# How do you write the standard form of a line given (-6, -3) and (-8, 4)?

Jan 19, 2017

$\textcolor{red}{7} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 48}$

#### Explanation:

First, we can use the point-slope formula to obtain the equation of the line. To do this we must first find the slope of the line which can be done because two points are provided.

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 problem gives:

$m = \frac{\textcolor{red}{4} - \textcolor{b l u e}{- 3}}{\textcolor{red}{- 8} - \textcolor{b l u e}{- 6}}$

$m = \frac{7}{-} 2 = - \frac{7}{2}$

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 calculate slope and the first point gives:

$\left(y - \textcolor{red}{- 3}\right) = \textcolor{b l u e}{- \frac{7}{2}} \left(x - \textcolor{red}{- 6}\right)$

$\left(y + \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{7}{2}} \left(x + \textcolor{red}{6}\right)$

Now we can transform this 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

$y + \textcolor{red}{3} = \left(\textcolor{b l u e}{- \frac{7}{2}} \times x\right) + \left(\textcolor{b l u e}{- \frac{7}{2}} \times \textcolor{red}{6}\right)$

$y + \textcolor{red}{3} = - \frac{7}{2} x - \frac{7 \times 6}{2}$

$\textcolor{red}{2} \left(y + 3\right) = \textcolor{red}{2} \left(- \frac{7}{2} x - \frac{42}{2}\right)$

$2 y + 6 = \left(\textcolor{red}{2} \times - \frac{7}{2} x\right) - \left(\textcolor{red}{2} \times \frac{42}{2}\right)$

$2 y + 6 = \left(\cancel{\textcolor{red}{2}} \times - \frac{7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x\right) - \left(\cancel{\textcolor{red}{2}} \times \frac{42}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right)$

$2 y + 6 = - 7 x - 42$

$2 y + 6 + \textcolor{red}{7 x} - \textcolor{b l u e}{6} = - 7 x - 42 + \textcolor{red}{7 x} - \textcolor{b l u e}{6}$

$\textcolor{red}{7 x} + 2 y + 6 - \textcolor{b l u e}{6} = - 7 x + \textcolor{red}{7 x} - 42 - \textcolor{b l u e}{6}$

$\textcolor{red}{7 x} + 2 y + 0 = 0 - 48$

$\textcolor{red}{7} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 48}$