# How do you write an equation in standard form given point (1,4) and (6,-1)?

Oct 4, 2016

$1 x + 1 y = 5$

#### Explanation:

The slope of the line through $\left(1 , 4\right)$ and $\left(6 , - 1\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{m} = \frac{\Delta y}{\Delta x} = \frac{4 - \left(- 1\right)}{1 - 6} = \textcolor{g r e e n}{- 1}$

The slope-point form for a line with slope $\textcolor{g r e e n}{m}$ through a point $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ is
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{b} = \textcolor{g r e e n}{m} \left(x - \textcolor{red}{a}\right)$

Using the previously determined slope $\textcolor{g r e e n}{m = - 1}$ and $\left(\textcolor{red}{1} , \textcolor{b l u e}{4}\right)$ as our point $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{4} = \textcolor{g r e e n}{- 1} \left(x - \textcolor{red}{1}\right)$

Our objective is to convert this into standard form:
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$

$y - 4 = - 1 \left(x - 1\right)$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y - 4 = - x + 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x + y = 5$

or, if we wish to be explicit in the variable coefficients:
$\textcolor{w h i t e}{\text{XXX}} 1 x + 1 y = 5$