}$ or $y = \textcolor{red}{\frac{1}{3}} x$