# How do you find a standard form equation for the line with (3,1), with slope of m=1/3?

Apr 16, 2018

$x - 3 y = 0$

#### Explanation:

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

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

$\text{where A is a positive integer and B,C are integers}$

$\text{we can obtain the equation in "color(blue)"point-slope form}$

•color(white)(x)y-y_1=m(x-x_1)

$\text{where m is the slope and "(x_1,y_1)" a point on the line}$

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

$\Rightarrow y - 1 = \frac{1}{3} \left(x - 3\right) \leftarrow \textcolor{b l u e}{\text{in point-slope form}}$

$\Rightarrow y - 1 = \frac{1}{3} x - 1 \leftarrow \textcolor{b l u e}{\text{distributing}}$

$\Rightarrow y = \frac{1}{3} x$

$\text{multiply through by 3}$

$\Rightarrow 3 y = x \Rightarrow x - 3 y = 0 \leftarrow \textcolor{red}{\text{in standard form}}$

Apr 16, 2018

$y = \frac{1}{3} x$

#### Explanation:

The standard equation is $y = m x + c$

So we have $y = \frac{1}{3} x + c$ and it passes through (3,1)

$1 = \frac{1}{3} \times 3 + c$ $\implies$ $1 = 1 + c$ so c=0