Slope measures the steepness of a line or the rise divided by the run. Another way to think of this is the comparison of the amount a line goes up or down compared to how much it changes left or right between two points. This is shown as a step. This is also known as the rate of change of a line. The following video will go over how to find the slope, especially when it’s hard to just look to see how far up and over the graph goes.

Video Source (07:26 mins) | Transcript

The following video shows another example of using the formula taught in the previous video.

Video Source (03:37 mins) | Transcript

\(\displaystyle slope = m = {\frac {rise}{run}} = {\frac {y2 - y1} {x2 - x1}} \)

Be sure to keep track of which point is point \( 1 = ( x1 , y1 ) \) and which is point \( 2 = ( x2 , y2 ) \) because it’s important that we are consistent with which one comes first in our formula.

Additional Resources

• Khan Academy: Slope from Two Points (07:11 mins, Transcript)
• Khan Academy: Slope (More Examples) (13:41 mins, Transcript)

Practice Problems

1. Find the slope of the line that contains the points \((10,−9)\) and \((7,6)\). (
   Solution

   x

   Solution:
   \(-5\)
   Details:
   We are trying to find the slope of the line that connects the points \((10,−9)\) and \((7,6)\).
   
   Remember, \(\displaystyle{\text{Slope}}={\text{m}}=\frac{\text{rise}}{\text{run}}=\frac{{\text{y}}_{2}-{\text{y}}_{1}}{{\text{x}}_{2}-{\text{x}}_{1}}\)
   
   Step 1: Decide which ordered pair is point 1, and which is point 2. For this problem \({\color{Red}(10,−9)}\) will be \({\color{Red}{\text{point 1}}}\) and \({\color{Blue}(7,6)}\) will be \({\color{Blue}{\text{point 2}}}\).
   
   (Note: It doesn’t matter which is which, just as long as you are consistent throughout the problem.)
   
   Step 2: Substitute the values of the two points into the formula and simplify:
   
   \(\displaystyle\frac{{\color{Blue}{\text{y}}_{2}}-{\color{Red}{\text{y}}_{1}}}{{\color{Blue}{\text{x}}_{2}}-{\color{Red}{\text{x}}_{1}}} = \frac{{\color{Blue}6}-{\color{Red}(-9)}}{{\color{Blue}7}-{\color{Red}10}} = \frac{15}{-3} = -5\)
   
   So the slope of the line is \(-5\).
   )
2. Find the slope of the line that contains the points \((10,−1)\) and \((14,−9)\). (
   Solution

   x

   Solution:
   \(-2\)
   )
3. Find the slope of the line that contains the points \((10,0)\) and \((17,−42)\). (
   Solution

   x

   Solution:
   \(-6\)
   )
4. Find the slope of the line that contains the points \((3,5)\) and \((13,20)\). (
   Solution

   x

   Solution:
   \(\dfrac{3}{2}\)
   Details:
   We are trying to find the slope of the line that connects the coordinate points \((3,5)\) and \((13,20)\).
   
   Remember, \(\displaystyle{\text{Slope}}={\text{m}}=\frac{\text{rise}}{\text{run}}=\frac{{\text{y}}_{2}-{\text{y}}_{1}}{{\text{x}}_{2}-{\text{x}}_{1}}\)
   
   Step 1: Decide which ordered pair is point 1, and which is point 2. For this problem \({\color{Red}(3, 5)}\) will be \({\color{Red}{\text{point 1}}}\) and \({\color{Blue}(13, 20)}\) will be \({\color{Blue}{\text{point 2}}}\).
   
   (Note: It doesn’t matter which is which, just as long as you are consistent throughout the problem.)
   
   Step 2: Substitute the values of the two points into the formula and simplify:
   
   \(\displaystyle\frac{\text{rise}}{\text{run}}=\frac{{\color{Blue}{\text{y}}_{2}}-{\color{Red}{\text{y}}_{1}}}{{\color{Blue}{\text{x}}_{2}}-{\color{Red}{\text{x}}_{1}}} = \frac{{\color{Blue}20}-{\color{Red}5}}{{\color{Blue}13}-{\color{Red}3}} = \frac{15}{10}\)
   
   We can divide the numerator and denominator by 5 since it is a common factor of both, so \(\dfrac{15}{10}\) simplifies to \(\dfrac{3}{2}\)
   
   So the slope of the line is \(\dfrac{3}{2}\)
   )
5. Find the slope of the line that contains the points \((−8,10)\) and \((−32,2)\). (
   Solution

   x

   Solution:
   \(\dfrac{1}{3}\)
   )
6. Find the slope of the line that contains the points \((8,−9)\) and \((−27,−30)\). (
   Solution

   x

   Solution:
   \(\dfrac{3}{5}\)
   )

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