# How do you find a standard form equation for the line with (-5, -16) with a slope of 2?

Jan 20, 2017

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

#### Explanation:

First, we use the point-slope formula to write 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.

$\left(y - \textcolor{red}{- 16}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{- 5}\right)$

$\left(y + \textcolor{red}{16}\right) = \textcolor{b l u e}{2} \left(x + \textcolor{red}{5}\right)$

Now we can transform to 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

$y + \textcolor{red}{16} = \left(\textcolor{b l u e}{2} \times x\right) + \left(\textcolor{b l u e}{2} \times \textcolor{red}{5}\right)$

$y + \textcolor{red}{16} = 2 x + 10$

$y + \textcolor{red}{16} - 16 - \textcolor{b l u e}{2 x} = 2 x + 10 - 16 - \textcolor{b l u e}{2 x}$

$- \textcolor{b l u e}{2 x} + y + \textcolor{red}{16} - 16 = 2 x - \textcolor{b l u e}{2 x} + 10 - 16$

$- \textcolor{b l u e}{2 x} + y + 0 = 0 + 10 - 16$

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

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

$2 x - y = 6$

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

Jan 20, 2017

$2 x - y - 6 = 0$

#### Explanation:

The $\textcolor{b l u e}{\text{standard form equation }}$ of a line is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{a x + b y + c = 0} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

Before this, we can write the equation in $\textcolor{b l u e}{\text{point-slope form}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

here $m = 2 \text{ and } \left({x}_{1} , {y}_{1}\right) = \left(- 5 , - 16\right)$

$\Rightarrow y - \left(- 16\right) = 2 \left(x - \left(- 5\right)\right)$

$\Rightarrow y + 16 = 2 \left(x + 5\right)$

$\Rightarrow y + 16 = 2 x + 10 \leftarrow \textcolor{red}{\text{ in point-slope form}}$

Rearranging in standard form.

$\Rightarrow 2 x - y - 6 = 0 \leftarrow \textcolor{red}{\text{ in standard form}}$