# How do you find a standard form equation for the line with (-5,7), (0, 1/2)?

Aug 8, 2017

See a 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:

$m = \frac{\textcolor{red}{\frac{1}{2}} - \textcolor{b l u e}{7}}{\textcolor{red}{0} - \textcolor{b l u e}{- 5}} = \frac{\textcolor{red}{\frac{1}{2}} - \textcolor{b l u e}{7}}{\textcolor{red}{0} + \textcolor{b l u e}{5}} = \frac{\textcolor{red}{\frac{1}{2}} - \textcolor{b l u e}{\frac{14}{2}}}{5} = \frac{- \frac{13}{2}}{\frac{5}{1}} = - \frac{13}{10}$

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

The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$

The $y$-intercept is of an equation in standard form is: $\frac{\textcolor{g r e e n}{C}}{\textcolor{b l u e}{B}}$

We know the slope so we can write:

$- \frac{13}{10} = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$

Therefore, $\textcolor{red}{A} = 13$ and $\textcolor{b l u e}{B} = 10$

From the point abobe we know the $y$ intercept is $\left(0 , \frac{1}{2}\right)$ and we know the value of $\textcolor{b l u e}{B}$ so we can write and solve for $\textcolor{g r e e n}{C}$

$\frac{\textcolor{g r e e n}{C}}{\textcolor{b l u e}{10}} = \frac{1}{2}$

$10 \times \frac{\textcolor{g r e e n}{C}}{\textcolor{b l u e}{10}} = 10 \times \frac{1}{2}$

$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{10}}} \times \frac{\textcolor{g r e e n}{C}}{\cancel{\textcolor{b l u e}{10}}} = 5$

$\textcolor{g r e e n}{C} = 5$

Substituting gives:

$\textcolor{red}{13} x + \textcolor{b l u e}{10} y = \textcolor{g r e e n}{5}$