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In This Chapter
3.1 The Rectangular Coordinate
System
3.2 Circles and Graphs
3.3 Equations of Lines
3.4 Variation
Chapter 3 Review Exercises
A Bit of History Every student of mathematics pays
the French mathematician René Descartes (1596–1650) hom- age whenever he or she sketches a graph. Descartes is consid- ered the inventor of analytic geometry, which is the melding of algebra and geometry—at the time thought to be completely unrelated fields of mathematics. In analytic geometry an equa- tion involving two variables could be interpreted as a graph in a two-dimensional coordinate system embedded in a plane. The rectangular or Cartesian coordinate system is named in his honor. The basic tenets of analytic geometry were set forth in La Géométrie, published in 1637. The invention of the Cartesian plane and rectangular coordinates contributed significantly to the subsequent development of calculus by its co-inventors Isaac Newton (1643–1727) and Gottfried Wilhelm Leibniz (1646–1716). René Descartes was also a scientist and wrote on optics, astronomy, and meteorology. But beyond his contributions to mathematics and science, Descartes is also remembered for his impact on philosophy. Indeed, he is often called the father of modern philosophy and his book Meditations on First Philosophy continues to be required reading to this day at some universities. His famous phrase cogito ergo sum (I think, there- fore I am) appears in his Discourse on the Method and Principles of Philosophy. Although he claimed to be a fervent Catholic, the Church was suspicious of Descartes’philosophy and writings on the soul, and placed all his works on the Index of Prohibited Books in 1693.
3 Rectangular Coordinate
System and Graphs
123
In Section 3.3 we will see that parallel lines have the same slope.
124 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
Introduction In Section 1.2 we saw that each real number can be associated with
exactly one point on the number, or coordinate, line. We now examine a correspondence between points in a plane and ordered pairs of real numbers.
The Coordinate Plane A rectangular coordinate system is formed by two per-
pendicular number lines that intersect at the point corresponding to the number 0 on each line. This point of intersection is called the origin and is denoted by the symbol O. The horizontal and vertical number lines are called the x -axis and the y -axis , respectively. These axes divide the plane into four regions, called quadrants , which are numbered as shown in FIGURE 3.1.1(a). As we can see in Figure 3.1.1(b), the scales on the x - and y -axes need not be the same. Throughout this text, if tick marks are not labeled on the coordinates axes, as in Figure 3.1.1(a), then you may assume that one tick corresponds to one unit. A plane containing a rectangular coordinate system is called an xy - plane , a coordinate plane , or simply 2-space.
The Rectangular Coordinate System
*This is the same notation used to denote an open interval. It should be clear from the context of the discussion whether we are considering a point ( a , b ) or an open interval ( a , b ).
II
(a) Four quadrants (b) Different scales on x - and y -axes
I
–100 100 III IV
Second quadrant
First quadrant
Third quadrant
Fourth quadrant
Origin
y
x O
y
x
2
4
FIGURE 3.1.1 Coordinate plane
The rectangular coordinate system and the coordinate plane are also called the Cartesian coordinate system and the Cartesian plane after the famous French mathematician and philosopher René Descartes (1596–1650).
Coordinates of a Point Let P represent a point in the coordinate plane. We asso-
ciate an ordered pair of real numbers with P by drawing a vertical line from P to the x -axis and a horizontal line from P to the y -axis. If the vertical line intersects the x -axis at the number a and the horizontal line intersects the y -axis at the number b , we asso- ciate the ordered pair of real numbers ( a , b ) with the point. Conversely, to each ordered pair ( a , b ) of real numbers, there corresponds a point P in the plane. This point lies at the intersection of the vertical line through a on the x -axis and the horizontal line pass- ing through b on the y -axis. Hereafter, we will refer to an ordered pair as a point and denote it by either P ( a , b ) or ( a , b ).* The number a is the x -coordinate of the point and the number b is the y -coordinate of the point and we say that P has coordinates ( a , b ). For example, the coordinates of the origin are (0, 0). See FIGURE 3.1.2.
126 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
We see from Figure 3.1.3 that for all points ( x , y ) in the second and fourth quad- rants. Hence we can represent the set of points for which by the shaded regions in FIGURE 3.1.6. The coordinate axes are shown as dashed lines to indicate that the points on these axes are not included in the solution set. (b) In Section 2.6 we saw that means that either Since x is not restricted in any way it can be any real number, and so the points ( x , y ) for which
can be represented by the two shaded regions in FIGURE 3.1.7. We use solid lines to represent the boundaries and of the region to indicate that the points on these boundaries are included in the solution set.
y 5 2 2 y 5 2
y $ 2 and 2, _x_ , or^ y^ # 22 and^ 2,^ _x_^ ,
(^0) y (^0) $ 2 y $ 2 or y # 22.
xy , 0
xy , 0
FIGURE 3.1.7 Region in the xy -plane satisfying condition in (b) of Example 3
x
y y ≥ 2 and
y ≤ –2 and
FIGURE 3.1.8 Distance between points P 1 and P 2
P 1 ( x 1 , y 1 )
P 3 ( x 1 , y 2 )
| x 2 – x 1 |
| y 2 – y 1 |
P 2 ( x 2 , y 2 )
d
x
y
THEOREM 3.1.1 Distance Formula
The distance between any two points P 1 ( x 1 , y 1 ) and P 2 ( x 2 , y 2 ) in the xy -plane is given by
d ( P 1 , P 2 ) 5 "( x 2 2 x 1 ) 2 1 ( y 2 2 y 1 ) 2. (2)
Distance Formula Suppose P 1 ( x 1 , y 1 ) and P 2 ( x 2 , y 2 ) are two distinct points in the
xy -plane that are not on a vertical line or on a horizontal line. As a consequence, P 1 , P 2 , and P 3 ( x 1 , y 2 ) are vertices of a right triangle as shown in FIGURE 3.1.8. The length of the side P 3 P 2 is , and the length of the side P 1 P 3 is If we denote the length of P 1 P 2 by d , then (1)
by the Pythagorean theorem. Since the square of any real number is equal to the square of its absolute values, we can replace the absolute-value signs in (1) with parentheses. The distance formula given next follows immediately from (1).
d^2 5 0 x 2 2 x 1 02 1 0 y 2 2 y 1 02
(^0) x 2 2 x 1 0 0 y 2 2 y 1 0.
FIGURE 3.1.6 Region in the xy -plane satisfying condition in (a) of Example 3
y
x
xy < 0
xy < 0
Although we derived this equation for two points not on a vertical or horizontal line, (2) holds in these cases as well. Also, because it makes no difference which point is used first in the distance formula, that is, d ( P 1 , P 2 ) 5 d ( P 2 , P 1 ).
( x 2 2 x 1 ) 2 5 ( x 1 2 x 2 ) 2 ,
3.1 The Rectangular Coordinate System 127
Distance Between Two Points
Find the distance between the points and B (3, 7).
Solution From (2) with A and B playing the parts of P 1 and P 2 :
The distance d is illustrated in FIGURE 3.1.9.
Three Points Form a Triangle
Determine whether the points and P 3 (4, 5) are the vertices of a right triangle.
Solution From plane geometry we know that a triangle is a right triangle if and only if the sum of the squares of the lengths of two of its sides is equal to the square of the length of the remaining side. Now, from the distance formula (2), we have
Since
we conclude that P 1 , P 2 , and P 3 are the vertices of a right triangle with the right angle at P 3. See FIGURE 3.1.10.
Midpoint Formula In Section 1.2 we saw that the midpoint of a line segment
between two numbers a and b on the number line is the average, In the xy - plane, each coordinate of the midpoint M of a line segment joining two points P 1 ( x 1 , y 1 ) and P 2 ( x 2 , y 2 ), as shown in FIGURE 3.1.11, is the average of the corresponding coordinates of the endpoints of the intervals [ x 1 , x 2 ] and [ y 1 , y 2 ]. To prove this, we note in Figure 3.1.11 that triangles P 1 CM and MDP 2 are congru- ent since corresponding angles are equal and Hence, or Solving the last equation for y gives Similarly, so that and therefore
x 5 We summarize the result.
x 1 1 x 2 2
y 5 d ( C , M ) 5 d ( D , P 2 ) x 2 x 1 5 x 2 2 x
y 1 1 y 2 2
d ( P 1 , C ) 5 d ( M , D ) y 2 y 1 5 y 2 2 y.
d ( P 1 ,^ M )^5 d ( M ,^ P 2 ).
( a 1 b )/2.
[ d ( P 3 , P 1 )] 2 1 [ d ( P 2 , P 3 )] 2 5 25 1 100 5 125 5 [ d ( P 1 , P 2 )] 2 ,
d ( P 3 , P 1 ) 5 "(7 2 4) 2 1 (1 2 5) 2
d ( P 2 , P 3 ) 5 "(4 2 ( 2 4)) 2 1 (5 2 ( 2 1)) 2
d ( P 1 , P 2 ) 5 "( 24 2 7) 2 1 ( 21 2 1) 2
P 1 (7, 1), P 2 ( 2 4, 2 1),
EXAMPLE 5
d ( A , B ) 5 "(3 2 8) 2 1 (7 2 ( 2 5)) 2
A (8, 2 5)
EXAMPLE 4
FIGURE 3.1.10 Triangle in Example 5
P 3 (4, 5)
P 1 (7, 1)
P 2 (–4, – 1)
y
x
FIGURE 3.1.11 M is the midpoint of the line segment joining P 1 and P 2
y
C
x
P 2 ( x 2 , y 2 )
P 1 ( x 1 , y 1 )
M ( x , y )
D
THEOREM 3.1.2 Midpoint Formula
The coordinates of the midpoint of the line segment joining the points P 1 ( x 1 , y 1 ) and P 2 ( x 2 , y 2 ) in the xy -plane are given by
a (3)
x 1 1 x 2 2
y 1 1 y 2 2
b.
FIGURE 3.1.9 Distance between two points in Example 4
d ( A , B )
A (8, –5)
B (3, 7)
y
x
3.1 The Rectangular Coordinate System 129
In Problems 33–36, determine whether the points A , B , and C are vertices of a right triangle.
**33. 34.
- 36.** A (4, 0), B (1, 1), C (2, 3) 37. Determine whether the points A (0, 0), B (3, 4), and C (7, 7) are vertices of an isosceles triangle. 38. Find all points on the y -axis that are 5 units from the point (4, 4). 39. Consider the line segment joining and B (3, 4). (a) Find an equation that expresses the fact that a point P ( x , y ) is equidistant from A and from B. (b) Describe geometrically the set of points described by the equation in part (a). 40. Use the distance formula to determine whether the points , and C (4, 10) lie on a straight line. 41. Find all points with x -coordinate 6 such that the distance from each point to is 42. Which point, or (0.25, 0.97), is closer to the origin?
In Problems 43–48, find the midpoint of the line segment joining the points A and B.
**43. 44.
- 48.**
In Problems 49–52, find the point B if M is the midpoint of the line segment joining points A and B.
**49. 50.
53.** Find the distance from the midpoint of the line segment joining and B (3, 5) to the midpoint of the line segment joining C (4, 6) and . 54. Find all points on the x -axis that are 3 units from the midpoint of the line segment joining (5, 2) and 55. The x -axis is the perpendicular bisector of the line segment through A (2, 5) and B ( x , y ). Find x and y. 56. Consider the line segment joining the points A (0, 0) and B (6, 0). Find a point C ( x , y ) in the first quadrant such that A , B , and C are vertices of an equilateral triangle. 57. Find points P 1 ( x 1 , y 1 ), P 2 ( x 2 , y 2 ), and P 3 ( x 3 , y 3 ) on the line segment joining A (3, 6) and B (5, 8) that divide the line segment into four equal parts.
Miscellaneous Applications
58. Going to Chicago Kansas City and Chicago are not directly connected by an interstate highway, but each city is connected to St. Louis and Des Moines. See FIGURE 3.1.15. Des Moines is approximately 40 mi east and 180 mi north of Kansas City, St. Louis is approximately 230 mi east and 40 mi south of Kansas City, and Chicago is approximately 360 mi east and 200 mi north of Kansas City. Assume that this part of the Midwest is a flat plane and that the connecting highways are straight lines. Which route from Kansas City to Chicago, through St. Louis or through Des Moines, is shorter?
D ( 2 2, 2 10)
A ( 2 1, 3)
A (5, 8), M ( 2 1, 2 1) A ( 2 10, 2), M (5, 1)
A ( 2 2, 1), M ( 32 , 0) A (4, (^12) ), M (7, (^252) )
A (2 a , 3 b ), B (4 a , 26 b ) A ( x , x ), B ( 2 x , x 1 2)
A ( 2 1, 0), B ( 2 8, 5) A ( 12 , 232 ), B ( 252 , 1)
A (4, 1), B ( 2 2, 4) A ( 23 , 1), B ( 73 , 2 3)
(1/ !2, 1/ !2 )
A ( 2 1, 2 5), B (2, 4)
A ( 2 1, 2)
A (2, 8), B (0, 2 3), C (6, 5)
A (8, 1), B ( 2 3, 2 1), C (10, 5) A ( 2 2, 2 1), B (8, 2), C (1, 2 11)
y
x St. Louis
Kansas City
Des Moines (^) Chicago
FIGURE 3.1.15 Map for Problem 58
130 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
For Discussion
59. The points A (1, 0), B (5, 0), C (4, 6), and D (8, 6) are vertices of a parallelogram. Discuss: How can it be shown that the diagonals of the parallelogram bisect each other? Carry out your ideas. 60. The points A (0, 0), B ( a , 0), and C ( a , b ) are vertices of a right triangle. Discuss: How can it be shown that the midpoint of the hypotenuse is equidistant from the vertices? Carry out your ideas.
Introduction In Chapter 2 we studied equations as an equality of two algebraic
quantities involving one variable. Our goal then was to find the solution set of the equa- tion. In this and subsequent sections that follow we study equations in two variables , say x and y. Such an equation is simply a mathematical statement that asserts two quan- tities involving these variables are equal. In the fields of the physical sciences, engineering, and business, equations in two (or more) variables are a means of communication. For example, if a physicist wants to tell someone how far a rock dropped from a great height travels in a certain time t , he or she will write A mathematician will look at and immediately classify it as a certain type of equation. The classification of an equation carries with it information about properties shared by all equations of that kind. The remainder of this text is devoted to examining different kinds of equations involv- ing two or more variables and studying their properties. Here is a sample of some of the equations in two variables that you will see:
Terminology A solution of an equation in two variables x and y is an ordered pair
of numbers ( a , b ) that yields a true statement when and are substituted into the equation. For example, is a solution of the equation because
is a true statement. We also say that the coordinates satisfy the equation. As in Chapter 2, the set of all solutions of an equation is called its solution set. Two equations are said to be equivalent if they have the same solution set. For example, we will see in Example 4 of this section that the equation is equiv- alent to In the list given in (1), you might object that the first equation does not involve two variables. It is a matter of interpretation! Because there is no explicit y dependence in the equation, the solution set of can be interpreted to mean the set
The solutions of are then ordered pairs (1, y ), where you are free to choose y arbitrarily so long as it is a real number. For example, (1, 0) and (1, 3) are solutions of the equation The graph of an equation is the visual representation in the rectangular coordinate system of the set of points whose coordinates ( a , b ) satisfy the equation. The graph of x 5 1 is the vertical line shown in FIGURE 3.2.1.
x 5 1.
x 5 1
(^5) ( x , y ) 0 x 5 1, where y is any real number (^6).
x 5 1
x 5 1
( x 1 5) 2 1 ( y 2 1) 2 5 3 2.
x^2 1 y^2 1 10 x 2 2 y 1 17 5 0
y 5 4 T T x 5 2 2
( 2 2, 4) y 5 x^2
x 5 a y 5 b
y 5 2 x , y 5 ln x , y^2 5 x 2 1,
x^2 4
y^2 9
x^2 2 y^2 5 1.
x 5 1, x^2 1 y^2 5 1, y 5 x^2 , y 5! x , y 5 5 x 2 1, y 5 x^3 2 3 x ,
s 5 16 t^2
s 5 16 t^2.
Circles and Graphs
(1, 0)
x = 1
(1, 3)
y
x
FIGURE 3.2.1 Graph of equation x 5 1
132 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
Since the point P (1, 4) lies on the circle as shown in FIGURE 3.2.4, its coordinates must satisfy equation (5). That is,
Thus the required equation in standard form is
Completing the Square If the terms and are expanded and the
like terms grouped together, an equation of a circle in standard form can be written as
(6)
Of course in this last form the center and radius are not apparent. To reverse the process, in other words, to go from (6) to the standard form (2), we must complete the square in both x and y. Recall from Section 2.3 that adding to a quadratic expression such as yields which is the perfect square By rearranging the terms in (6),
and then adding and to both sides of the last equation
we obtain the standard form of the equation of a circle:
You should not memorize the last equation; we strongly recommend that you work through the process of completing the square each time.
Completing the Square
Find the center and radius of the circle whose equation is
(7)
Solution To find the center and radius we rewrite equation (7) in the standard form (2). First, we rearrange the terms,
Then, we complete the square in x and y by adding, in turn, in the first set of parentheses and in the second set of parentheses. Proceed carefully here because we must add these numbers to both sides of the equation:
From the last equation we see that the circle is centered at and has radius 3. See FIGURE 3.2.5.
( x 1 5) 2 1 ( y 2 1) 2 5 3 2.
( x^2 1 10 x 1 25) 1 ( y^2 2 2 y 1 1) 5 9
[ x^2 1 10 x 1 ( 102 ) 2 ] 1 [ y^2 2 2 y 1 (^222 ) 2 ] 5 2 17 1 ( 102 ) 2 1 (^222 ) 2
( x^2 1 10 x ) 1 ( y^2 2 2 y ) 5 217.
x^2 1 y^2 1 10 x 2 2 y 1 17 5 0.
EXAMPLE 4
a x 1
a 2
b
2 1 a y 1
b 2
b
2 5
( a^2 1 b^2 2 4 c ).
a x^2 1 ax 1 a
a 2
b
2 b 1 a y^2 1 by 1 a
b 2
b
2 b 5 a
a 2
b
2 1 a
b 2
b
2 2 c ,
( a /2) 2 ( b /2) 2
( x^2 1 ax ) 1 ( y^2 1 by ) 5 2 c ,
x^2 1 ax x^2 1 ax 1 ( a /2) 2 , ( x 1 a /2) 2.
( a /2) 2
x^2 1 y^2 1 ax 1 by 1 c 5 0.
( x 2 h ) 2 ( y 2 k ) 2
( x 2 4) 2 1 ( y 2 3) 2 5 10.
(1 2 4) 2 1 (4 2 3) 2 5 r^2 or 10 5 r^2.
The terms in color added inside the parenthe- ses on the left-hand side are also added to the right-hand side of the equality. This new equation is equivalent to (6).
P (1, 4)
C (4, 3)
y
x FIGURE 3.2.4 Circle in Example 3
(–5, 1)
y
x
3
FIGURE 3.2.5 Circle in Example 4
3.2 Circles and Graphs 133
It is possible that an expression for which we must complete the square has a lead- ing coefficient other than 1. For example,
Note
is an equation of a circle. As in Example 4 we start by rearranging the equation:
Now, however, we must do one extra step before attempting completion of the square; that is, we must divide both sides of the equation by 3 so that the coefficients of x^2 and y^2 are each 1:
.
At this point we can now add the appropriate numbers within each set of parentheses and to the right-hand side of the equality. You should verify that the resulting standard form is
Semicircles If we solve (3) for y we get or. This
last expression is equivalent to the two equations and In like manner if we solve (3) for x we obtain and By convention, the symbol denotes a nonnegative quantity, thus the y -values defined by an equation such as are nonnegative. The graphs of the four equations highlighted in color are, in turn, the upper half, lower half, right half, and the left half of the circle shown in Figure 3.2.3. Each graph in FIGURE 3.2.6 is called a semicircle.
y 5 " r^2 2 x^2
x 5 " r^2 2 y^2 x 5 2" r^2 2 y^2.
y 5 " r^2 2 x^2 y 5 2" r^2 2 x^2.
y^2 5 r^2 2 x^2 y 5 ; " r^2 2 x^2
( x 2 3) 2 1 ( y 1 1) 2 5 283.
( x^2 2 6 x ) 1 ( y^2 1 2 y ) 5 2
(3 x^2 2 18 x ) 1 (3 y^2 1 6 y ) 5 22.
3 x^2 1 3 y^2 2 18 x 1 6 y 1 2 5 0
c c
y
x
(0, r )
(0, – r )
(– r , 0) ( r , 0)
y = √ r^2 – x^2
y
x (– r , 0) ( r , 0)
y = – √ r^2 – x^2
(a) Upper half (b) Lower half
y
x
(0, r )
(0, – r )
( r , 0)
x = √ r^2 – y^2
(c) Right half
y
x
(0, r )
(0, – r )
(– r , 0)
x = – √ r^2 – y^2
(d) Left half FIGURE 3.2.6 Semicircles
3.2 Circles and Graphs 135
( b ) Setting yields This is a quadratic equation and can be solved either by factoring or by the quadratic formula. Factoring gives
and so and The x -intercepts of the graph are the points and Now, setting in the equation immediately gives The y -intercept of the graph is the point
Example 4 Revisited
Let’s return to the circle in Example 4 and determine its intercepts from equation (7). Setting y 0 in and using the quadratic formula to solve shows the x -intercepts of this circle are and If we let , then the quadratic formula shows that the roots of the equation are complex numbers. As seen in Figure 3.2.5, the circle does not cross the y -axis.
Symmetry A graph can also possess symmetry. You may already know that the graph
of the equation is called a parabola. FIGURE 3.2.8 shows that the graph of is symmetric with respect to the y -axis since the portion of the graph that lies in the sec- ond quadrant is the mirror image or reflection of that portion of the graph in the first quad- rant. In general, a graph is symmetric with respect to the y - axis if whenever ( x , y ) is a point on the graph, is also a point on the graph. Note in Figure 3.2.8 that the points (1, 1) and (2, 4) are on the graph. Because the graph possesses y -axis symmetry, the points and must also be on the graph. A graph is said to be sym- metric with respect to the x - axis if whenever ( x , y ) is a point on the graph, is also a point on the graph. Finally, a graph is symmetric with respect to the origin if whenever ( x , y ) is on the graph, is also a point on the graph. FIGURE 3.2. illustrates these three types of symmetries.
( 2 x , 2 y )
( x , 2 y )
( 2 x , y )
y 5 x^2 y 5 x^2
y^2 2 2 y 1 17 5 0
( 25 1 2 !2, 0). x 5 0
x^2 1 10 x 1 17 5 0 ( 25 2 2 !2, 0)
x^2 1 y^2 1 10 x 2 2 y 1 17 5 0
EXAMPLE 6
y 5 212. (0, 2 12).
( 32 , 0). x 5 0 y 5 2 x^2 1 5 x 2 12
x 5 2 4 x 5 32. ( 2 4, 0)
( x 1 4)(2 x 2 3) 5 0
y 5 0 2 x^2 1 5 x 2 12 5 0.
FIGURE 3.2.8 Graph with y -axis symmetry
FIGURE 3.2.9 Symmetries of a graph
Observe that the graph of the circle given in Figure 3.2.3 possesses all three of these symmetries. As a practical matter we would like to know whether a graph possesses any sym- metry in advance of plotting its graph. This can be done by applying the following tests to the equation that defines the graph.
(–2, 4) (2, 4)
(–1, 1) (1, 1)
y y^ =^ x^2
x
(– x , y ) ( x , y )
( x , y )
( x , – y )
( x , y )
(– x , – y )
(a) Symmetry with respect to the y -axis
(b) Symmetry with respect to the x -axis
(c) Symmetry with respect to the origin
y
x
y
x
y
x
136 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
The advantage of using symmetry in graphing should be apparent: If say, the graph of an equation is symmetric with respect to the x -axis, then we need only produce the graph for since points on the graph for are obtained by taking the mirror images, through the x -axis, of the points in the first and second quadrants.
Test for Symmetry
By replacing x by – x in the equation and using we see that
This proves what is apparent in Figure 3.2.8; that is, the graph of is symmetric with respect to the y -axis.
Intercepts and Symmetry
Determine the intercepts and any symmetry for the graph of (8)
Solution Intercepts : Setting in equation (8) immediately gives The graph of the equation has a single x -intercept, (10, 0). When we get , which implies that Thus there are two y -intercepts, and Symmetry : If we replace x by in the equation we get This is not equivalent to equation (8). You should also verify that replacing x and y by and in (8) does not yield an equivalent equation. However, if we replace y by , we find that
Thus, the graph of the equation is symmetric with respect to the x -axis. Graph : In the graph of the equation given in FIGURE 3.2.10 the intercepts are indi- cated and the x -axis symmetry should be apparent.
In Problems 1–6, find the center and the radius of the given circle. Sketch its graph.
**1. 2.
5.** (^) ( x (^2 12) ) 2 1 ( y (^2 32) ) 2 5 1 6. ( x 1 3) 2 1 ( y 2 5) 2 5 25
x^2 1 ( y 2 3) 2 5 49 ( x 1 2) 2 1 y^2 5
x^2 1 y^2 5 5 x^2 1 y^2 5
Exercises Answers to selected odd-numbered problems
begin on page ANS-5.
x 1 ( 2 y ) 2 5 10 is equivalent to x 1 y^2 5 10.
2 y
2 x 2 y
2 x x 1 y^2 5 10, 2 x 1 y^2 5 10.
(0, 2 !10) (0, !10 ).
y 5 2!10 or y 5 !10.
x 5 0, y^2 5
y 5 0 x 5 10.
x 1 y^2 5 10.
EXAMPLE 8
y 5 x^2
y 5 ( 2 x ) 2 is equivalent to y 5 x^2.
y 5 x^2 ( 2 x ) 2 5 x^2 ,
EXAMPLE 7
y $ 0 y , 0
THEOREM 3.2.1 Tests for Symmetry
The graph of an equation is symmetric with respect to the ( i ) y - axis if replacing x by results in an equivalent equation; ( ii ) x - axis if replacing y by results in an equivalent equation; ( iii ) origin if replacing x and y by 2 x and 2 y results in an equivalent equation.
2 y
2 x
FIGURE 3.2.10 Graph of equation in Example 8
(10, 0)
(0, √10)
(0, –√10)
y
x
138 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
In Problems 63–66, state all the symmetries of the given graph.
0 x 0 1 0 y 0 5 4 x 1 3 5 0 y 2 50
y 5 0 x 2 9 0 x 5 0 y 0 2 4
y 5! x 2 3 y 5 2 2! x 1 5
y 5
( x 1 2)( x 2 8) x 1 1
y 5
x^2 2 x 2 20 x 1 6
y
x
FIGURE 3.2.11 Graph for Problem 63
y
x
FIGURE 3.2.12 Graph for Problem 64
y
x
FIGURE 3.2.13 Graph for Problem 65
y
x
FIGURE 3.2.14 Graph for Problem 66
In Problems 67–72, use symmetry to complete the given graph.
67. The graph is symmetric 68. The graph is symmetric with respect to the y -axis. with respect to the x -axis. 69. The graph is symmetric 70. The graph is symmetric with respect to the origin. with respect to the y -axis.
y
x
FIGURE 3.2.15 Graph for Problem 67
y
x
FIGURE 3.2.16 Graph for Problem 68
y
x
FIGURE 3.2.17 Graph for Problem 69
x
y
FIGURE 3.2.18 Graph for Problem 70
3.3 Equations of Lines 139
71. The graph is symmetric 72. The graph is symmetric with respect to the x - and y -axes. with respect to the origin.
For Discussion
73. Determine whether the following statement is true or false. Defend your answer. If a graph has two of the three symmetries defined on page 135, then the graph must necessarily possess the third symmetry. 74. (a) The radius of the circle in FIGURE 3.2.21(a) is r. What is its equation in standard form? (b) The center of the circle in Figure 3.2.21(b) is ( h , k ). What is its equation in standard form? 75. Discuss whether the following statement is true or false. Every equation of the form x^2 1 y^2 1 ax 1 by 1 c 5 0 is a circle.
y
x
FIGURE 3.2.19 Graph for Problem 71
y
x
FIGURE 3.2.20 Graph for Problem 72
y
(a) (b)
x
y
x
FIGURE 3.2.21 Circles in Problem 74
Introduction Any pair of distinct points in the xy -plane determines a unique
straight line. Our goal in this section is to find equations of lines. Fundamental to find- ing equations of lines is the concept of slope of a line.
Slope If P 1 ( x 1 , y 1 ) and P 2 ( x 2 , y 2 ) are two points such that , then the number
is called the slope of the line determined by these two points. It is customary to call the change in y or rise of the line; is the change in x or the run of the line. Therefore, the slope (1) of a line is
See FIGURE 3.3.1(a). Any pair of distinct points on a line will determine the same slope. To see why this is so, consider the two similar right triangles in Figure 3.3.1(b). Since we know that the ratios of corresponding sides in similar triangles are equal we have
Hence the slope of a line is independent of the choice of points on the line.
y 2 2 y 1 x 2 2 x 15
y 4 2 y 3 x 4 2 x 3.
m 5
rise run
y 2 2 y 1 x 2 2 x 1
m 5
y 2 2 y 1 x 2 2 x 1
x 1 2 x 2
Equations of Lines
FIGURE 3.3.1 Slope of a line
y
x x 2
x 2 – x 1
y 2 – y 1
x 1
( x 2 , y 2 )
( x 1 , y 1 ) (^) Run
Rise
y
x
Run x 2 – x 1
Run x 4 – x 3
Rise y 2 – y 1
Rise y 4 – y 3
( x 2 , y 2 )
( x 1 , y 1 )
( x 4 , y 4 )
( x 3 , y 3 )
(a) Rise and run
(b) Similar triangles
3.3 Equations of Lines 141
Point-Slope Equation
Find an equation of the line with slope 6 and passing through
Solution Letting , and we obtain from (3)
Simplifying gives
Point-Slope Equation
Find an equation of the line passing through the points (4, 3) and Solution First we compute the slope of the line through the points. From (1),
The point-slope equation (3) then gives the distributive law T T
Slope-Intercept Equation Any line with slope (that is, any line that is not ver-
tical) must cross the y -axis. If this y -intercept is (0, b ), then with the point-slope form (3) gives The last equation simplifies to the next result.
y 2 b 5 m ( x 2 0).
x 1 5 0, y 1 5 b ,
y 2 3 5 2
( x 2 4) or y 5 2
x 1
m 5
EXAMPLE 3
y 2 2 5 6( x (^1 12) ) or y 5 6 x 1 5.
y 2 2 5 6[ x (^2) ( (^212) )].
m 5 6, x 1 5 2^12 y 1 5 2
(^212 , 2).
EXAMPLE 2
THEOREM 3.3.2 Slope-Intercept Equation
The slope-intercept equation of the line with slope m and y -intercept (0, b ) is
y 5 mx 1 b. (4)
FIGURE 3.3.4 Lines through the origin are y 5 mx
y
x
y = – x y^ =^ x y = x
y = –2 x y = –4 x y = 4 xy = 2 x
1 y = – 12 x 2
The distributive law
is the source of many errors on students’ papers. A common error goes something like this:
The correct result is:
5 2 2 x 1 3.
5 ( 2 1)2 x 2 ( 2 1)
2 (2 x 2 3) 5 ( 2 1)(2 x 2 3)
2 (2 x 2 3) 5 2 2 x 2 3.
a ( b 1 c ) 5 ab 1 ac
When in (4), the equation represents a family of lines that pass through the origin (0, 0). In FIGURE 3.3.4 we have drawn a few of the members of that family.
Example 3 Revisited
We can also use the slope-intercept form (4) to obtain an equation of the line through the two points in Example 3. As in that example, we start by finding the slope The equation of the line is then By substituting the coordinates of either point (4, 3) or into the last equation enables us to determine b. If we use and , then and so The equation of the line is
Horizontal and Vertical Lines We saw in Figure 3.3.2(c) that a horizontal line
has slope An equation of a horizontal line passing through a point ( a , b ) can be obtained from (3), that is, y 2 b 5 0( x 2 a )or y 5 b.
m 5 0.
y 5 2^13 x 1 133.
y 5 3 3 5 2^13 #^4 1 b b 5 3 1 43 5 133.
( 2 2, 5) x 5 4
y 5 2^13 x^^1 b.
m 5 2^13.
EXAMPLE 4
b 5 0 y 5 mx
142 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS
THEOREM 3.3.3 Equation of Horizontal Line
The equation of a horizontal line with y -intercept (0, b ) is y 5 b. (5)
A vertical line through ( a , b ) has undefined slope and all points on the line have the same x -coordinate. The equation of a vertical line is then
THEOREM 3.3.4 Equation of Vertical Line
The equation of a vertical line with x -intercept ( a , 0) is
x 5 a. (6)
Vertical and Horizontal Lines
Find equations for the vertical and horizontal lines through Graph these lines. Solution Any point on the vertical line through has x -coordinate 3. The equation of this line is then Similarly, any point on the horizontal line through has y -coordinate 1. The equation of this line is Both lines are graphed in FIGURE 3.3.5.
Linear Equation The equations (3), (4), (5), and (6) are special cases of the
general linear equation in two variables x and y (7)
where a and b are real constants and not both zero. The characteristic that gives (7) its name linear is that the variables x and y appear only to the first power. Observe that
d horizontal line
d vertical line
. d line with nonzero slope
Slope and y -intercept
Find the slope and the y -intercept of the line Solution We solve the linear equation for y :
Comparing the last equation with (4) we see that the slope of the line is and the y -intercept is
If the x - and y -intercepts are distinct, the graph of the line can be drawn through the corresponding points on the x - and y -axes.
(0,^57 ).^
m (^5 )
y 5
x 1
7 y 5 3 x 1 5
3 x 2 7 y 1 5 5 0
3 x 2 7 y 1 5 5 0.
EXAMPLE 6
a 2 0, b 2 0, gives y 5 2
a b
x 2
c b
a 2 0, b 5 0, gives x 5 2
c a ,
a 5 0, b 2 0, gives y 5 2
c b
ax 1 by 1 c 5 0,
(3, 2 1) y 5 21.
x 5 3.
EXAMPLE 5
FIGURE 3.3.5 Horizontal and vertical lines in Example 5
y
x
y = –
x = 3
(3, –1)
