# How do you write the standard form of a line given (-5,1) and (3,-3)?

Mar 9, 2017

See the entire solution process below:

#### Explanation:

First, we must determine the slope. 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:

$m = \frac{\textcolor{red}{- 3} - \textcolor{b l u e}{1}}{\textcolor{red}{3} - \textcolor{b l u e}{- 5}} = \frac{\textcolor{red}{- 3} - \textcolor{b l u e}{1}}{\textcolor{red}{3} + \textcolor{b l u e}{5}} = - \frac{4}{8} = - \frac{1}{2}$

We can now use the point slope formula to get 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 the first point from the problem gives:

$\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- \frac{1}{2}} \left(x - \textcolor{red}{- 5}\right)$

$\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- \frac{1}{2}} \left(x + \textcolor{red}{5}\right)$

We can now transform this equation into the standard form. 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

First, we will multiply each side of the equation by $2$ to eliminate the fraction and allow each of the coefficients to be an integer.

$2 \left(y - \textcolor{red}{1}\right) = 2 \times \textcolor{b l u e}{- \frac{1}{2}} \left(x + \textcolor{red}{5}\right)$

$2 y - 2 = \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}} \times \textcolor{b l u e}{- \frac{1}{\cancel{2}}} \left(x + \textcolor{red}{5}\right)$

$2 y - 2 = - \left(x + \textcolor{red}{5}\right)$

$2 y - 2 = - x - 5$

Next, add $\textcolor{red}{x}$ and $\textcolor{b l u e}{2}$ to each side of the equation to move both the $x$ and $y$ term to the left side of the equation and the constants to the right side of the equation:

$\textcolor{red}{x} + 2 y - 2 + \textcolor{b l u e}{2} = \textcolor{red}{x} - x - 5 + \textcolor{b l u e}{2}$

$\textcolor{red}{x} + 2 y - 0 = 0 - 3$

$\textcolor{red}{1} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 3}$