# How do you write the standard form of a line given Slope = -2; containing the point (3, -1)?

Aug 22, 2017

See a solution process below:

#### Explanation:

First, we can use the point-slope formula to write and equation for the line. The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.

Substituting the slope from the problem and the values from the point in the problem gives:

$\left(y - \textcolor{b l u e}{- 1}\right) = \textcolor{red}{- 2} \left(x - \textcolor{b l u e}{3}\right)$

$\left(y + \textcolor{b l u e}{1}\right) = \textcolor{red}{- 2} \left(x - \textcolor{b l u e}{3}\right)$

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 solve for this format as follows:

$y + \textcolor{b l u e}{1} = \left(\textcolor{red}{- 2} \times x\right) - \left(\textcolor{red}{- 2} \times \textcolor{b l u e}{3}\right)$

$y + \textcolor{b l u e}{1} = - 2 x - \left(- 6\right)$

$y + \textcolor{b l u e}{1} = - 2 x + 6$

$\textcolor{red}{2 x} + y + \textcolor{b l u e}{1} - 1 = \textcolor{red}{2 x} - 2 x + 6 - 1$

$2 x + y + 0 = 0 + 5$

$\textcolor{red}{2} x + \textcolor{b l u e}{1} y = \textcolor{g r e e n}{5}$