Analytic Geometry

Lesson

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

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

Finding the slope 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 slope 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 slope 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 slope:

$\frac{\text{rise }}{\text{run }}=\frac{-7}{35}$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 slope if we are given 2 points.

Slope $m=\frac{y_2-y_1}{x_2-x_1}$`m`=`y`2−`y`1`x`2−`x`1

- Sketch the two points to see whether the slope should be positive or negative
- Identify one of the points as $\left(x_1,y_1\right)$(
`x`1,`y`1) and the other as $\left(x_2,y_2\right)$(`x`2,`y`2) - Substitute the values into the slope formula and evaluate
- Check that the sign of the slope is as expected

Slope

Description of slope: $\text{Slope }=\frac{\text{rise }}{\text{run }}$Slope =rise run

Slope of Vertical Line is undefined

Slope of Horizontal Line = 0

Slope formula $m=\frac{y_2-y_1}{x_2-x_1}$`m`=`y`2−`y`1`x`2−`x`1

Here are some worked solutions relating to finding the slope.

**Find the slope 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 slope equal to $-4$−4.**

Find the value of $t$`t`.

Determine, through investigation, various formulas for the slope of a line segment or to determine the slope of a line segment or a line