# How do you write the standard form of a line given (-6,-5) and has slope -2?

##### 1 Answer
Jan 22, 2017

See the entire solution process below:

#### Explanation:

First, because we are given a point and the slope we can find an equation using the point-slope formula.

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 values from the problem gives:

$\left(y - \textcolor{red}{- 5}\right) = \textcolor{b l u e}{- 2} \left(x - \textcolor{red}{- 6}\right)$

$\left(y + \textcolor{red}{5}\right) = \textcolor{b l u e}{- 2} \left(x + \textcolor{red}{6}\right)$

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 now transform our equation into this format.

$y + \textcolor{red}{5} = \left(\textcolor{b l u e}{- 2} \times x\right) + \left(\textcolor{b l u e}{- 2} \times \textcolor{red}{6}\right)$

$y + \textcolor{red}{5} = - 2 x - 12$

$y + \textcolor{red}{5} - 5 + \textcolor{b l u e}{2 x} = - 2 x - 12 - 5 + \textcolor{b l u e}{2 x}$

$y + \textcolor{red}{5} - 5 + \textcolor{b l u e}{2 x} = - 2 x - 12 - 5 + \textcolor{b l u e}{2 x}$

$2 x + y + 0 = 0 - 17$

$\textcolor{red}{2} x + \textcolor{b l u e}{1} y = \textcolor{g r e e n}{- 17}$