# What is the equation of the line passing through (13,-4) and (14,-9)?

Dec 23, 2017

$y + 4 = - 5 \left(x - 13\right)$

#### Explanation:

I'm not sure which form of equation you want it to be in, but going to show the simplest, or point-slope form, which is $y - {y}_{1} = m \left(x - {x}_{1}\right)$.

First, we need to find the slope of the line, $m$.

To find the slope, we use the formula $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$, also known as "rise over run", or change of $y$ over change of $x$.

Our two coordinates are $\left(13 , - 4\right)$ and $\left(14 , - 9\right)$. So let's plug those values into the slope equation and solve:
$m = \frac{- 9 - \left(- 4\right)}{14 - 13}$
$m = - \frac{5}{1}$
$m = - 5$

Now, we need a set of coordinates from the given or the graph. Let's use the point $\left(13 , - 4\right)$

So our equation is:
$y - \left(- 4\right) = - 5 \left(x - 13\right)$

Simplified...
$y + 4 = - 5 \left(x - 13\right)$

Dec 23, 2017

$y = - 5 x + 61$

#### Explanation:

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

•color(white)(x)y=mx+b

$\text{where m is the slope and b the y-intercept}$

$\text{to calculate m use the "color(blue)"gradient formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{1} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{let "(x_1,y_1)=(13,-4)" and } \left({x}_{2} , {y}_{2}\right) = \left(14 - 9\right)$

$\Rightarrow m = \frac{- 9 - \left(- 4\right)}{14 - 13} = - 5$

$\Rightarrow y = - 5 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b use either of the two given points}$

$\text{using } \left(13 , - 4\right)$

$- 4 = - 65 + b \Rightarrow b = 61$

$\Rightarrow y = - 5 x + 61 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$