# How do you find a standard form equation for the line with P= (-5,-5) and Q= (-3,-2)?

Feb 5, 2017

$3 x - 2 y = - 5$
(see below for method)

#### Explanation:

Step 1: Develop a slope-point form for the line
The slope of a line between two points is defined as:
$\textcolor{w h i t e}{\text{XX}}$The difference between the $y$ coordinate values
$\textcolor{w h i t e}{\text{XX}}$divided by the corresponding difference between the $x$ coordinates.
$\textcolor{w h i t e}{\text{XXXXX}} m = \frac{\Delta y}{\Delta x} = \frac{- 2 - \left(- 5\right)}{- 3 - \left(- 5\right)} = \frac{3}{2}$

This relationship must hold for any arbitrary point $\left(x , y\right)$ on the line;
so
$\textcolor{w h i t e}{\text{XXXXX}} m = \frac{y - \left(- 5\right)}{x - \left(- 5\right)} = \frac{y + 5}{x + 5}$

Therefore
$\textcolor{w h i t e}{\text{XXXXX}} \frac{y + 5}{x + 5} = \frac{3}{2}$
or
$\textcolor{w h i t e}{\text{XXXXX}} y + 5 = \frac{3}{2} \left(x + 5\right)$

Step 2: Convert the slope-point form into standard form
Note that the standard form for a linear equation is
$\textcolor{w h i t e}{\text{XX}} A x + B y = C$ with constants $A , B , C$ and $A \ge 0$

If
$\textcolor{w h i t e}{\text{XXXXX}} y + 5 = \frac{3}{2} \left(x + 5\right)$
then
$\textcolor{w h i t e}{\text{XXXXX}} 2 \left(y + 5\right) = 3 \left(x + 5\right)$
$\Rightarrow$
$\textcolor{w h i t e}{\text{XXXXX}} 2 y + 10 = 3 x + 15$
$\Rightarrow$
$\textcolor{w h i t e}{\text{XXXXX}} - 5 = 3 x - 2 y$
or
$\textcolor{w h i t e}{\text{XXXXX}} 3 x - 2 y = - 5$

Here is the graph for verification purposes: