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Section 13.1 The Line in the Sand

In the previous section, we used intuition to find solutions to linear equations in two variables. It is not always the case that we can easily find solutions by inspection. So we are going to develop a more systematic approach to graphing that does not rely on intuition. To do this, we're going to build off of the solutions of lines that we looked at in the previous section.

You might have noticed that when we had charts of values that you can find other charts by changing the \(x\) and \(y\) values in a consistent manner. This sets up the idea that points on a line satisfy a certain ratio with regards to how the coordinates change. This can be formalized as the concept of the slope of a line.

Definition 13.1.1. Slope.

The slope of a line is defined by the following ratio for any two distinct points on the line:

\begin{equation*} m = \frac{\Delta y}{\Delta x} = \frac{\text{The change of $y$}}{\text{The change of $x$}} \end{equation*}

There is a logical reason for not trying to apply this to vertical lines. In a vertical line, there is no change in the \(y\)-coordinate which means that we would be dividing by zero, and that leads to an undefined expression.

Activity 13.1.1. Calculating Slopes from Points.

The slope can be calculated between any two points on the line and it will result in the same value. The consistent ratio of the changes of the two variables is what makes the line straight.

\begin{equation*} \begin{array}{c|c|c} x \amp y \amp (x,y) \\ \hline -5 \amp -2 \amp (-5, -2) \\ -3 \amp -1 \amp (-3, -1) \\ -1 \amp 0 \amp (-1, 0) \\ 1 \amp 1 \amp (1,1) \\ 3 \amp 2 \amp (3,2) \\ 5 \amp 3 \amp (5,3) \end{array} \end{equation*}

Try it!

Calculate the slope of this line using three different pairs of points. Do not always pick consecutive points for this exercise.

Solution.

The selected points may vary, but the slope will always be \(\frac{1}{2}\text{.}\)

\begin{equation*} \begin{array}{ll} \text{From $(-5, -2)$ to $(-3, -1)$:} \amp m = \frac{\Delta y}{\Delta x} = \frac{1}{2} \\ \text{From $(-3, -1)$ to $(1, 1)$:} \amp m = \frac{\Delta y}{\Delta x} = \frac{2}{4} = \frac{1}{2} \\ \text{From $(-1, 0)$ to $(5, 3)$:} \amp m = \frac{\Delta y}{\Delta x} = \frac{3}{6} = \frac{1}{2} \end{array} \end{equation*}

Activity 13.1.2. Identifying the Slope from a Graph.

When working with graphs of lines, we often use a different language to represent the same concept. The "rise" of a function is the change in the variable between two points (up is positive, down is negative), and the "run" of a function is the change in the variable between two points (right is positive, left is negative). This leads us to sometimes say "the slope is the rise over the run."

Try it!

Determine the slope of the line in the diagram above.

Solution.
\begin{equation*} m = \frac{\Delta y}{\Delta x} = \frac{4}{6} = \frac{2}{3} \end{equation*}

If the slope is positive, the line points from the lower-left to upper-right, and if the slope is negative, the line points from upper-left to lower-right. If the line is horizontal, then the slope is 0.

There is a formula for the slope of the line if you have the coordinates of two points. The formula is commonly given to students, but it can sometimes be a distraction. If you are able to construct a meaningful picture of the line, you will not need to memorize this formula. There are some practical uses from an algebraic perspective, but it's far better to have an understanding of the concepts than simply trying to apply formulas blindly.

The slope is one of the two parameters in the slope-intercept form of a line. The other parameter is the of the \(y\)-intercept. The \(y\)-intercept is the point where the line crosses the When this point is not a point on the grid, students often try to estimate the value. This is somewhat acceptable when sketching a graph, but it is usually not acceptable when reading a graph. The challenge is that is can be very difficult to identify them correctly.

Try to estimate the of the in the two graphs below, and you'll understand why estimation is not a reliable technique.

This highlights the importance of developing an algebraic method for working with lines so that we can avoid needing to make estimates. Specifically, if we're given the equation of a line, it is a common goal to write it in slope-intercept form.

Definition 13.1.2. Slope-Intercept Form.

The slope-intercept form of a line is \(y = mx + b\text{.}\) In this form, \(m\) represents the slope and \(b\) represents the \(y\)-coordinate of the \(y\)-intercept of the line.

The reason this works is because the of a line is the solution when \(x = 0\text{.}\) But in this form, setting \(x = 0\) makes the first term disappear, and you're left with \(y = b\text{,}\) which tells us that the line must pass through the point \((x,y) = (0,b)\text{.}\)

Activity 13.1.3. Writing a Line in Slope-Intercept Form.

Writing an equation in slope-intercept form is simply a matter of solving for in a linear equation.

\begin{equation*} \begin{aligned} 3x + 4y \amp = 8 \\ 4y \amp = -3x + 8 \amp \eqnspacer \amp \text{Subtract $3x$ from both sides} \\ y \amp = -\frac{3}{4} x + 2 \amp \amp \text{Divide both sides by $4$} \end{aligned} \end{equation*}

Try it!

Write the equation \(2x - 5y = 10\) in slope-intercept form.

Solution.
\begin{equation*} \begin{aligned} 2x - 5y \amp = 10 \\ -5y \amp = -2x + 10 \amp \amp \text{Subtract $3x$ from both sides} \\ y \amp = \frac{2}{5} x - 2 \amp \amp \text{Divide both sides by $-2$} \end{aligned} \end{equation*}

Once you have the equation written in slope-intercept form, you have the information to sketch the graph. You can immediately identify the and then you can use the slope to find a second point.

Activity 13.1.4. Graphing a Line from Slope-Intercept Form.

The line \(y = -\frac{1}{2} x + 1\) has its at the point \((0, 1)\text{.}\) Since the slope is \(-\frac{1}{2}\) we can find a second point on the line by moving to the right 2 spaces and down 1 space. This gives us enough information to sketch the line.

Try it!

Graph the line \(y = -\frac{1}{2} x + 1\) on the empty grid.

Solution.