4.1: Use the Rectangular Coordinate System
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 Contributed by Lynn Marecek
 Professor (Mathematics) at Santa Ana College
 Publisher: OpenStax CNX
Learning Objectives
By the end of this section, you will be able to:
 Plot points in a rectangular coordinate system
 Verify solutions to an equation in two variables
 Complete a table of solutions to a linear equation
 Find solutions to a linear equation in two variables
Note
Before you get started, take this readiness quiz.
Plot Points on 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 xyplane or the ‘coordinate plane’.
The horizontal number line is called the xaxis. The vertical number line is called the yaxis. The xaxis and the yaxis 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. See Figure \(\PageIndex{1}\).
In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the xcoordinate of the point, and the second number is the ycoordinate of the point.
ORDERED PAIR
An ordered pair, (x,y)(x,y), gives the coordinates of a point in a rectangular coordinate system.
The first number is the xcoordinate.
The second number is the ycoordinate.
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.
THE ORIGIN
The point \((0,0)\) is called the origin. It is the point where the xaxis and yaxis intersect.
We use the coordinates to locate a point on the xyplane. Let’s plot the point \((1,3)\) as an example. First, locate 1 on the xaxis and lightly sketch a vertical line through x=1x=1. Then, locate 3 on the yaxis and sketch a horizontal line through y=3y=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 \((1,3)\).
Exercise \(\PageIndex{1}\)
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:
 (−5,4)
 (−3,−4)
 (2,−3)
 (−2,3)
 \((3, \frac{5}{2})\)
 Answer

The first number of the coordinate pair is the xcoordinate, and the second number is the ycoordinate.
 Since x=−5, the point is to the left of the yaxis. Also, since y=4, the point is above the xaxis. The point (−5,4) is in Quadrant II.
 Since x=−3, the point is to the left of the yaxis. Also, since y=−4, the point is below the xaxis. The point (−3,−4) is in Quadrant III.
 Since x=2, the point is to the right of the yaxis. Since y=−3, the point is below the xaxis. The point (2,−3) is in Quadrant lV.
 Since x=−2, the point is to the left of the yaxis. Since y=3, the point is above the xaxis. The point (−2,3) is in Quadrant II.
 Since x=3, the point is to the right of the yaxis. Since \(y = \frac{5}{2}\), the point is above the xaxis. (It may be helpful to write \(\frac{5}{2}\) as a mixed number or decimal.) The point \((3, \frac{5}{2})\) is in Quadrant I.
Exercise \(\PageIndex{2}\)
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
 (−2,1)
 (−3,−1)
 (4,−4)
 (−4,4)
 \((4, \frac{3}{2})\)
 Answer
Exercise \(\PageIndex{3}\)
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
 (−4,1)
 (−2,3)
 (2,−5)
 (−2,5)
 \((3, \frac{5}{2})\)
 Answer
How do the signs affect the location of the points? You may have noticed some patterns as you graphed the points in the previous example.
For the point in Figure \(\PageIndex{4}\) in Quadrant IV, what do you notice about the signs of the coordinates? What about the signs of the coordinates of points in the third quadrant? The second quadrant? The first quadrant?
Can you tell just by looking at the coordinates in which quadrant the point (−2,5) is located? In which quadrant is (2,−5) located?
QUADRANTS
We can summarize sign patterns of the quadrants in this way.
\[\begin{array}{ccc}{\text { Quadrant I }} & {\text { Quadrant II }} & {\text { Quadrant III }} & {\text { Quadrant IV }} \\ {(x, y)} & {(x, y)} & {(x, y)} & {(x, y)} \\ {(+,+)} & {(,+)} & {(,)} & {(+,)}\end{array}\]
What if one coordinate is zero as shown in Figure \(\PageIndex{6}\)? Where is the point (0,4) located? Where is the point (−2,0) located?
The point (0,4) is on the yaxis and the point (−2,0) is on the xaxis.
POINTS ON THE AXES
Points with a ycoordinate equal to 0 are on the xaxis, and have coordinates (a,0).
Points with an xcoordinate equal to 0 are on the yaxis, and have coordinates (0,b).
Exercise \(\PageIndex{4}\)
Plot each point:
 (0,5)
 (4,0)
 (−3,0)
 (0,0)
 (0,−1)
 Answer

 Since x=0, the point whose coordinates are (0,5) is on the yaxis.
 Since y=0, the point whose coordinates are (4,0) is on the xaxis.
 Since y=0, the point whose coordinates are (−3,0) is on the xaxis.
 Since x=0 and y=0, the point whose coordinates are (0,0) is the origin.
 Since x=0, the point whose coordinates are (0,−1) is on the yaxis.
Figure \(\PageIndex{7}\)
Exercise \(\PageIndex{5}\)
Plot each point:
 (4,0)
 (−2,0)
 (0,0)
 (0,2)
 (0,−3).
 Answer
Exercise \(\PageIndex{6}\)
Plot each point:
 (−5,0)
 (3,0)
 (0,0)
 (0,−1)
 (0,4).
 Answer
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 xcoordinate of a point on a graph, read the number on the xaxis directly above or below the point. To identify the ycoordinate of a point, read the number on the yaxis directly to the left or right of the point. Remember, when you write the ordered pair use the correct order, (x,y).
Exercise \(\PageIndex{7}\)
Name the ordered pair of each point shown in the rectangular coordinate system.
 Answer

Point A is above −3 on the xaxis, so the xcoordinate of the point is −3.
The point is to the left of 3 on the yaxis, so the ycoordinate of the point is 3.  The coordinates of the point are (−3,3).

Point B is below −1 on the xaxis, so the xcoordinate of the point is −1.
The point is to the left of −3 on the yaxis, so the ycoordinate of the point is −3.  The coordinates of the point are (−1,−3).
Point C is above 2 on the xaxis, so the xcoordinate of the point is 2.
The point is to the right of 4 on the yaxis, so the ycoordinate of the point is 4.  The coordinates of the point are (2,4).
 Point D is below 4 on the xaxis, so the xcoordinate of the point is 4.
 The point is to the right of −4 on the yaxis, so the ycoordinate of the point is −4.
 The coordinates of the point are (4,−4).

Point E is on the yaxis at y=−2. The coordinates of point E are (0,−2).
Point F is on the xaxis at x=3. The coordinates of point F are (3,0).
Exercise \(\PageIndex{8}\)
Name the ordered pair of each point shown in the rectangular coordinate system.
 Answer

A: (5,1) B: (−2,4) C: (−5,−1) D: (3,−2) E: (0,−5) F: (4,0)
Exercise \(\PageIndex{9}\)
Name the ordered pair of each point shown in the rectangular coordinate system.
 Answer

A: (4,2) B: (−2,3) C: (−4,−4) D: (3,−5) E: (−3,0) F: (0,2)
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.)
Here’s an example of an equation in one variable, and its one solution.
\[\begin{aligned} 3 x+5 &=17 \\ 3 x &=12 \\ x &=4 \end{aligned}\]
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.
LINEAR EQUATION
An equation of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.
Notice the word line in linear. Here is an example of a linear equation in two variables, x and y.
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.
\(\begin{array}{llll} {} &{y} &{=} &{3x + 5} \\ {\text{Add to both sides.}} &{y + 3x } &{=} &{3x + 5 + 3x} \\{\text{Simplify.}} &{y + 3x} &{=} &{5} \\{\text{Use the Commutative Property to put it in}} &{3x + y} &{=} &{5} \\{Ax+By = C\text{ form.}} &{} &{} &{} \end{array}\)
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.
STANDARD FORM OF LINEAR EQUATION
A linear equation is in standard form when it is written Ax+By=C.
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.
SOLUTION OF A LINEAR EQUATION IN TWO VARIABLES
An ordered pair (x,y) is a solution of the linear equation Ax+By=C, if the equation is a true statement when the x and yvalues of the ordered pair are substituted into the equation.
Exercise \(\PageIndex{10}\)
Determine which ordered pairs are solutions to the equation x+4y=8.
(a) (0,2)
(b) (2,−4)
(c) (−4,3)
 Answer

Substitute the x and yvalues from each ordered pair into the equation and determine if the result is a true statement.
Exercise \(\PageIndex{11}\)
Which of the following ordered pairs are solutions to 2x+3y=6?
 (3,0)
 (2,0)
 (6,−2)
 Answer

1, 3
Exercise \(\PageIndex{12}\)
Which of the following ordered pairs are solutions to the equation 4x−y=8?
 (0,8)
 (2,0)
 (1,−4)
 Answer

2, 3
Exercise \(\PageIndex{13}\)
Which of the following ordered pairs are solutions to the equation y=5x−1?
(a) (0,−1)
(b) (1,4)
(c) (−2,−7)
 Answer

Substitute the x and yvalues from each ordered pair into the equation and determine if it results in a true statement.
Exercise \(\PageIndex{14}\)
Which of the following ordered pairs are solutions to the equation y=4x−3?
 (0,3)
 (1,1)
 (−1,−1)
 Answer

2
Exercise \(\PageIndex{15}\)
Which of the following ordered pairs are solutions to the equation y=−2x+6?
 (0,6)
 (1,4)
 (−2,−2)
 Answer

1, 2
Complete a Table of Solutions to a Linear Equation in Two Variables
In the examples above, we substituted the x and yvalues 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 xx and then solve the equation for yy. Or, pick a value for yy and then solve for xx.
We’ll start by looking at the solutions to the equation y = 5x−1 that we found in Exercise \(\PageIndex{13}\). We can summarize this information in a table of solutions, as shown in Table \(\PageIndex{1}\).
y=5x−1  
x  y  (x,y) 
0  −1  (0,−1) 
1  4  (1,4) 
Table \(\PageIndex{1}\)
To find a third solution, we’ll let x=2 and solve for y.
The ordered pair (2,9) is a solution to y=5x−1. We will add it to Table \(\PageIndex{2}\).
y=5x−1  
x  y  (x,y) 
0  −1  (0,−1) 
1  4  (1,4) 
2  9  (2,9) 
Table \(\PageIndex{2}\)
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.
Exercise \(\PageIndex{16}\)
Complete Table to find three solutions to the equation y=4x−2.
y=4x−2  
x  y  (x,y) 
0  
−1  
2 
Table \(\PageIndex{3}\)
 Answer

Substitute x=0, x=−1, and x=2 into y=4x−2.

The results are summarized in Table \(\PageIndex{4}\).
y=4x−2 x y (x,y) 0 −2 (0,−2) −1 −6 (−1,−6) 2 6 (2,6) Table \(\PageIndex{4}\)
Exercise \(\PageIndex{17}\)
Complete the table to find three solutions to this equation: y=3x−1.
y=3x−1  
x  y  (x,y) 
0  
−1  
2 
Table \(\PageIndex{5}\)
 Answer

y=3x−1 x y (x,y) 0 1 (0, 1) −1 4 (1, 4) 2 5 (2, 5) Table \(\PageIndex{6}\)
Exercise \(\PageIndex{18}\)
Complete the table to find three solutions to this equation: y=6x+1.
y=6x+1  
x  y  (x,y) 
2 
Table \(\PageIndex{7}\)
 Answer

y=6x+1 x y (x,y) 0 1 (0,1) 1 7 (1,7) −2 −11 (−2,−11) Table \(\PageIndex{8}\)
Exercise \(\PageIndex{19}\)
Complete Table to find three solutions to the equation 5x−4y=20.
5x−4y=20  
x  y  (x,y) 
0  
5 
Table \(\PageIndex{9}\)
 Answer

Substitute the given value into the equation 5x−4y=20 and solve for the other variable. Then, fill in the values in the table.

The results are summarized in Table \(\PageIndex{10}\).
5x−4y=20 x y (x,y) 0 −5 (0,−5) 4 0 (4,0) 8 5 (8,5) Table \(\PageIndex{10}\)
Exercise \(\PageIndex{20}\)
Complete the table to find three solutions to this equation: 2x−5y=20.
2x−5y=20  
x  y  (x,y) 
5 
Table \(\PageIndex{11}\)
 Answer

2x−5y=20 x y (x,y) 0 −4 (0,−4) 10 0 (10,0) −5 −6 (−5,−6) Table \(\PageIndex{12}\)
Exercise \(\PageIndex{21}\)
Complete the table to find three solutions to this equation: 3x−4y=12.
3x−4y=12  
x  y  (x,y) 
4 
Table \(\PageIndex{13}\)
 Answer

3x−4y=12 x y (x,y) 0 −3 (0,−3) 4 0 (4,0) −4 −6 (−4,−6) Table \(\PageIndex{14}\)
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 yform, 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.
Exercise \(\PageIndex{22}\)
Find three solutions to the equation y=−3x+2.
 Answer

We can substitute any value we want for x or any value for y. Since the equation is in yform, it will be easier to substitute in values of x. Let’s pick x=0, x=1, and x=−1.
Substitute the value into the equation. Simplify. Simplify. Write the ordered pair. (0, 2) (1, 1) (1, 5) Check. y=−3x+2 y=−3x+2 y=−3x+2 \(2 \stackrel{?}{=} 3 \cdot 0 + 2\) \(1 \stackrel{?}{=} 3 \cdot 1 + 2\) \(5 \stackrel{?}{=} 3 (1) + 2\) \(2 \stackrel{?}{=} 0 + 2\) \(1 \stackrel{?}{=} 3 + 2\) \(5 \stackrel{?}{=} 3 + 2\) \(2 = 2\checkmark\) \(1 = 1\checkmark\) \(5 = 5\checkmark\)  Table \(\PageIndex{15}\)

So, (0,2), (1,−1) and (−1,5) are all solutions to y=−3x+2. We show them in Table \(\PageIndex{16}\) .
y=−3x+2 x y (x,y) 0 2 (0,2) 1 −1 (1,−1) −1 5 (−1,5) Table \(\PageIndex{16}\)
Exercise \(\PageIndex{23}\)
Find three solutions to this equation: y=−2x+3.
 Answer

Answers will vary.
Exercise \(\PageIndex{24}\)
Find three solutions to this equation: y=−4x+1.
 Answer

Answers will vary.
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.
Exercise \(\PageIndex{25}\)
Find three solutions to the equation 3x+2y=6.
 Answer

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.
Substitute the value into the equation. Simplify. Solve. Write the ordered pair. (0, 3) (2, 0) \((1,\frac{3}{2})\) Check. 3x+2y=6 3x+2y=6 3x+2y=6 \(3\cdot 0 + 2\cdot 3 \stackrel{?}{=} 6\) \(3\cdot 2 + 2\cdot 0 \stackrel{?}{=} 6\) \(3\cdot 1 + 2\cdot \frac{3}{2} \stackrel{?}{=} 6\) \(0 + 6 \stackrel{?}{=} 6\) \(6 + 0 \stackrel{?}{=} 6\) \(3 + 3 \stackrel{?}{=} 6\) \(6 = 6\checkmark\) \(6 = 6\checkmark\) \(6 = 6\checkmark\) Table \(\PageIndex{17}\)
So (0,3), (2,0), and \((1,\frac{3}{2})\) are all solutions to the equation 3x+2y=6. We can list these three solutions in Table \(\PageIndex{18}\) .
3x+2y=63x+2y=6 x y (x,y) 0 3 (0,3) 2 0 (2,0) 1 \(\frac{3}{2}\) \((1, \frac{3}{2})\) Table \(\PageIndex{18}\)
Exercise \(\PageIndex{26}\)
Find three solutions to the equation 2x+3y=6.
 Answer

Answers will vary.
Exercise \(\PageIndex{27}\)
Find three solutions to the equation 4x+2y=8.
 Answer

Answers will vary.
Key Concepts
 Sign Patterns of the Quadrants
\(\begin{array}{ll}{\text { Quadrant I }} & {\text { Quadrant II }} & {\text { Quadrant III }} & {\text { Quadrant IV }} \\ {(x, y)} & {(x, y)} & {(x, y)} & {(x, y)} \\ {(+,+)} & {(,+)} & {(,)} & {(+,)}\end{array}\)  Points on the Axes
 On the xaxis, y=0. Points with a ycoordinate equal to 0 are on the xaxis, and have coordinates (a,0).
 On the yaxis, x=0. Points with an xcoordinate equal to 0 are on the yaxis, and have coordinates (0,b).
 Solution of a Linear Equation
 An ordered pair (x,y) is a solution of the linear equation Ax+By=C, if the equation is a true statement when the x and y values of the ordered pair are substituted into the equation.
Glossary
 linear equation
 A linear equation is of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.
 ordered pair
 An ordered pair (x,y) gives the coordinates of a point in a rectangular coordinate system.
 origin
 The point (0,0)(0,0) is called the origin. It is the point where the xaxis and yaxis intersect.
 quadrant
 The xaxis and the yaxis divide a plane into four regions, called quadrants.
 rectangular coordinate system
 A grid system is used in algebra to show a relationship between two variables; also called the xyplane or the ‘coordinate plane’.
 xcoordinate
 The first number in an ordered pair (x,y).
 ycoordinate
 The second number in an ordered pair (x,y).