# How do you write an equation of a line given (-1, 2) and (3, -4)?

##### 2 Answers
Oct 17, 2017

$3 x + 2 y + 1 = 0$

#### Explanation:

Standard form of equation with two points given is
$\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$
$\frac{y - \left(- 4\right)}{2 - \left(- 4\right)} = \frac{x - 3}{- 1 - 3}$

$\frac{y + 4}{6} = \frac{x - 3}{- 4}$
$- 4 y - 16 = 6 x - 18$

$6 x + 4 y = 2$
$3 x + 2 y = 1$

Oct 17, 2017

$y = - \frac{3}{2} x + \frac{1}{2}$

#### Explanation:

Okay so there are 3 main forms you can use:
- slope-intercept form [$y = m x + b$]
- standard form [$A x + B y = C$]
- point-slope form [${y}_{1} - {y}_{2} = m \left({x}_{1} - {x}_{2}\right)$]

Since you did not specify which form you wanted it in, I am going to use slope-intercept form because that's the easiest to understand, in my opinion. (:

$y = m x + b$

SLOPE ($m$)
To find $m$ (slope), you need to find $\frac{r i s e}{r u n}$, or which is the change in $y$ divided by the change in $x$. Use this formula: $\frac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}$
$\left(- 1 , 2\right) = \left({x}_{1} , {y}_{1}\right)$
$\left(3 , - 4\right) = \left({x}_{2} , {y}_{2}\right)$

It doesn't matter which coordinate pair you choose to be $\left({x}_{1} , {y}_{1}\right)$ or $\left({x}_{2} , {y}_{2}\right)$. Just stay consistent!

$m = \frac{2 - \left(- 4\right)}{- 1 - 3} = \frac{2 + 4}{- 1 - 3} = \frac{6}{-} 4 = - \frac{3}{2}$
$m$= -3/2#

Y-INTERCEPT ($b$)
Choose one of the coordinates to substitute into $y = m x + b$, which will be substituting for $x$ and $y$.
I chose $\left(- 1 , 2\right)$. Substitute $m$ for $- \frac{3}{2}$.
$\left(2\right) = \left(- \frac{3}{2}\right) \left(- 1\right) + b$
Solve for $b$.
$2 = \frac{3}{2} + b$
$b = \frac{1}{2}$

FINAL FORM
Substitute $m$ and $b$ for their values. This is your answer!
$y = - \frac{3}{2} x + \frac{1}{2}$