# How do you find slope of (1,-1); (-2,-6)?

Apr 7, 2015
• $\left[S l o p e\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{a} l \ge b r \frac{a}{g} r a p h s - o f - l \in e a r - e q u a t i o n s - \mathmr{and} - f u n c t i o n \frac{s}{s} l o p e\right)$ = $\frac{R i s e}{R u n}$
The $R i s e$ is the Difference of the Y coordinates of any two points on the line
And the $R u n$ is the Difference of the X coordinates of those two points

• If the coordinates of the points are $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$, then $S l o p e = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
Here, the coordinates are $\left(1 , - 1\right)$ and $\left(- 2 , - 6\right)$

$S l o p e = \frac{- 6 - \left(- 1\right)}{- 2 - 1} = \frac{- 6 + 1}{-} 3 = \frac{- 5}{-} 3 = \frac{5}{3}$

The slope of the line passing through points (1,-1) and (-2,-6) is $\frac{5}{3}$

• The graph of the line will look like this:
graph{y=(5x/3)-(8/3) [-10, 10, -5, 5]}