# How do you write an equation in standard form of the lines passing through (-1,5) and (0,8)?

Jun 30, 2015

$3 x - y = - 8$

#### Explanation:

$\textcolor{w h i t e}{\text{XXXX}}$(y-8)/(x-0) = (8-5)/(0-(-1)
Which simplifies as
$\textcolor{w h i t e}{\text{XXXX}}$$y - 8 = 3 x$

Standard form of a linear equation is
$\textcolor{w h i t e}{\text{XXXX}}$$A x + B y = C$ with $A , B , C \epsilon \mathbb{Z}$ and $A \ge 0$

Converting $y - 8 = 3 x$ into this form:
$\textcolor{w h i t e}{\text{XXXX}}$$3 x - y = - 8$

Jun 30, 2015

$- 3 x + y = 8$

#### Explanation:

The standard form of an equation is given by;
$A x + B y = C$

To find the equation of line passing through the points (-1,5) and (0,8), we need to used given formula;

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$..........equation 1
where m = slope and given by the formula;

$m = \setminus \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

Now, Let's assume that $\left({x}_{1} , {y}_{1}\right)$ is (-1,5) and $\left({x}_{2} , {y}_{2}\right)$ is (0,8).
First find the slope of line using slope formula, we get;

$m = \setminus \frac{8 - 5}{0 - \left(- 1\right)} = \setminus \frac{3}{1} = 3$

Now, plug $\left({x}_{1} , {y}_{1}\right)$ is (-1,5) and m = 3 in equation 1, we get
$\left(y - 5\right) = 3 \left(x - \left(- 1\right)\right)$
or, $y - 5 = 3 \left(x + 1\right)$
or, $y - 5 = 3 x + 3$

Add 5 on both side, we get,
or, $y = 3 x + 3 + 5$
or, $y = 3 x + 8$

Subtract 3x on both side, we get
or, $- 3 x + y = 8$
This is our required equation in standard form.