Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

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

$\text{Gradient }=\frac{rise}{run}$Gradient =`r``i``s``e``r``u``n` formula.

In this lesson we will add 2 more ways of calculating gradient to our linear toolkit.

Pictured here is a straight line on the Cartesian plane.

If we wanted to find the gradient, we could look at $\frac{\text{rise }}{\text{run }}$rise run , which in this line is $\frac{AB}{CB}$`A``B``C``B`.

Look at angle $\theta$`θ`, what is $\tan\theta$`t``a``n``θ`?

$\tan\theta=\frac{AB}{CB}$`t``a``n``θ`=`A``B``C``B`. This is the value of the gradient we worked out just before!

*This means that $\tan\theta$ tanθ, or specifically the tangent of the angle with the $x$x-axis in the positive direction, is equal to the gradient.*

$\theta$`θ` is called the **angle of inclination**.

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 segment 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.

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 $-3-4=-7$−3−4=−7

We denote this more generally as $y_2-y_1$`y`2−`y`1: 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 $x_2-x_1$`x`2−`x`1: 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 = $y_2-y_1$`y`2−`y`1

run = $x_2-x_1$`x`2−`x`1

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`=`y`2−`y`1`x`2−`x`1

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`=`y`2−`y`1`x`2−`x`1

Angle of Inclination the line makes with the positive $x$`x`-axis.

Gradient is also calculated using the angle of inclination $\theta$`θ` : $m=\tan\theta$`m`=`t``a``n``θ`

Here are some worked solutions relating to finding the gradient.

**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=y2−y1x2−x1.**

**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`.