# How do you write an equation in standard form for a line passing through (5,22) and (3,12)?

Mar 3, 2018

Use the points to write the equation in slope-intercept form.
Then write that equation in standard form.

Standard form for these points is

$\textcolor{w h i t e}{\ldots . .}$$5 x - y = 3$

#### Explanation:

To write an equation in standard form when you have two points, first write the equation in slope-intercept form.

$\text{Write the equation in slope-intercept form}$
$\textcolor{w h i t e}{\ldots \ldots . .}$$y = m x + b$

$\textcolor{w h i t e}{\ldots \ldots \ldots}$―――――――

First find $m$, the slope

Slope = $\frac{\left(y - y '\right)}{\left(x - x '\right)}$

1) Decide which point will be $\left(x , y\right)$

Assign $\left(5 , 22\right)$ to be $\left(x , y\right)$
Assign $\left(3 , 12\right)$ to be $\left(x ' , y '\right)$

You can choose either point to be $\left(x ' , y '\right)$ and it will come out the same, but if you select wisely, you can sometimes avoid working with negative numbers.

2) Sub in the values for the variables

$\frac{\left(y - y '\right)}{\left(x - x '\right)}$

$\frac{\left(22 - 12\right)}{\left(5 - 3\right)}$

3) Do the subtractions to find the slope $m$

$\textcolor{w h i t e}{\ldots . .}$$\frac{10}{2}$ $= 5$ $\leftarrow$ value for slope $m$

4) So now the equation so far is

$\textcolor{w h i t e}{\ldots . .}$$y = 5 x + b$

$\textcolor{w h i t e}{\ldots \ldots \ldots}$―――――――

Next find $b$, the y intercept

1) Using either given point, sub in the values for $\left(x , y\right)$ and solve for $b$

$\textcolor{w h i t e}{\ldots \ldots .}$ y = 5  x  + b
$\textcolor{w h i t e}{\ldots . .}$$12 = 5 \left(3\right) + b$

2) Clear the parentheses
$12 = 15 + b$

3) Subtract $15$ from both sides to isolate $b$
$- 3 = b$

So the equation in point-slope form is
$\textcolor{w h i t e}{\ldots . .}$$y = 5 x - 3$

$\textcolor{w h i t e}{\ldots \ldots \ldots}$―――――――

Now put the equation in standard form

$\textcolor{w h i t e}{\ldots \ldots .}$$a x + b y = c$
where $a$ is a positive integer

$\textcolor{w h i t e}{\ldots . .}$$y = 5 x - 3$

1) Subtract $5 x$ from both sides to isolate the constant

$- 5 x + y = - 3$

2) Multiply through by $- 1$ to clear the minus sign on $x$

$5 x - y = 3$ $\leftarrow$ standard form

Mar 3, 2018

$5 x - y = 3$ same as $y = 5 x - 3$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

$m = \left(y - \left(\text{known value for "y))/(x-("known value of } x\right)\right) . . E q u a t i o n \left(1\right)$

Method
Step 1: determine $m$ from the given points.
Step 2: use one of the given points to substitute values for $x \mathmr{and} y$ into $E q u a t i o n \left(1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{b r o w n}{\text{Step 1:}}$

Always read left to right on the x-axis for this:

Set the left most point as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(3 , 12\right)$
Set the rightmost point as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(5 , 22\right)$

Set the gradient (slope) as $m$

$m = \left(\text{change in the y-axis")/("change in the x-axis}\right) \to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{22 - 12}{5 - 3} = \frac{10}{2}$

But $\frac{10}{2}$ is the same as $\frac{5}{1} \to 5$

$m = \frac{5}{1} \leftarrow \text{ Deliberately written this way}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Step 2:}}$

Using ${P}_{1}$

$m = \left(\text{change in the y-axis")/("change in the x-axis}\right) \to m = \frac{5}{1} = \frac{y - 12}{x - 3}$

Cross multiply giving:

$5 \left(x - 3\right) = 1 \left(y - 12\right)$

$5 x - 15 = y - 12$

$5 x - y = 15 - 12$

$5 x - y = 3$