# How do you write an equation in slope-intercept form for a line containing (9,-3) and (5, 5)?

Jun 7, 2017

$y = - 2 x + 15$

#### Explanation:

The general form of a linear equation is $y = m x + c$ where $m$ is the gradient (slope) and $\left(0 , c\right)$ is the $y$-intercept.

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ for a line containing $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$.

For $\left(9 , - 3\right)$ and $\left(5 , 5\right)$,
$m = \frac{5 - \left(- 3\right)}{5 - 9}$
$m = - 2$

Therefore, $y = - 2 x + c$. To find c, substitute $\left(5 , 5\right)$ into this equation.
$5 = - 2 \left(5\right) + c$
$c = 5 + 10$
$c = 15$

Substitute $c$ value back into equation,
$y = - 2 x + 15$

Jun 7, 2017

$y = - 2 x + 15$

#### Explanation:

We first begin with finding out the slope using the slope formula, which

is $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$. Plug in our points $\frac{5 - \left(- 3\right)}{5 - 9}$=$- \frac{8}{2}$=$- 2$. This is our slope now we use the point slope formula $y - {y}_{1} = m \left(x - {x}_{1}\right)$.

You can pick any of the two points, $m$ is our slope which is $- 2$.

$y - 5 = - 2 \left(x - 5\right)$ go ahead and solve.

You should arrive to the answer of $y = - 2 x + 15$.

When you do $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ it doesn't matter which point is your $\left({y}_{2} , {x}_{2}\right)$ or $\left({y}_{1} , {x}_{1}\right)$.