Linear Equations
UK Secondary (7-11)
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The Gradient Formula II
Lesson

In our previous lesson we looked at finding the gradient through creating right triangles and by using the formula:

$\text{Gradient }=\frac{rise}{run}$Gradient =riserun

 

Gradient Formula

Finding the gradient using the idea of $\frac{\text{rise }}{\text{run }}$rise run is a really critical skill for our further studies in linear relationships and graphing.  

Consider the interval joining point A$\left(1,1\right)$(1,1) and point B$\left(4,4\right)$(4,4). We can read off a variety of information from this diagram. 

Remembering that we move from left to right, we can see that the rise is 3, and the run is 3. So we could use the rule:  $\frac{\text{rise }}{\text{run }}$rise run  ,

and calculate the gradient to be $\frac{3}{3}=1$33=1.

We could also look for how far the line rises, for every 1 horizontal unit increase.  We can see that this is also 1.

If we don't have such a nice grid, where the rise and run are so easy to read off, then this information is a bit more complicated to find.

Example

What if the points A and B were instead something more obscure like  A$\left(-12,4\right)$(12,4) and B$\left(23,-3\right)$(23,3)?

In this case, good mathematical practice is certainly to draw yourself a quick sketch of where the points are on the plane.

From this sketch you can identify if the gradient of the line will be positive or negative  (see how it will be negative?)  

To find the rise and run we could draw a right triangle on our sketch and carry on as we did before, but there is another way to think about it.

 

 

  • The rise is the difference in the $y$y values.  The difference in the $y$y values is $-3-4=-7$34=7 

We denote this more generally as $y_2-y_1$y2y1: the $y$y coordinate of the second point minus the $y$y coordinate of the first point.

  •  The run is the difference in the $x$x values.  The difference in the $x$x values is $23-\left(-12\right)=23+12$23(12)=23+12 = $35$35.  

We denote this more generally as $x_2-x_1$x2x1: the $x$x coordinate of the second point minus the $x$x coordinate of the first point.

BE CAREFUL - the most common error here is when students are not consistent with which point is the first point and which point is the second.  

Now that we have the rise and the run we can calculate the gradient: 

$\frac{\text{rise }}{\text{run }}=\frac{-7}{35}$rise run =735  = $\frac{-1}{5}$15 (negative as we suspected from our sketch!)

Let's just remind ourselves how we calculated the rise and run again.

rise = $y_2-y_1$y2y1

run = $x_2-x_1$x2x1

This means we can generate a new rule for finding the gradient if we are given 2 points.

Gradient     $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1

 

  • Sketch the two points to see whether the gradient should be positive or negative
  • Identify one of the points as $\left(x_1,y_1\right)$(x1,y1) and the other as $\left(x_2,y_2\right)$(x2,y2) 
  • Substitute the values into the gradient formula and evaluate
  • Check that the sign of the gradient is as expected

 

Gradient

Description of gradient:  $\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run

Gradient of Vertical Line is undefined

Gradient of Horizontal Line = 0

Gradient formula     $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1

 

Here are some worked solutions relating to finding the gradient.

Question 1

Find the gradient of the line that passes through Point A $\left(3,5\right)$(3,5) and Point B $\left(1,8\right)$(1,8), using $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1.

Question 2

A line passing through the points $\left(5,3\right)$(5,3) and $\left(2,t\right)$(2,t) has a gradient equal to $-4$4.

Find the value of $t$t.

 

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