Use the Rectangular Coordinate System

Plot points in a rectangular coordinate system

Just like 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. The rectangular coordinate system is also called the $xy$-plane or the “coordinate plane.”

The horizontal number line is called the $x$-axis. The vertical number line is called the $y$-axis. The $x$-axis and the $y$-axis together form the rectangular coordinate system. These axes divide a plane into four regions, 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.

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

We use the coordinates to locate a point on the $xy$-plane. Let’s plot the point $(1, 3)$ as an example. First, locate $1$ on the $x$-axis and lightly sketch a vertical line through $x = 1$. Then locate $3$ on the $y$-axis and sketch a horizontal line through $y = 3$. Now, find the point where these two lines meet — that is the point with coordinates $(1, 3)$.

Notice that the vertical line through $x = 1$ and the horizontal line through $y = 3$ are not part of the graph. We just used them to help us locate the point $(1, 3)$.

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

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

(a) Since $x = -5$, the point is to the left of the $y$-axis. Also, since $y = 4$, the point is above the $x$-axis. The point $(-5, 4)$ is in Quadrant II.

(b) Since $x = -3$, the point is to the left of the $y$-axis. Also, since $y = -4$, the point is below the $x$-axis. The point $(-3, -4)$ is in Quadrant III.

(c) Since $x = 2$, the point is to the right of the $y$-axis. Since $y = -3$, the point is below the $x$-axis. The point $(2, -3)$ is in Quadrant IV.

(d) Since $x = -2$, the point is to the left of the $y$-axis. Since $y = 3$, the point is above the $x$-axis. The point $(-2, 3)$ is in Quadrant II.

(e) Since $x = 3$, the point is to the right of the $y$-axis. Since $y = \tfrac{5}{2}$, the point is above the $x$-axis. (It may be helpful to write $\tfrac{5}{2}$ as a mixed number or decimal — it is halfway between $2$ and $3$.) The point $\left(3, \tfrac{5}{2}\right)$ is in Quadrant I.

We can summarize the sign patterns of the quadrants this way.

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.

Example. Plot each point: (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.

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, when you write the ordered pair, use the correct order $(x, y)$.

Example. Name the ordered pair of each point shown, where points $A, B, C, D, E, F$ are plotted in the rectangular coordinate system: $A$ is above $-3$ on the $x$-axis and to the left of $3$ on the $y$-axis; $B$ is below $-1$ on the $x$-axis and to the left of $-3$ on the $y$-axis; $C$ is above $2$ on the $x$-axis and to the left of $4$ on the $y$-axis; $D$ is below $4$ on the $x$-axis and to the right of $-4$ on the $y$-axis; $E$ is on the $y$-axis at $y = -2$; $F$ is on the $x$-axis at $x = 3$.

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 $(4, -4)$.

Point $E$ is on the $y$-axis at $y = -2$. The coordinates of point $E$ are $(0, -2)$.

Point $F$ is on the $x$-axis at $x = 3$. The coordinates of point $F$ are $(3, 0)$.

Verify solutions to an equation in two variables

Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. The process of solving an equation ended with a statement like $x = 4$. (Then, you checked the solution by substituting back into the equation.)

But equations can have more than one variable. Equations with two variables may be of the form $Ax + By = C$. Equations of this form are called linear equations in two variables.

Notice the word line 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 $C = 8$.

The equation $y = -3x + 5$ is also a linear equation. But it does not appear to be in the form $Ax + By = C$. We can use the Addition Property of Equality and rewrite it in $Ax + By = C$ form.

By rewriting $y = -3x + 5$ as $3x + y = 5$, we can easily see that it is a linear equation in two variables because it is of the form $Ax + By = C$. When an equation is in the form $Ax + By = C$, we say it is in standard form.

Most people prefer to have $A$, $B$, and $C$ be integers and $A \geq 0$ when writing a linear equation in standard form, although it is not strictly necessary.

Linear equations 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 to 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. Which of the following ordered pairs are solutions to $y = 5x - 1$: (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 two variables

In the examples above, 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 you find the ordered pairs if they are not given? It’s easier than you might think — you can just pick a value for $x$ and then solve the equation for $y$. Or, pick a value for $y$ and then solve for $x$.

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

To find a third solution, we’ll 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 add it to the table.

We can find more solutions to the equation by substituting in 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 infinitely many solutions of 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 $x = 0$, $x = -1$, and $x = 2$ 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 $y = 5$.

Substitute the given value into the equation $5x - 4y = 20$ and solve for the other variable.

When $x = 0$: $\ 5(0) - 4y = 20$, so $0 - 4y = 20$, then $-4y = 20$, and $y = -5$; the ordered pair is $(0, -5)$.

When $y = 0$: $\ 5x - 4(0) = 20$, so $5x - 0 = 20$, then $5x = 20$, 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 a linear equation

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for $x$ or $y$. But since you’ll need to use that number to solve for the other variable, it’s a good idea to choose a number that’s easy to work with.

When the equation is in $y$-form, with the $y$ by itself on one side of the equation, it is usually easier to choose values of $x$ and then solve for $y$.

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

We can substitute any value we want for $x$ or any value for $y$. Since the equation is in $y$-form, it will be easier to substitute in values of $x$. Let’s pick $x = 0$, $x = 1$, and $x = -1$.

When $x = 0$: $\ y = -3 \cdot 0 + 2 = 0 + 2 = 2$; the ordered pair is $(0, 2)$. Check: $2 \stackrel{?}{=} -3 \cdot 0 + 2$, so $2 = 2$ ✓.

When $x = 1$: $\ y = -3 \cdot 1 + 2 = -3 + 2 = -1$; the ordered pair is $(1, -1)$. Check: $-1 \stackrel{?}{=} -3 \cdot 1 + 2$, so $-1 = -1$ ✓.

When $x = -1$: $\ y = -3(-1) + 2 = 3 + 2 = 5$; the ordered pair is $(-1, 5)$. Check: $5 \stackrel{?}{=} -3(-1) + 2$, so $5 = 5$ ✓.

So $(0, 2)$, $(1, -1)$, and $(-1, 5)$ are all solutions to $y = -3x + 2$. We show them in a table.

We have seen how using zero as one value of $x$ makes finding the value of $y$ easy. When an equation is in standard form, with both the $x$ and $y$ on the same side of the equation, it is usually easier to first find one solution when $x = 0$, find a second solution when $y = 0$, and then find a third solution.

Example. Find three solutions 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 $y$. Since the equation is in standard form, let’s pick first $x = 0$, then $y = 0$, and then find a third point.

Step 2: Substitute that value into the equation. Solve for the other variable.

When $x = 0$: $\ 3(0) + 2y = 6$, so $0 + 2y = 6$, then $2y = 6$, and $y = 3$.

When $y = 0$: $\ 3x + 2(0) = 6$, so $3x + 0 = 6$, then $3x = 6$, and $x = 2$.

When $x = 1$: $\ 3(1) + 2y = 6$, so $3 + 2y = 6$, then $2y = 3$, and $y = \tfrac{3}{2}$.

Step 3: Write the solution as an ordered pair. So the solutions are $(0, 3)$, $(2, 0)$, and $\left(1, \tfrac{3}{2}\right)$.

Step 4: Check. Substitute each pair into $3x + 2y = 6$:

$(0, 3)$: $\ 3(0) + 2(3) \stackrel{?}{=} 6$, so $0 + 6 \stackrel{?}{=} 6$, and $6 = 6$ ✓.

$(2, 0)$: $\ 3(2) + 2(0) \stackrel{?}{=} 6$, so $6 + 0 \stackrel{?}{=} 6$, and $6 = 6$ ✓.

$\left(1, \tfrac{3}{2}\right)$: $\ 3(1) + 2 \cdot \tfrac{3}{2} \stackrel{?}{=} 6$, so $3 + 3 \stackrel{?}{=} 6$, and $6 = 6$ ✓.

Key terms

rectangular coordinate system — a grid formed by a horizontal $x$-axis and a vertical $y$-axis, used to show a relationship between two variables; also called the $xy$-plane. 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; the first number is the $x$-coordinate and the second is the $y$-coordinate. origin — the point $(0, 0)$, where the $x$-axis and $y$-axis intersect. linear equation in two variables — an equation of the form $Ax + By = C$, where $A$ and $B$ are not both zero. standard form — a linear equation is in standard form when it is written $Ax + By = C$. solution of a linear equation in two variables — an ordered pair $(x, y)$ that makes the equation a true statement when its $x$- and $y$-values are substituted in for $x$ and $y$.

This section is adapted from Elementary Algebra 2e, Section 4.1: 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 quadrant and plotted-point figures as accessible inline graphics and the rewriting/solution steps as tables; omitted the Be Prepared quiz, Media links, and Section Exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.