# How do you write the equation of the line that passes through (2, 2) and (6, 3) in standard form?

Feb 19, 2017

$\textcolor{red}{1} x + \textcolor{b l u e}{- 4} y = \textcolor{g r e e n}{- 6}$

#### Explanation:

First, we need to determine the slope of the line passing through the two points. 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:

$m = \frac{\textcolor{red}{3} - \textcolor{b l u e}{2}}{\textcolor{red}{6} - \textcolor{b l u e}{2}} = \frac{1}{4}$

Next, 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 calculated and the first point from the problem gives:

$\left(y - \textcolor{red}{2}\right) = \textcolor{b l u e}{\frac{1}{4}} \left(x - \textcolor{red}{2}\right)$

The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1. We can convert the equation in point-slope form to standard form as follows:

First, we need to remove the fraction as all coefficients must be integers. We will multiply each side of the equation by $\textcolor{red}{4}$:

$\textcolor{red}{4} \left(y - 2\right) = \textcolor{red}{4} \times \frac{1}{4} \left(x - 2\right)$

$\left(\textcolor{red}{4} \times y\right) - \left(\textcolor{red}{4} \times 2\right) = \cancel{\textcolor{red}{4}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} \left(x - 2\right)$

$4 y - 8 = 1 x - 2$

Next, we will add $\textcolor{red}{8}$ and subtract $\textcolor{b l u e}{1 x}$ from each side of the equation to isolate the $x$ and $y$ terms on the left side of the equation:

$- \textcolor{b l u e}{1 x} + 4 y - 8 + \textcolor{red}{8} = - \textcolor{b l u e}{1 x} + 1 x - 2 + \textcolor{red}{8}$

$- 1 x + 4 y - 0 = 0 + 6$

$- 1 x + 4 y = 6$

Now, we will multiply each side of the equation by $\textcolor{red}{- 1}$ to make the $x$ coefficient a positive integer:

$\textcolor{red}{- 1} \left(- 1 x + 4 y\right) = \textcolor{red}{- 1} \times 6$

$\textcolor{red}{1} x + \textcolor{b l u e}{- 4} y = \textcolor{g r e e n}{- 6}$