Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

We have already been working with the idea of the slope of a line.

We have seen that some lines have increasing slopes.

And some have decreasing slopes. (we sometimes call this a negative rise)

This applet will show you positive and negative gradients.

The slope of a line is a measure of how steep it is. In mathematics we call this the **GRADIENT**.

A gradient is a single value that describes:

- if a line is increasing (has positive gradient)
- if a line is decreasing (has negative gradient)
- how far up or down the line moves (change in the $y$
`y`value) with every unit step to the right (for every 1 unit increase in $x$`x`)

Take a look at this line for example. I've highlighted the horizontal and vertical steps.

We call the horizontal measurement the **RUN **and the vertical measurement the **RISE**.

Here, for every 1 step across (run of 1), the line goes up 2 (rise of 2). This line has a gradient of 2.

Sometimes it is difficult to measure how far the line goes up or down (how much the $y$`y` value changes) in 1 horizontal unit. In this case we calculate the gradient by using a formula.

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

Where you take any two points whose coordinates are known or can be easily found, and look for the rise and run between them.

If you have a graph, like this one, you can find the rise and run by drawing a right triangle created by any two points on the line. The line itself becomes the hypotenuse.

This line has a gradient of $\frac{\text{rise }}{\text{run }}=\frac{4}{3}$rise run =43

In this case, the gradient is positive because, over the $3$3 unit increase in the $x$`x` values, the $y$`y` value has * increased*.

This applet allows you to see the rise and run between two points on a line of your choosing.

If you have a pair of coordinates, such as $\left(3,6\right)$(3,6) and $\left(7,-2\right)$(7,−2) we can find the gradient of the line between these points also using the rule.

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

Firstly, it is good mathematical practice to draw a quick sketch of the points. This helps us quickly identify what the line looks like (we can see that it will be decreasing in nature, so it will have negative gradient).

The rise is the difference in the $y$`y` values of the points. In this case the difference between $-2$−2 and $6$6 is $-8$−8

($-2-6=-8$−2−6=−8).

The run is the difference in the $x$`x` values of the points. In this case the difference between $7$7 and $3$3 is $4$4

($7-3=4$7−3=4)

Notice that we subtracted the $x$`x` values and the $y$`y` values in the same order.

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

$=$= | $\frac{-8}{4}$−84 | |

$=$= | $-2$−2 |

Gradient of horizontal and vertical lines

Horizontal lines have NO rise value. The $rise=0$`r``i``s``e`=0.

So the gradient of a **horizontal **line is $\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run =$\frac{0}{run}$0`r``u``n` = $0$0

It doesn't matter what the run is, the gradient will be **0**.

Vertical lines have NO run value. The $run=0$`r``u``n`=0.

So the gradient of a **vertical **line is $\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run =$\frac{\text{rise }}{0}$rise 0

It doesn't matter what the rise is, any division by 0 results in the value being **undefined**.

Gradient

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

Gradient of Vertical Line = undefined

Gradient of Horizontal Line = 0

Let's have a look at some worked examples.

What is the gradient of the line shown in the graph, given that Point A $\left(3,3\right)$(3,3) and Point B $\left(6,5\right)$(6,5) both lie on the line?

Loading Graph...

What is the gradient of the line going through A $\left(-1,1\right)$(−1,1) and B $\left(5,2\right)$(5,2)?

Loading Graph...

The gradient of interval AB is $3$3. A is the point ($-2$−2, $4$4), and B lies on $x=3$`x`=3. What is the $y$`y`-coordinate of point B, denoted by $k$`k`?