An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly's location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers: the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes' grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis. The Cartesian System, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2.1.
The center where the axes cross is the origin $(0, 0)$. Positive numbers increase to the right on the x-axis and up the y-axis; negatives go left and down. Each point is given as an ordered pair $(x, y)$.
Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair, indicating the combined distance from the origin in the form $(x,y)$. An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point $(3,-4)$ in the plane by moving three units to the right of the origin in the horizontal direction, then four units down in the vertical direction. See Figure 2.2:
When dividing the axes into equally spaced increments, note that the x-axis may be considered separate from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, 10, or even 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
A two-dimensional place where the:
A point in the plane is defined as an ordered pair, $(x,y)$, such that $x$ is the horizontal distance from the origin and $y$ is the vertical distance from the origin.
Plot the points $(-2,4)$, $(3,3)$, and $(0,-3)$ in the plane, and identify the quadrant or axis each point lies on.
Now that we've covered how to plot points in the coordinate plane, we will next talk about how to plot equations by plotting points in the coordinate plane. Graphically, an equation is a collection of multiple points used to create a picture of sorts. That picture can also be represented by a "model" or equation. In a linear model, the equation contains both an $x$ and a $y$ variable, thus it is called an equation in two variables. Likewise, its graph is also called a graph in two variables. Any graph in two variables is a graph in a two-dimensional plane.
Suppose we want to graph the equation $y=2x-1$. We can begin by choosing $x$ values that can be substituted into the equation for the variable $x$ to solve for the $y$ variable. By doing this, we find ourselves an ordered pair that can be plotted. When plotting an equation using this method, ultimately the decision is yours as to what $x$ values you would like to substitute in. For this example, we are going to use the integer values from $-3$ to $3$. The table below shows the ordered pairs we'd end up with.
| $x$ | $y=2x-1$ | $(x,y)$ |
|---|---|---|
| $-3$ | $y=2(-3)-1$ | $(-3,-7)$ |
| $-2$ | $y=2(-2)-1$ | $(-2,-5)$ |
| $-1$ | $y=2(-1)-1$ | $(-1,-3)$ |
| $0$ | $y=2(0)-1$ | $(0,-1)$ |
| $1$ | $y=2(1)-1$ | $(1,1)$ |
| $2$ | $y=2(2)-1$ | $(2,3)$ |
| $3$ | $y=2(3)-1$ | $(3,5)$ |
Using those points, we can then plot each ordered pair and connect the points using a line. Technically speaking, you only need two points to form a line. If you're a country music fan, think about the Dixie Chicks. They had an entire song where they said "...the shortest distance between two points is a straight line...". However, the more points you plot, the more accurate you can draw the line.
Given an equation in the form $y=mx+b$, graph the equation by following the steps below.
Graph the equation $y=-x+2$ by plotting points.
First, you will need to select 5 to 7 x-values to plug into an x/y table to calculate $y$ and find the coordinate pair.
| $x$ | $y=-x+2$ | $(x,y)$ |
|---|---|---|
| $-3$ | $y=-(-3)+2$ | $(-3,5)$ |
| $-2$ | $y=-(-2)+2$ | $(-2,4)$ |
| $-1$ | $y=-(-1)+2$ | $(-1,3)$ |
| $0$ | $y=-(0)+2$ | $(0,2)$ |
| $1$ | $y=-(1)+2$ | $(1,1)$ |
| $2$ | $y=-(2)+2$ | $(2,0)$ |
| $3$ | $y=-(3)+2$ | $(3,-1)$ |
Next, you'll need to take those ordered pairs in your last column, plot them, and draw a line that connects all the dots.
Construct a table and graph the equation by plotting points for $y=\frac{1}{2}x+2$.
The Intercepts of a graph are the points at which the graph crosses the $x$ and $y$ axis. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero and the point is $(x,0)$. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero and the point is $(0,y)$.
To determine the x-intercept, you would substitute $y=0$ into the equation and solve for x. For example, let's find the intercepts of the equation $y=3x-1$.
| Algebra | Explanation |
|---|---|
| $y=3x-1$ | |
| $0=3x-1$ | 1. Substitute $y=0$ into the equation. |
| $1=3x$ | 2. Add $1$ to both sides. |
| $\frac{1}{3}=x$ | 3. Divide both sides by $3$. |
| 4. The x-intercept is $\left(\frac{1}{3}, 0\right)$. |
To find the y-intercept, we do similarly the same thing, except we substitute $x=0$ into the equation and then solve for $y$.
| Algebra | Explanation |
|---|---|
| $y=3x-1$ | |
| $y=3(0)-1$ | 1. Substitute $x=0$ into the equation. |
| $y=0-1$ | 2. Multiply $3*0=0$. |
| $y=-1$ | 3. Simplify the right side to get $-1$. |
| 4. This would give us the y-intercept of $(0,-1)$. |
Looking at the graph below of $y=3x-1$, we can confirm these solutions and see that the point where it crosses the x-axis is indeed $\left(\frac{1}{3},0\right)$ and the point where it crosses the y-axis is indeed $(0,-1)$.
Find the intercepts of the equation $y=-\frac{1}{3}x-4$. Then, use a graphing utility to verify your answer.
Finding the x-intercept:
To find our x-intercept, we substitute $y=0$ in and solve for $x$.
\begin{align*} y&=-\frac{1}{3}x-4\\ 0&=-\frac{1}{3}x-4\\ 4&=-\frac{1}{3}x\\ -\frac{3}{1}*4&=x\\ -12&=x \end{align*}Finding the y-intercept:
To find our y-intercept, we substitute $x=0$ in and solve for $y$.
\begin{align*} y&=-\frac{1}{3}x-4\\ y&=-\frac{1}{3}(0)-4\\ y&=0-4\\ y&=-4 \end{align*}This means our x-intercept is $(-12,0)$ and our y-intercept is $(0,-4)$. We can verify that answer graphically below.
In Exercises 1-8, plot each point on the coordinate axis and identify the quadrant or axis it falls in or on.
In Exercises 9-16, graph each equation by creating a table.
In Exercises 17-24, find the $x$ and $y$-intercepts. Verify your answer using a graphing utility.