# How do you find the slope given (1,2) and (2,5)?

Mar 6, 2018

$\setminus q \quad \setminus q \quad \text{slope of line between" \ ( 1, 2 ) \quad "and} \setminus \quad \left(2 , 5\right) \setminus = \setminus 3 \setminus .$

#### Explanation:

$\text{Recall the definition of the slope of a line between two points: }$

$\setminus \quad \text{slope of line between" \ ( x_1, y_1 ) \quad "and} \setminus \quad \left({x}_{2} , {y}_{2}\right) \setminus = \setminus \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} .$

$\text{Applying this definition to our two given points, we get:}$

$\setminus \quad \text{slope of line between" \ ( 1, 2 ) \quad "and} \setminus \quad \left(2 , 5\right) \setminus = \setminus \frac{\left(5\right) - \left(2\right)}{\left(2\right) - \left(1\right)}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \frac{3}{1} \setminus = \setminus 3.$

$\text{So, we conclude:}$

$\setminus q \quad \setminus q \quad \text{slope of line between" \ ( 1, 2 ) \quad "and} \setminus \quad \left(2 , 5\right) \setminus = \setminus 3 \setminus .$