Use the Rectangular Coordinate System

Plot points on a rectangular coordinate system

Many maps use a grid system to identify locations. Imagine a campus map with the numbers $1, 2, 3,$ and $4$ across the top and bottom, and the letters $A, B, C,$ and $D$ along the sides. Every location on the map can be identified by a number and a letter — the Student Center might sit in grid section $2B$: in the grid section above the number $2$ and next to the letter $B$.

Just as maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. To create a rectangular coordinate system, start with a horizontal number line. Show both positive and negative numbers, using a convenient scale unit. This horizontal number line is called the $x$-axis.

Now make a vertical number line passing through the $x$-axis at $0$. Put the positive numbers above $0$ and the negative numbers below $0$. This vertical line is called the $y$-axis.

The $x$-axis and the $y$-axis form the rectangular coordinate system. These axes divide a plane into four areas, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise.

In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the $x$-coordinate of the point, and the second number is the $y$-coordinate of the point.

So how do the coordinates of a point help you locate a point on the $x\text{-}y$ plane? Let’s try locating the point $(2, 5)$. In this ordered pair, the $x$-coordinate is $2$ and the $y$-coordinate is $5$. Start by locating the $x$ value, $2$, on the $x$-axis, and lightly sketch a vertical line through $x = 2$. Then locate the $y$ value, $5$, on the $y$-axis, and sketch a horizontal line through $y = 5$. The point where these two lines meet is the point with coordinates $(2, 5)$.

Example. Plot $(1, 3)$ and $(3, 1)$ in the same rectangular coordinate system.

The coordinate values are the same for both points, but the $x$ and $y$ values are reversed. For $(1, 3)$, locate $1$ on the $x$-axis and sketch a vertical line through $x = 1$; locate $3$ on the $y$-axis and sketch a horizontal line through $y = 3$. Where the two lines meet, plot the point $(1, 3)$. For $(3, 1)$, locate $3$ on the $x$-axis and $1$ on the $y$-axis; where those lines meet, plot the point $(3, 1)$.

Notice that the order of the coordinates does matter, so $(1, 3)$ is not the same point as $(3, 1)$.

Example. Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located: (a) $(-1, 3)$ (b) $(-3, -4)$ (c) $(2, -1)$ (d) $\left(3, \tfrac{5}{2}\right)$.

The first number of the coordinate pair is the $x$-coordinate, and the second number is the $y$-coordinate.

(a) Since $x = -1, y = 3$, the point $(-1, 3)$ is in Quadrant II.

(b) Since $x = -3, y = -4$, the point $(-3, -4)$ is in Quadrant III.

(c) Since $x = 2, y = -1$, the point $(2, -1)$ is in Quadrant IV.

(d) Since $x = 3, y = \tfrac{5}{2}$, the point $\left(3, \tfrac{5}{2}\right)$ is in Quadrant I. It may be helpful to write $\tfrac{5}{2}$ as the mixed number $2\tfrac{1}{2}$, or decimal $2.5$. Then we know that the point is halfway between $2$ and $3$ on the $y$-axis.

We can summarize the sign patterns of the quadrants as follows.

Example. How do the signs affect the location of the points? Plot each point: (a) $(-5, 2)$ (b) $(-5, -2)$ (c) $(5, 2)$ (d) $(5, -2)$.

As we locate the $x$-coordinate and the $y$-coordinate, we must be careful with the signs. Point (a) lands in Quadrant II (left of the $y$-axis, above the $x$-axis). Point (b) lands in Quadrant III (left of the $y$-axis, below the $x$-axis). Point (c) lands in Quadrant I (right of the $y$-axis, above the $x$-axis). Point (d) lands in Quadrant IV (right of the $y$-axis, below the $x$-axis).

You may have noticed some patterns as you graphed the points in the two previous examples. For each point in Quadrant IV, the $x$-coordinate is positive and the $y$-coordinate is negative. In Quadrant III, both coordinates are negative. In Quadrant II, the $x$-coordinate is negative and the $y$-coordinate is positive. In Quadrant I, both coordinates are positive.

What if one coordinate is zero? The point $(0, 4)$ is on the $y$-axis, and the point $(-2, 0)$ is on the $x$-axis.

What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is $(0, 0)$. This point has a special name — it is called the origin.

Example. Plot each point on a coordinate grid: (a) $(0, 5)$ (b) $(4, 0)$ (c) $(-3, 0)$ (d) $(0, 0)$ (e) $(0, -1)$.

(a) Since $x = 0$, the point whose coordinates are $(0, 5)$ is on the $y$-axis. (b) Since $y = 0$, the point whose coordinates are $(4, 0)$ is on the $x$-axis. (c) Since $y = 0$, the point whose coordinates are $(-3, 0)$ is on the $x$-axis. (d) Since $x = 0$ and $y = 0$, the point whose coordinates are $(0, 0)$ is the origin. (e) Since $x = 0$, the point whose coordinates are $(0, -1)$ is on the $y$-axis.

Identify points on a graph

In algebra, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the $x$-coordinate of a point on a graph, read the number on the $x$-axis directly above or below the point. To identify the $y$-coordinate of a point, read the number on the $y$-axis directly to the left or right of the point. Remember to write the ordered pair using the correct order $(x, y)$.

Example. Name the ordered pair of each point shown on the graph.

Point $A$ is above $-3$ on the $x$-axis, so the $x$-coordinate of the point is $-3$. The point is to the left of $3$ on the $y$-axis, so the $y$-coordinate of the point is $3$. The coordinates of the point are $(-3, 3)$.

Point $B$ is below $-1$ on the $x$-axis, so the $x$-coordinate of the point is $-1$. The point is to the left of $-3$ on the $y$-axis, so the $y$-coordinate of the point is $-3$. The coordinates of the point are $(-1, -3)$.

Point $C$ is above $2$ on the $x$-axis, so the $x$-coordinate of the point is $2$. The point is to the right of $4$ on the $y$-axis, so the $y$-coordinate of the point is $4$. The coordinates of the point are $(2, 4)$.

Point $D$ is below $4$ on the $x$-axis, so the $x$-coordinate of the point is $4$. The point is to the right of $-4$ on the $y$-axis, so the $y$-coordinate of the point is $-4$. The coordinates of the point are $(4, -4)$.

Example. Name the ordered pair of each point shown, where points $A, B, C, D$ lie on the axes: $A$ is on the $x$-axis at $x = -4$; $B$ is on the $y$-axis at $y = -2$; $C$ is on the $x$-axis at $x = 3$; $D$ is on the $y$-axis at $y = 1$.

Point $A$ is on the $x$-axis at $x = -4$, so the coordinates of point $A$ are $(-4, 0)$. Point $B$ is on the $y$-axis at $y = -2$, so the coordinates of point $B$ are $(0, -2)$. Point $C$ is on the $x$-axis at $x = 3$, so the coordinates of point $C$ are $(3, 0)$. Point $D$ is on the $y$-axis at $y = 1$, so the coordinates of point $D$ are $(0, 1)$.

Verify solutions to an equation in two variables

All the equations we’ve solved so far have been equations with one variable. In almost every case, when we solved the equation we got exactly one solution — the process ended with a statement such as $x = 4$, checked by substituting back into the equation.

But equations can have more than one variable. Equations with two variables can be written in the general form $Ax + By = C$. An equation of this form is called a linear equation in two variables.

Notice that the word “line” is in linear. Here is an example of a linear equation in two variables, $x$ and $y$: $x + 4y = 8$, where $A = 1$, $B = 4$, and $C = 8$.

Is $y = -5x + 1$ a linear equation? It does not appear to be in the form $Ax + By = C$. But we could rewrite it in this form. Add $5x$ to both sides: $y + 5x = -5x + 1 + 5x$. Simplify: $y + 5x = 1$. Use the Commutative Property to put it in $Ax + By = C$ form: $5x + y = 1$.

By rewriting $y = -5x + 1$ as $5x + y = 1$, we can see that it is a linear equation in two variables because it can be written in the form $Ax + By = C$.

Linear equations in two variables have infinitely many solutions. For every number that is substituted for $x$, there is a corresponding $y$ value. This pair of values is a solution to the linear equation and is represented by the ordered pair $(x, y)$. When we substitute these values of $x$ and $y$ into the equation, the result is a true statement because the value on the left side is equal to the value on the right side.

Example. Determine which ordered pairs are solutions of the equation $x + 4y = 8$: (a) $(0, 2)$ (b) $(2, -4)$ (c) $(-4, 3)$.

Substitute the $x$- and $y$-values from each ordered pair into the equation and determine if the result is a true statement.

(a) $x = 0, y = 2$: $0 + 4 \cdot 2 \stackrel{?}{=} 8$, so $0 + 8 \stackrel{?}{=} 8$, and $8 = 8$ ✓. $(0, 2)$ is a solution.

(b) $x = 2, y = -4$: $2 + 4(-4) \stackrel{?}{=} 8$, so $2 + (-16) \stackrel{?}{=} 8$, and $-14 \neq 8$. $(2, -4)$ is not a solution.

(c) $x = -4, y = 3$: $-4 + 4 \cdot 3 \stackrel{?}{=} 8$, so $-4 + 12 \stackrel{?}{=} 8$, and $8 = 8$ ✓. $(-4, 3)$ is a solution.

Example. Determine which ordered pairs are solutions of the equation $y = 5x - 1$: (a) $(0, -1)$ (b) $(1, 4)$ (c) $(-2, -7)$.

(a) $x = 0, y = -1$: $-1 \stackrel{?}{=} 5(0) - 1$, so $-1 \stackrel{?}{=} 0 - 1$, and $-1 = -1$ ✓. $(0, -1)$ is a solution.

(b) $x = 1, y = 4$: $4 \stackrel{?}{=} 5(1) - 1$, so $4 \stackrel{?}{=} 5 - 1$, and $4 = 4$ ✓. $(1, 4)$ is a solution.

(c) $x = -2, y = -7$: $-7 \stackrel{?}{=} 5(-2) - 1$, so $-7 \stackrel{?}{=} -10 - 1$, and $-7 \neq -11$. $(-2, -7)$ is not a solution.

Complete a table of solutions to a linear equation

In the previous examples, we substituted the $x$- and $y$-values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do we find the ordered pairs if they are not given? One way is to choose a value for $x$ and then solve the equation for $y$. Or, choose a value for $y$ and then solve the equation for $x$.

We’ll start by looking at the solutions to the equation $y = 5x - 1$ we found above. We can summarize this information in a table of solutions.

To find a third solution, let $x = 2$ and solve for $y$: substituting $x = 2$ gives $y = 5(2) - 1$, so $y = 10 - 1$, and $y = 9$. The ordered pair $(2, 9)$ is a solution to $y = 5x - 1$. We can find more solutions to the equation by substituting any value of $x$ or any value of $y$ and solving the resulting equation to get another ordered pair that is a solution. There are an infinite number of solutions for this equation.

Example. Complete the table to find three solutions to the equation $y = 4x - 2$, using $x = 0$, $x = -1$, and $x = 2$.

Substitute each value of $x$ into $y = 4x - 2$: when $x = 0$, $y = 4 \cdot 0 - 2 = 0 - 2 = -2$; when $x = -1$, $y = 4(-1) - 2 = -4 - 2 = -6$; when $x = 2$, $y = 4 \cdot 2 - 2 = 8 - 2 = 6$. The results are summarized in the table.

Example. Complete the table to find three solutions to the equation $5x - 4y = 20$, given $x = 0$, $y = 0$, and $y = 5$.

When $x = 0$: $5(0) - 4y = 20$, so $-4y = 20$, and $y = -5$; the ordered pair is $(0, -5)$. When $y = 0$: $5x - 4(0) = 20$, so $5x = 20$, and $x = 4$; the ordered pair is $(4, 0)$. When $y = 5$: $5x - 4(5) = 20$, so $5x - 20 = 20$, then $5x = 40$, and $x = 8$; the ordered pair is $(8, 5)$. The results are summarized in the table.

Find solutions to linear equations in two variables

To find a solution to a linear equation, we can choose any number we want to substitute into the equation for either $x$ or $y$. We could choose $1$, $100$, $1{,}000$, or any other value we want. But it’s a good idea to choose a number that’s easy to work with. We’ll usually choose $0$ as one of our values.

Example. Find a solution to the equation $3x + 2y = 6$.

Step 1: Choose any value for one of the variables in the equation. We can substitute any value we want for $x$ or any value for $y$. Let’s pick $x = 0$. What is the value of $y$ if $x = 0$?

Step 2: Substitute that value into the equation. Solve for the other variable. Substitute $0$ for $x$: $3(0) + 2y = 6$. Simplify: $0 + 2y = 6$, so $2y = 6$. Divide both sides by $2$: $y = 3$.

Step 3: Write the solution as an ordered pair. So, when $x = 0$, $y = 3$. This solution is represented by the ordered pair $(0, 3)$.

Step 4: Check. Substitute $x = 0, y = 3$ into the equation $3x + 2y = 6$: $3(0) + 2(3) \stackrel{?}{=} 6$, so $0 + 6 \stackrel{?}{=} 6$, and $6 = 6$ ✓.

We said that linear equations in two variables have infinitely many solutions, and we’ve just found one of them. Let’s find some other solutions to the equation $3x + 2y = 6$.

Example. Find three more solutions to the equation $3x + 2y = 6$.

To find solutions to $3x + 2y = 6$, choose a value for $x$ or $y$. Remember, we can choose any value we want. Let’s choose $y = 0$, $x = 1$, and $y = -3$.

When $y = 0$: $3x + 2(0) = 6$, so $3x + 0 = 6$, then $3x = 6$, and $x = 2$; the ordered pair is $(2, 0)$.

When $x = 1$: $3(1) + 2y = 6$, so $3 + 2y = 6$, then $2y = 3$, and $y = \tfrac{3}{2}$; the ordered pair is $\left(1, \tfrac{3}{2}\right)$.

When $y = -3$: $3x + 2(-3) = 6$, so $3x - 6 = 6$, then $3x = 12$, and $x = 4$; the ordered pair is $(4, -3)$.

So $(2, 0)$, $\left(1, \tfrac{3}{2}\right)$, and $(4, -3)$ are all solutions to the equation $3x + 2y = 6$. Together with $(0, 3)$ found above, we can list these solutions in a table.

Example. Find three solutions to the equation $x - 4y = 8$.

Choose a value for $x$ or $y$. Let’s use $x = 0$, $y = 0$, and $y = 3$.

When $x = 0$: $0 - 4y = 8$, so $-4y = 8$, and $y = -2$; the ordered pair is $(0, -2)$. When $y = 0$: $x - 4(0) = 8$, so $x - 0 = 8$, and $x = 8$; the ordered pair is $(8, 0)$. When $y = 3$: $x - 4(3) = 8$, so $x - 12 = 8$, then $x = 20$; the ordered pair is $(20, 3)$.

So $(0, -2)$, $(8, 0)$, and $(20, 3)$ are three solutions to the equation $x - 4y = 8$. Remember, there are an infinite number of solutions to each linear equation. Any point you find is a solution if it makes the equation true.

Key terms

rectangular coordinate system — a grid formed by a horizontal $x$-axis and a vertical $y$-axis, used to show the relationship between two variables. quadrant — one of the four regions the $x$-axis and $y$-axis divide the plane into, numbered I through IV counterclockwise starting from the upper right. ordered pair — a pair of numbers $(x, y)$ that gives the coordinates of a point in a rectangular coordinate system. origin — the point $(0, 0)$, where the $x$-axis and $y$-axis intersect. linear equation — an equation of the form $Ax + By = C$, where $A$ and $B$ are not both zero. solution to a linear equation in two variables — an ordered pair $(x, y)$ that makes the equation a true statement when substituted in for $x$ and $y$.

Practice

Plot points on a rectangular coordinate system

Identify points on a graph

Verify solutions to an equation in two variables

Complete a table of solutions to a linear equation

Find solutions to linear equations in two variables

This section is adapted from Prealgebra 2e, Section 11.1: Use the Rectangular Coordinate System by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the coordinate-grid and quadrant figures as accessible inline graphics and the solution tables as markdown tables; omitted the Be Prepared quiz, campus-map figure, Media links, Everyday Math, Writing Exercises, and Self Check blocks; converted the practice problems (“Try Its”) into interactive exercises with instant feedback; and adapted selected end-of-section exercises into the interactive Practice block, recreating two exercise graphs as accessible inline graphics, expanding multipart items part by part, and restating the “plot each point” and “which ordered pairs are solutions” prompts as gradable quadrant and yes/no questions.