# Line k passes through A(-3, -5) and has a slope of -1/3. What is the standard form of the equation of line k?

Jun 14, 2017

$x + 3 y = - 18$

#### Explanation:

$\text{ the equation of a line in "color(blue)"standard form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y = C} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where A is a positive integer and B, C are integers.

$\text{firstly express the equation in "color(blue)"point slope form}$

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

$\text{here " m=-1/3" and } \left({x}_{1} , {y}_{1}\right) = \left(- 3 , - 5\right)$

$\Rightarrow y - \left(- 5\right) = - \frac{1}{3} \left(x - \left(- 3\right)\right)$

$\Rightarrow y + 5 = - \frac{1}{3} \left(x + 3\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$

$\text{rearranging into standard form}$

$y + 5 = - \frac{1}{3} x - 1 \leftarrow \text{ distributing}$

$\text{add " 1/3x" to both sides}$

$\frac{1}{3} x + y + 5 = - 1$

$\text{subtract 5 from both sides}$

$\frac{1}{3} x + y = - 6$

$\text{multiply through by 3}$

$x + 3 y = - 18 \leftarrow \textcolor{red}{\text{ in standard form}}$