# Slope Between Points on Cartesian Plane

DEFINITION

The Slope Between Two Points

For the slope (gradient) of the line that joins two points, A(x1, y1) and B(x2, y2) we have the equation,

change in y
change in x
=
y2 - y1
x2 - x1
=
Δ y
Δ x

For slopes between two points it is important to note that,

If the change from x1 to x2 is zero i.e. x1 = x2, then x2 - x1 = 0, as that would results in us dividing by zero, the results would be undefined as it is not possible to divide by zero, this would mean we are dealing with a vertical (“straight up and down”) line. This makes sense, as if the y-value changed, but the x-value stayed the same, the line would be going straight up.

If the change from y1 to y2 is zero i.e. y2 - y1 = 0 then, when we divide 0 by (x2 - x1)we will always be left with 0, no matter what x2 - x1 is. This means we will be dealing with a horizontal line (“straight across”) with a slope of 0. This also logically makes sense, if the x-value changed, but the y-value stayed the same, there would be no change in the vertical height, and the straight line between the two points would be a straight horizontal line, with no slope.

## Example: Slope Between Two Points

Question: For the points A(-2, 3) and B(6, -13)

a) Find Δ x and Δ y

b) Using your answer from a), find the slope of the line that A and B lie on.

So, recall

Δ x is the change in the x-value between two points,

Δ x = x2 - x1

With A as our first point, x1 = -2 and B as our second point x2 = 6

So, Δ x = 6 - (-2) = 8

Likewise, Δ y is the change in the y-value,

A is our first point so y1 = 3 and B is our second point y2 = -13

So, Δ y = -13 - 3 = -16

b) now we want to calculate the slop between these two points,

We know the slope between two points is,

M =
Δ y
Δ x
=

So, from a)

M =
-16
8
= -2

So our slope between the two points is -2

Like all math, the method works as well for when we have variables in our questions, stick to the methods when we have extra variables thrown in!

## Example: Slope between two points, with variables

Question: Find the slope between the two points U(4, -1) and V(4 - h, 3h - 1)

Don't be thrown off by the point V with variables in it, stick to the method, we have the tools to solve it.

With U as our first point, x1 = 4 and y1 = -1

V as our second point, x2 = 4 - h and y2 = 3h - 1

We use,

Slope =
Δ y
Δ x
=
y2 - y1
x2 - x1
Slope =
3h - 1 + 1
4 - 4 - h
Slope =
3h
- h
=
3
-1
= -3

So our slope between the two points is -3

Often working with variables, the working will cancel out the variables, however not always.

## Example: Average rate of change (Slope) between two points

Below is a graphed line that represents the temperature of a pie after it has been taken out of the oven as time passes. The temperature (Fahrenheit) is on the y-axis and the time passed (minutes) is on the x-axis.

The average rate of cooling at a point in time (minutes) after the pie is taken out of the oven, is represented by the slope of the tangent at that point in time on the x - axis.

Using the units of degrees per minute to the nearest tenth decimal place and finding the temperature at a given time by looking the graph, what is the average rate of cooling between the times of: a) 0 minutes and 13 minutes after the pie has been taken out of the oven

Remember, in this case, the average rate of cooling between two points, is represented by the slope of the straight line between those two point.

The equation for this is
change in y
change in x
=
y2 - y1
x2 - x1

Or in this case,

change in temperature
change in time

So, looking at the graph, the temperature (F) at minute (m) 0 is 130 °F and at minute (m) 13 is 70 °F

So, our average rate of change is given by the equation for the slope of the straight line between these two points,

70°F - 130°F
13 m - 0 m
=
-60°F
13 m
=

-4.6 °F /m =

The average rate the pie cools between 0 and 13 minutes is -4.6 degrees Fahrenheit per minute.

[we have a negative because the temperature is decreasing]

b) 1 minutes and 5 minutes after the pie has been taken out the oven

Looking at our values of temperature on the graph,

At 1 minutes the temperature is 130F

At 5 minutes the temperature is 120F

So, remembering our equation for the slope of a line between two points, that gives us the rate of change between those two points,

120°F - 130°F
5 min. - 1 min.
=
-10°F
4 min.
=

-2.5 °F /minute =

The average rate the pie cools between 1 and 5 minutes is -2.5 degrees Fahrenheit per minute.