Table 1.1 Toppings x
Cost y
0 $ 5
1 7
2 9
3 11
4 13
corresponding cost. The ordered pair 13, 112 indicates that a three-topping pizza costs $11. The order of the numbers matters. In this context, 111, 32 would indicate that an 11-topping pizza costs $3.
To picture ordered pairs of real numbers we use the rectangular coordinate system or Cartesian coordinate system, named after the French mathematician René Descartes (1596–1650). The Cartesian coordinate system consists of two num- ber lines drawn perpendicular to one another, intersecting at zero on each number line as shown in Fig. 1.7. The point of intersection of the number lines is called the origin. The horizontal number line is the x-axis and its positive numbers are to the right of the origin. The vertical number line is the y-axis and its positive numbers are above the origin. The two number lines divide the plane into four regions called quadrants, numbered as shown in Fig. 1.7. The quadrants do not include any points on the axes. We call a plane with a rectangular coordinate system the coordinate plane or the xy-plane.
Just as every real number corresponds to a point on the number line, every ordered pair of real numbers 1a, b2 corresponds to a point P in the xy-plane. For this reason, or- dered pairs of numbers are often called points. So a and b are the coordinates of 1a, b2 or the coordinates of the point P. Locating the point P that corresponds to 1a, b2 in the xy-plane is referred to as plotting or graphing the point, and P is called the graph of 1a, b2. In general, a graph is a set of points in the rectangular coordinate system.
Figure 1.7
– 1 – 3 –4
– 5 –1
–2 –3 –4
1 2 3
– 2 2 1 4 5
4 5 3
–5
Quadrant I Quadrant II
Quadrant IV Quadrant III
x-axis
Origin
y-axis y
–1 x –3 – 4
–5 – 1
– 2 – 3 – 4 –2 3
2 1 4 5
4 5 3
– 5
(3, 5) (– 3, 4)
(0, 2)
(– 2, – 5) (4, – 5)
+3 +5
Figure 1.8
EXAMPLE 1 Plotting points
Plot the points 13, 52, 14, -52, 1-3, 42, 1-2, -52, and 10, 22 in the xy-plane.
Solution
The point 13, 52 is located three units to the right of the origin and five units above the x-axis as shown in Fig. 1.8. The point 14, -52 is located four units to the right of the origin and five units below the x-axis. The point 1-3, 42 is located three units to the left of the origin and four units above the x-axis. The point 1-2, -52 is located two units to the left of the origin and five units below the x-axis. The point 10, 22 is on the y-axis because its first coordinate is zero.
TRY THIS. Plot 1-3, -22, 1-1, 32, 15>2, 02, and 12, -32. ■ Note that for points in quadrant I, both coordinates are positive. In quadrant II the first coordinate is negative and the second is positive, while in quadrant III, both coordinates are negative. In quadrant IV the first coordinate is positive and the second is negative.
H I S T O R I C A L N O T E
Pythagoras of Samos (582 B.C.–507 B.C.) was a Greek mathematician and philoso- pher. He founded the mystic, religious, and scientific society called Pythagore- ans. He is best known for the Pythago- rean theorem. Pythagoras made influ- ential contributions to philosophy and religious teaching in the late 6th cen- tury B.C. Pythagoras and his students believed that everything was related to mathematics, and that through math- ematics everything could be predicted and measured in rhythmic patterns or cycles.
H I S T O R I C A L N O T E
René Descartes (1596–1650) was a noted French philosopher, mathemati- cian, and scientist. He has been called the “Founder of Modern Philosophy”
and the “Father of Modern Mathemat- ics.” He ranks as one of the most important and influential thinkers of modern times.
The Pythagorean Theorem and the Distance Formula
You have probably studied the Pythagorean theorem in an algebra or a trigonometry course that you have taken. This theorem states that the sum of the squares of the legs of any right triangle is equal to the square of the hypotenuse.
The Pythagorean Theorem The triangle shown here is a right triangle if and only if a2+ b2 =c2.
a
b c
legs hypotenuse
We will not prove the Pythagorean theorem here. If you want to see a proof, try an Internet search for the Pythagorean theorem. You will get many sites with numer- ous proofs. This author found one site with 79 proofs, including one by a former presi- dent of the United States. Here we will show how the Pythagorean theorem leads to a formula for the distance between two points in the plane and the equation for a circle.
If a and b are real numbers, then the distance between them on the number line is 0a - b0. Now consider the points A1x1, y12 and B1x2, y22 shown in Fig. 1.9. Let AB represent the length of line segment AB. Now AB is the hypotenuse of the right triangle in Fig. 1.9. Since A and C lie on a horizontal line, the distance between them is 0x2- x10. Likewise CB = 0 y2- y10. Since the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse (the Pythagorean theorem) we have
d 2= 0x2 - x102 + 0y2- y102.
Since the distance between two points is a nonnegative real number, we have d = 21x2 - x122+ 1y2- y122. The absolute value symbols are replaced with parentheses, because 0a - b02= 1a -b22 for any real numbers a and b.
Figure 1.9 y
x C(x2, y1) d
⎫⎪
⎪⎪
⎬⎪
⎪⎪
⎭ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎫ ⎭
⏐x2 – x1⏐ A(x1, y1)
B(x2, y2)
⏐y2 – y1⏐
The Distance Formula The distance d between the points 1x1, y12 and 1x2, y22 is given by the formula d = 21x2 - x122 + 1y2 -y122.
EXAMPLE 2 Finding the distance between two points
Find the exact distance between each pair of points.
a. 15, -32, 1-1, -62 b. 1p>2, 02, 1p>3, 12
Solution
a. Let 1x1, y12 = 15, -32 and 1x2, y22 =1-1, -62. These points are shown on the graph in Fig. 1.10. Substitute these values into the distance formula:
d= 21-1 - 522 +1-6- 1-3222
= 21-622+ 1-322
= 136 + 9= 145 = 315 The exact distance between the points is 315.
Figure 1.10 (5, – 3)
(–1, – 6) y
–1 x – 1 – 2 – 3 – 4
1 2 3
– 2 1
4 5
– 5 d
b. Use the distance formula as follows.
d = Bap 2 - p
3b2+ 10 -122
= Ba3p 6 - 2p
6 b2 +1-122 Get common denominators.
= Bap
6b2+ 1= B
p2 36 +1 =
B p2+ 36
36 = 3p2+ 36 6 The exact distance between the points is 2p2 + 36>6.
TRY THIS. Find the distance between 1-3, -22 and 1-1, 42. ■ Note that the distance between two points is the same regardless of which point is chosen as 1x1, y12 or 1x2, y22.
The Midpoint Formula
When you average two test scores (by finding their sum and dividing by 2), you are finding a number midway between the two scores. Likewise, the midpoint of the line segment with endpoints -1 and 7 in Fig. 1.11 is 1-1 + 72>2 or 3. In gen- eral, 1a+ b2>2 is the midpoint of the line segment with endpoints a and b shown in Fig. 1.12.
Figure 1.12
a a + b—— b
2 Midpoint
⎫ ⎪ ⎪ ⎪⎪⎪ ⎬ ⎪ ⎪⎪⎪ ⎪⎪ ⎭
⎫ ⎪⎪⎪⎪ ⎪ ⎪⎬ ⎪⎪ ⎪ ⎪⎪ ⎭
3 4 5 6 7
2 1 0 –1
4 4
Midpoint
⎪ ⎪ ⎪ ⎪⎪⎪⎪⎪ ⎪⎪
Figure 1.11
Theorem:
Midpoint on a Number Line If a and b are real numbers, then a+2 b is midway between them on the num- ber line.
PROOF The distance between two numbers on the number line is the absolute value of their difference. So
`a - a + b 2 ` = `2a
2 - a + b
2 ` = `a -b
2 ` = 0a- b0 2 and
`b - a +b 2 ` = `2b
2 - a +b
2 ` = `b -a
2 ` = 兩b -a兩
2 .
Since 0a -b0 = 0b -a0, the distances from a+2 b to a and from a +2 b to b are equal. Since a+2 b is equidistant from a and b, it must be between a and b. ■ We can find the midpoint of a line segment in the xy-plane in the same manner.
Theorem:
The Midpoint Formula
The midpoint of the line segment with endpoints 1x1, y12 and 1x2, y22 is ax1+ x2
2 , y1 +y2 2 b.
PROOF Start with a line segment with endpoints A1x1, y12 and B1x2, y22, as shown in Fig. 1.13. Let M be the midpoint of AB. Draw horizontal and vertical line segments that form three right triangles, as shown in Fig. 1.13. Since M is the mid- point of AB, the two small right triangles are congruent. So D is the midpoint of AC and E is the midpoint of BC. Since the midpoint on a number line is found by add- ing and dividing by two, the x-coordinate of D is x1 +2 x2 and the y-coordinate of E is y1 +2 y2. So the midpoint M is 1x1+2 x2, y1 +2 y22. ■
EXAMPLE 3 Finding the midpoint
Find the midpoint of the line segment with the given endpoints.
a. 15, -32, 1-1, -62 b. 1p>2, 02, 1p>3, 12
Solution
a. To get the midpoint, add the corresponding coordinates and divide by 2:
a5 + 1-12
2 , -3+ 1-62
2 b = a4
2, -9
2 b = a2, - 9 2b The midpoint is 12, -9>22.
b. To get the midpoint, add the corresponding coordinates and divide by 2:
£ p2 + p3
2 , 0+ 1 2 ≥ = £
3p6 + 2p6 2 , 1
2≥ = £ 5p6
2, 1
2≥ = a5p 12, 1
2b The midpoint is 15p>12, 1>22.
TRY THIS. Find the midpoint of the line segment with endpoints 14, -12 and
13, 1>22. ■
EXAMPLE 4 Using the midpoint formula
Prove that the diagonals of the parallelogram with vertices 10, 02, 11, 32, 15, 02, and 16, 32 bisect each other.
Solution
The parallelogram is shown in Fig. 1.14. The midpoint of the diagonal from 10, 02 to 16, 32 is
a0 +6 2 , 0+ 3
2 b,
or 13, 1.52. The midpoint of the diagonal from 11, 32 to 15, 02 is a1 +5
2 , 3+ 0 2 b,
or 13, 1.52. Since the diagonals have the same midpoint, they bisect each other.
TRY THIS. Prove that the diagonals of the square with vertices 10, 02, 14, 12,
13, 52, and 1-1, 42 bisect each other. ■
The Circle
An ordered pair is a solution to or satisfies an equation in two variables if the equa- tion is correct when the variables are replaced by the coordinates of the ordered pair.
For example, 13, 112 satisfies y = 2x + 5 because 11 = 2132+ 5 is correct. The Figure 1.13
x y
B(x2, y2)
E M
C(x2, y1) A(x1, y1)
D
Figure 1.14 y
x –1
1 2 3
1
4 6
2 3
(5, 0) (1, 3)
(0, 0)
(6, 3)
solution set to an equation in two variables is the set of all ordered pairs that satisfy the equation. The graph of (the solution set to) an equation is a geometric object that gives us a visual image of an algebraic object. Circles provide a nice example of this relationship between algebra and geometry.
A circle is the set of all points in a plane that lie a fixed distance from a given point in the plane. The fixed distance is called the radius, and the given point is the center. The distance formula can be used to write an equation for the circle shown in Fig. 1.15 with center 1h, k2 and radius r for r 7 0. A point 1x, y2 is on the circle if and only if it satisfies the equation
21x- h22+ 1y -k22 =r.
Since both sides of 21x - h22 +1y - k22= r are positive, we can square each side to get the following standard form for the equation of a circle.
Figure 1.15
Theorem: Equation for a Circle in Standard Form
The equation for a circle with center 1h, k2 and radius r for r 7 0 is 1x - h22 +1y - k22 = r2.
A circle centered at the origin has equation x2+ y2= r2. y
x (x, y)
r (h, k)
Note that squaring both sides of an equation produces an equivalent equation only when both sides are positive. If we square both sides of 1x = -3, we get x = 9. But 19 ⬆ -3 since the square root symbol always represents the nonnega- tive square root. Raising both sides of an equation to a power is discussed further in Section 3.4.
EXAMPLE 5 Graphing a circle
Sketch the graph of the equation 1x - 122 + 1y + 222 = 9.
Solution
To determine the center and radius of this circle, compare its equation to the stan- dard form:
1x - h22 + 1y- k22= r2 1x - 122 + 1y+ 222= 9 1x- 122+ 1y -1-2222 =32
It is clear that h= 1, but you must rewrite (or at least think of) y+ 2 as y - 1-22 to determine that k= -2. So the center of the circle is 11, -22 and the radius is 3.
You can draw the circle as in Fig. 1.16 with a compass or computer. To draw a circle by hand, locate the points that lie 3 units above, below, right, and left of the center, as shown in Fig. 1.16. Then sketch a circle through these points.
To support these results with a graphing calculator you must first solve the equation for y:
1x - 122 + 1 y + 222 = 9
1 y + 222 = 9- 1x- 122 y +2 = {29 - 1x- 122
y = -2 { 29- 1x -122
Now enter y1 and y2 as in Fig. 1.17(a) on the next page. Set the viewing window as in Fig. 1.17(b). The graph in Fig. 1.17(c) supports our previous conclusion. A circle looks round only if the same unit distance is used on both axes. Some calculators Figure 1.16
(x – 1)2 + (y + 2)2 = 9 y
–1 x –1
–3 –2
–4
1 2 3
–2 1
4
–5
(1, –2)
automatically draw the graph with the same unit distance on both axes given the cor- rect command (Zsquare on a TI-83).
TRY THIS. Sketch the graph of 1x+ 222+ 1y -422= 25. ■ Note that an equation such as 1x -122 +1y + 222 = -9 is not satisfied by any pair of real numbers, because the left-hand side is a nonnegative real number, while the right-hand side is negative. The equation 1x - 122 +1y + 222 = 0 is satisfied only by 11, -22. Since only one point satisfies 1x -122+ 1y +222= 0, its graph is sometimes called a degenerate circle with radius zero. We study circles again later in this text when we study the conic sections.
In the next example we start with a description of a circle and write its equation.
EXAMPLE 6 Writing an equation of a circle
Write the standard equation for the circle with the center 1-3, 52 and passing through 11, 82 as shown in Fig. 1.18.
Solution
The radius of this circle is the distance between 1-3, 52 and 11, 82:
r = 211- 1-3222+ 18 -522= 216 + 9 =5
Now use h= -3, k = 5, and r = 5 in the standard equation of the circle 1x -h22+ 1y - k22 = r2.
1x - 1-3222 +1y - 522 = 52 So the equation of the circle is 1x + 322 +1y - 522 = 25.
TRY THIS. Write the standard equation for the circle with center 12, -12 and
passing through 13, 62. ■
A circle in standard form could be rewritten as follows:
1x + 322 +1y - 522 =4 Standard form for a circle x2 +6x +9 + y2 - 10y +25 =4 Square the binomials.
x2 + 6x + y2 - 10y = -30
To find the center and radius of the circle that is given by the last equation, we go back to standard form by completing the square.
Completing the square means finding the third term of a perfect square trino- mial when given the first two. That is, if we start with x2+ bx, then what third term will make a perfect square trinomial? Since
ax + b
2b2 = x2+ 2#b
2#x + ab
2b2 =x2 + bx + ab 2b2
adding 1b222 to x2 + bx completes the square. For example, the perfect square tri- nomial that starts with x2 + 6x is x2+ 6x+ 9. Note that 9 can be found by taking one-half of 6 and squaring.
Figure 1.17
⫺2 4
⫺5 1
y2
y1
(a) (b) (c)
Figure 1.18 y
– 2 x – 2 – 4 2
4 2 8 10
6 (1, 8) (–3, 5)
– 8 – 6 – 10
Rule for Completing the Square of x2 ⴙ bx ⴙ ?
The last term of a perfect square trinomial 1with a =12 is the square of one- half of the coefficient of the middle term. In symbols, the perfect square trino- mial whose first two terms are x2+ bx is
x2 +bx + ab 2b2.
EXAMPLE 7 Changing an equation of a circle to standard form
Graph the equation x2+ 6x+ y2- 5y = - 14. Solution
Complete the square for both x and y to get the standard form.
x2+ 6x +9 + y2 - 5y + 25 4 = - 1
4 + 9+ 25
4 112#622=9, 112#522=254 1x + 322 + ay - 5
2b2= 15 Factor the trinomials on the left side.
The graph is a circle with center 1-3, 522 and radius 115. See Fig. 1.19.
TRY THIS. Graph the equation x2 +3x +y2 -2y =0. ■
Figure 1.19 y
–1 x –3
– 4 – 1
– 2 1 2 –2
2 1 4 5
3
–6 –7 –8
6 7
5– (x + 3)2 + y –⎛ 2 = 15
⎝ ⎛
⎝
2
H I S T O R I C A L N O T E
Euclid of Alexandria (325–265 B.C.) was a Greek mathematician who lived in Alexandria, Egypt, during the reign of Ptolemy I (323 B.C.–283 B.C.). He is considered to be the “father of geome- try.” His most popular work, Elements, is one of the most successful text- books in the history of mathematics.
Within it, the properties of geometrical objects are deduced from a small set of axioms, thereby founding the axi- omatic method of mathematics.
Whether an equation of the form x2 +y2 +Ax + By =C is a circle depends on the values of A, B, and C. We can always complete the squares for x 2+ Ax and y 2 + By as in Example 7. Since a circle must have a positive radius, this equation is the equation of a circle only if the right side of the equation turns out to be posi- tive after completing the squares. If the right side turns out to be zero, then only one point satisfies the equation. If the right side turns out to be negative, then no points satisfy the equation.
The Line
For the circle, we started with the geometric definition and developed the algebraic equation. We would like to do the same thing for lines, but geometric definitions of lines are rather vague. Euclid’s definition was “length without breadth.” A modern geometry textbook states that a line is “a straight set of points extending infinitely in both directions.” Another modern textbook defines lines algebraically as the graph
of an equation of the form Ax + By= C. So we will accept the following theorem without proof.
Theorem: Equation of a Line in Standard Form
If A, B, and C are real numbers, then the graph of the equation Ax+ By = C
is a straight line, provided A and B are not both zero. Every straight line in the coordinate plane has an equation in the form Ax + By= C, the standard form for the equation of a line.
An equation of the form Ax +By = C is called a linear equation in two vari- ables. The equations
2x + 3y= 5, x =4, and y = 5
are linear equations in standard form. An equation such as y =3x - 1 that can be rewritten in standard form is also called a linear equation.
There is only one line containing any two distinct points. So to graph a linear equation we simply find two points that satisfy the equation and draw a line through them. We often use the point where the line crosses the x-axis, the x-intercept, and the point where the line crosses the y-axis, the y-intercept. Since every point on the x-axis has y-coordinate 0, we find the x-intercept by replacing y with 0 and then solv- ing the equation for x. Since every point on the y-axis has x-coordinate 0, we find the y-intercept by replacing x with 0 and then solving for y. If the x- and y-intercepts are both at the origin, then you must find another point that satisfies the equation.
EXAMPLE 8 Graphing lines and showing the intercepts
Graph each equation. Be sure to find and show the intercepts.
a. 2x -3y =9 b. y =40 -x Solution
a. Since the y-coordinate of the x-intercept is 0, we replace y by 0 in the equation:
2x -3102 = 9 2x = 9 x = 4.5
To find the y-intercept, we replace x by 0 in the equation:
2102 - 3y = 9 -3y = 9 y = -3
The x-intercept is 14.5, 02 and the y-intercept is 10, -32. Locate the intercepts and draw the line as shown in Fig. 1.20. To check, locate a point such as 13, -12, which also satisfies the equation, and see if the line goes through it.
b. If x= 0, then y = 40 - 0= 40 and the y-intercept is 10, 402. If y =0, then 0 =40 -x or x =40. The x-intercept is 140, 02. Draw a line through these points as shown in Fig. 1.21. Check that 110, 302 and 120, 202 also satisfy y = 40 - x and the line goes through these points.
The calculator graph shown in Fig. 1.22 is consistent with the graph in Fig. 1.21. Note that the viewing window is set to show the intercepts.
TRY THIS. Graph 2x + 5y = 10 and determine the intercepts. ■ Figure 1.20
y-intercept
x-intercept
(0, – 3) (4.5, 0)
(3, –1) y
–1 x –3 – 4
–5 – 1
– 2
– 4
1 2 3
–2 2 1
4 5
– 5 – 6
2x – 3y = 9
Figure 1.21 y
10 x 20 10 40 30
20 30 40 50 (0, 40)
(10, 30) (20, 20)
60 50
–10 –10
y = 40 – x (40, 0)
Figure 1.22
⫺10 60
⫺20 60
EXAMPLE 9 Graphing horizontal and vertical lines
Sketch the graph of each equation in the rectangular coordinate system.
a. y =3 b. x = 4 Solution
a. The equation y = 3 is equivalent to 0#x + y =3. Because x is multiplied by 0, we can choose any value for x as long as we choose 3 for y. So ordered pairs such as 1-3, 32, 1-2, 32, and 14, 32 satisfy the equation y =3. The graph of y= 3 is the horizontal line shown in Fig. 1.23.
b. The equation x =4 is equivalent to x +0#y =4. Because y is multiplied by 0, we can choose any value for y as long as we choose 4 for x. So ordered pairs such as 14, -32, 14, 22, and 14, 52 satisfy the equation x = 4. The graph of x = 4 is the vertical line shown in Fig. 1.24.
Note that you cannot graph x =4 using the Y= key on your calculator.
You can graph it on a calculator using polar coordinates or parametric equations (discussed later in this text).
TRY THIS. Graph y = 5 in the rectangular coordinate system. ■
In the context of two variables the equation x = 4 has infinitely many solu- tions. Every ordered pair on the vertical line in Fig. 1.24 satisfies x =4. In the context of one variable, x = 4 has only one solution, 4.
Using a Graph to Solve an Equation
Graphing and solving equations go hand in hand. For example, the graph of y= 2x -6 in Fig. 1.25 has x-intercept (3, 0), because if x =3 then y= 0. Of course, 3 is also the solution to the corresponding equation 2x - 6= 0 (where y is replaced by 0). For this reason, the solution to an equation is also called a zero or root of the equation. Every x-intercept on a graph provides us a solution to the cor- responding equation. However, an x-intercept on a graph may not be easy to iden- tify. In the next example we see how a graphing calculator identifies an x-intercept and thus gives us the approximate solution to an equation.
EXAMPLE 10 Using a graph to solve an equation
Use a graphing calculator to solve 0.551x - 3.452 +13.98 = 0.
Solution
First graph y = 0.551x- 3.452 + 13.98 using a viewing window that shows the x-intercept as in Fig. 1.26(a). Next press ZERO or ROOT on the CALC menu. The calculator can find a zero between a left bound and a right bound, which you must Figure 1.23
y
– 1 x – 3 – 4
– 5 – 2 –1 1 2 3
2 1 4 5
4 5 (4, 3) (–3, 3)
(–2, 3)
y = 3
Figure 1.24 y
–1 x – 1 – 2 – 3 – 4
1 2 3
2 1 4 5
5 3
– 5
(4, 5)
(4, 2)
(4, – 3) x = 4
Figure 1.25 (3, 0) y
x –2
–4
2 –2
4 2
4
–8
(0, –6) Solution to
2x – 6 = 0 y = 2x – 6
(a) (b) (c) Figure 1.26
20
⫺30 10
⫺20 20
⫺30 10
⫺20 20
⫺30 10
⫺20
enter. The calculator also asks you to make a guess. See Fig. 1.26(b). The more ac- curate the guess, the faster the calculator will find the zero. The solution to the equa- tion rounded to two decimal places is -21.97. See Fig. 1.26(c). As always, consult your calculator manual if you are having difficulty.
TRY THIS. Use a graphing calculator to solve 0.341x -2.32 +4.5 = 0. ■
1. The point 12, -32 is in quadrant II.
2. The point 14, 02 is in quadrant I.
3. The distance between 1a, b2 and 1c, d2 is 21a -b22+ 1c -d22.
4. The equation 3x2 + y= 5 is a linear equation.
5. The solution to 7x -9 = 0 is the x-coordinate of the x-intercept of y = 7x - 9.
6. 272 +92 =7 + 9
7. The origin lies midway between 11, 32 and 1-1, -32.
8. The distance between 13, -72 and 13, 32 is 10.
9. The x-intercept for the graph of 3x - 2y= 7 is 17>3, 02.
10. The graph of 1x+ 222+ 1y -122= 5 is a circle centered at 1-2, 12 with radius 5.
FOR thought... True or False? Explain.
Fill in the blank.
1. If x and y are real numbers, then 1x, y2 is an pair of real numbers.
2. The first coordinate in 1x, y2 is the and the second coordinate is the .
3. Ordered pairs are graphed in the rectangular coordinate system or the coordinate system.
4. The intersection of the x-axis and the y-axis is the . 5. The set of all points in a plane that lie a fixed distance from a
given point in the plane is a .
6. Finding the third term of a perfect square trinomial when given the first two is .
7. An equation of the form Ax+By=C is a in two variables.
8. The point where a line crosses the y-axis is the .
In Exercises 9–18, for each point shown in the xy-plane, write the corresponding ordered pair and name the quadrant in which it lies or the axis on which it lies.