audio read-through Practice with Slope

Want some basic practice first?  Introduction to the Slope of a Line

In a rush? Need to skip the concepts and jump right to the slope formula? (Please come back and explore the beautiful ideas at a later time.)

Consider the equation $\,y = mx + b\,.$ As we'll see in this section, every equation of this form graphs as a non-vertical line in the coordinate plane.

In this equation, $\,m\,$ is the coefficient of the $\,x$-term, and $\,b\,$ is a constant term. For example, in the equation $\,y = 2x + 3\,,$ we have $\,m = 2\,$ and $\,b = 3\,.$

The number $\,m\,$ is called the slope of the line, and gives information about the ‘slant’ of the line. It answers questions like:  “Is it an uphill or downhill line?   How steep is it?”  The purpose of this section is to explain why this is true.

Why does $\,y = mx + b\,$ graph as a line? Why does the number $\,m\,$ give info about the slant of this line?

The key idea (which we'll prove below) is that the equation $\,y = mx + b\,$ defines a very special relationship between $\,x\,$ and $\,y\,$: equal changes in $\,x\,$ (the input) give rise to equal changes in $\,y\,$ (the output).

Indeed, if $\,x\,$ changes by an amount $\,\Delta x\,$ (read aloud as ‘delta ex’ or ‘change in ex’), then $\,y\,$ changes by $\,m\Delta x\,.$ That is, $\,y\,$ changes $\,m\,$ times as fast as $\,x\,.$ This is a bit symbol-intensive, so let's look at a concrete example.

Suppose that $\,m\,$ is $\,2\,.$ (See the sketch below.) Then, $\,y\,$ changes $\,2\,$ times as fast as $\,x\,$ (that is, twice as fast as $\,x\,$). If (say) $\,x\,$ changes by $\,1\,,$ then $\,y\,$ will change by $\,2(1) = 2\,.$ If (say) $\,x\,$ changes by $\,5\,,$ then $\,y\,$ will change by $\,2(5) = 10\,,$ and so on.

a line with slope 2

Here's another way to look at it. Imagine you're ‘standing on’ a point in the coordinate plane. Let's force $\,y\,$ to change twice as fast as $\,x\,$, and see what new points result:

If you take $\,1\,$ step to the right (i.e., let $\,x\,$ change by $\,1\,$), then you'll have to take $\,2\,$ steps up (i.e., let $\,y\,$ change by $\,2\,$). Do it again—one step to the right, two steps up. Do it again—one step to the right, two steps up. (Repeat, if you want, with different changes in $\,x\,$.) Hmmm$\,\ldots\,$ what pattern are these points creating? A line!!

If we repeat the previous exercise with $\,m=100\,,$ think about the line that would result: one step to the right, $\,100\,$ steps up. Pretty steep line! So, clearly, the number $\,m\,$ gives information about the steepness or ‘slant’ of the line.

The next part is a bit technical, so perhaps a break is in order first!

Proof That, in the Equation $\,y=mx+b\,,$ Equal Changes in $\,x\,$ Give Rise to Equal Changes in $\,y\,$

Let $\,(x_1,y_1)\,$ be a point on the graph of $\,y = mx + b\,,$ so that $\,y_1 = mx_1 + b\,.$ That is, when you substitute $\,x_1\,$ for $\,x\,$ and $\,y_1\,$ for $\,y\,,$ the resulting equation is true.

Let $\,x_1\,$ change by an amount $\,\Delta x\,,$ to get a new $\,x\,$-value:   $\,x_2 = x_1 + \Delta x\,.$ Let $\,y_2\,$ represent the $\,y$-value corresponding to $\,x_2\,.$ Then,

$y_2 = mx_2 + b$ find the $\,y$-value corresponding to $\,x_2\,$;   call it $\,y_2\,$
$\quad\, = m(x_1 + \Delta x) + b$ substitute $\,x_1 + \Delta x\,$ for $\,x_2\,$
$\quad\, = mx_1 + m\Delta x + b$ multiply out; i.e., use the Distributive Law
$\quad\, = (mx_1 + b) + m\Delta x$ re-order and re-group
$\quad\, = y_1 + m\Delta x$ look back:  $\,mx_1 + b\,$ is precisely $\,y_1\,$

Thus, $\,y\,$ has changed by $\,m\Delta x\,.$

Make sure you see this!

The starting point is: $($ $x_1$ $,$ $y_1$ $)$
The ending point is: $($ $x_1 + \Delta x$ $,$ $y_1 + m\Delta x$ $)$

So, when $\,x\,$ changes by $\,\Delta x\,,$ $\,y\,$ changes by $\,m\Delta x\,.$

A Formula for Slope

Let's summarize things from the previous proof.

Solving this last equation for the slope, $\,m\,,$ gives us an important formula:

SLOPE OF A LINE
Let $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$ be two different points on a non-vertical line. Then, the slope of this line is: $$ \begin{align} \cssId{s115}{m} &\cssId{s116}{= \text{slope}}\cr\cr &\cssId{s117}{= \frac{y_2-y_1}{x_2-x_1}}\cr\cr &\cssId{s118}{= \frac{\Delta y}{\Delta x}}\cr\cr &\cssId{s119}{= \frac{\text{change in } y}{\text{change in } x}}\cr\cr &\cssId{s120}{= \frac{\text{rise}}{\text{run}}} \end{align} $$

Analysis of the Slope Formula

There are lots of important things you should know about the slope formula:

horizontal line, zero slope vertical line, no slope
horizontal line,
zero slope
vertical line,
no slope
Uphill Lines, Positive Slopes
gradual uphill, small positive slope steep uphill, large positive slope
Gradual uphill,
small positive slope
Steep uphill,
large positive slope
Downhill Lines, Negative Slopes
gradual downhill, small negative slope steep downhill, large negative slope
Gradual downhill,
small negative slope
Steep downhill,
large negative slope

Examples

Question: Find the slope of the line through $\,(-1,3)\,$ and $\,(2,-5)\,.$
Solution:
$$ \cssId{s229}{\text{slope}} \cssId{s230}{= \frac{-5-3}{2-(-1)}} \cssId{s231}{= \frac{-8}{3}} \cssId{s232}{= -\frac83} $$ The answers in this exercise are reported as fractions, in simplest form.
Question: Find the slope of the line through $\,(2,5)\,$ and $\,(-7,5)\,.$
Solution: $$ \cssId{s236}{\text{slope}} \cssId{s237}{= \frac{5-5}{-7-2}} \cssId{s238}{= \frac{0}{-9}} \cssId{s239}{= 0} $$ Recall that zero, divided by any nonzero number, is zero. This is a horizontal line.
Question: Find the slope of the line through $\,(2,5)\,$ and $\,(2,-7)\,.$
Solution: $$ \begin{align} \cssId{s245}{\text{slope}}\ &\cssId{s246}{= \frac{-7-5}{2-2}} \cssId{s247}{= \frac{-12}{0}}\cr\cr &\cssId{s248}{=\text{can't continue}} \end{align} $$ Division by zero is not allowed. This line has no slope. It is a vertical line.
Question: Horizontal lines have (circle one):

no slope         zero slope

Solution: zero slope
Question: Vertical lines have (circle one):

no slope         zero slope

Solution: no slope
Question: Suppose a fraction has a zero in the numerator, and a nonzero denominator. What is the value of the fraction?
Solution: zero
Question: Suppose a fraction has a zero in the denominator, and a nonzero numerator. What is the value of the fraction?
Solution: it is not defined
Question: Start at a point $\,(x,y)\,$ on a line. To get to a new point, move up $\,3\,$ and to the right $\,4\,.$ What is the slope of the line?
Solution: Rise is $\,3\,$;  run is $\,4\,.$ $$ \cssId{s281}{\text{slope}} \cssId{s282}{= \frac{\text{rise}}{\text{run}}} \cssId{s283}{= \frac{3}{4}} $$
Question: Start at a point $\,(x,y)\,$ on a line. To get to a new point, move down $\,2\,$ and to the left $\,6\,.$ What is the slope of the line?
Solution: Rise is $\,-2\,$;  run is $\,-6\,.$ $$ \cssId{s291}{\text{slope}} \cssId{s292}{= \frac{\text{rise}}{\text{run}}} \cssId{s293}{= \frac{-2}{-6}} \cssId{s294}{= \frac13} $$
Question: Start at a point $\,(x,y)\,$ on a line. To get to a new point, move straight up $\,5\,$ units. What is the slope of the line?
Solution: no slope (vertical line)
Question: Start at a point $\,(x,y)\,$ on a line. To get to a new point, move directly to the left $\,5\,$ units. What is the slope of the line?
Solution: zero slope (horizontal line)
Question: Suppose you are walking along a line, moving from left to right. You are going uphill. Then, the slope of the line is (circle one):

positive         negative

Solution: positive

Concept Practice

Answers are reported as fractions in simplest form.