math iv circles and lines

8/11/2019 Math IV Circles and Lines

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Mathematics IV: Advanced Algebra

Chapter 2: Circles and Lines

Lesson 2.1: The Rectangular Coordinate System

Each of the pairs of numbers and is an example of an orderedpair; that is, a pair of numbers written within parentheses in which the order of the numbers is

important. The two numbers are the components of the ordered pair. an ordered pair is

graphed using two real number lines that intersect at right angles at the zero points, as shown

in Figure 2.1. The common zero point is called the origin. The horizontal line, the ,represents the second. The and the make up a rectangular coordinatesystem, or the . The axes form four quadrants, numbered I, II, III, and IV asshown in Figure 2.1. (A point on an axis is not considered to be in any of the four quadrants.)

Figure 2.1

We locate, orplot, the point on the coordinate system that corresponds to the ordered

pair by going one unit from zero to the right alone the axis, and then two units upparallel to the axis. The phrase the point corresponding to the ordered pair often isabbreviated the point . The numbers in an ordered pair are calledthe coordinatesof thecorresponding point. In particular, the first coordinate is referred to as the ,and the second coordinate is referred to as the . If the point is labeledas point , we often write . The set of all ordered pairs of real numbers is denoted by; that is * +.

The parentheses used to represent an ordered pair are also used to represent an open

interval (introduced in an earlier chapter). In general, there should be no confusion between

these symbols because the context of the discussion tells us whether we are discussing ordered

pair or open intervals.


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The French philosopher Ren Descartes (1595 1650) is the person usually credited

with the invention of the rectangular coordinate system. In honor of Descartes, the rectangular

coordinate system is often referred to as the Cartesian Coordinate System. As a philosopher,

Descartes is responsible for the statement I think, therefore I am. Until Descartes invented his

coordinate system in 1637, algebra and geometry were treated as separate subjects. The

rectangular (or Cartesian) coordinate system allows us to connect algebra and geometry byassociating geometric shapes with algebraic equations. In this sense, the Cartesian coordinate

system is being superimposed on the Euclidean plane so that the definitions, axioms and the

theorems of Euclidean geometry apply.

Figure 2.2

Suppose that we wish to find the distance || between two points and . The Pythagorean Theorem allows us to do this. See figure 2.2. Let be the point withcoordinates

. It is easy to see that |

|

|

| and |

|

|

|. ByPythagorean Theorem, we have

|| || || | | | |.

We have just proven the following theorem.

THEOREM 2.1.1 DISTANCE FORMULA

The distance between the points and is given by||

EXAMPLE 2.1.2 Determine whether the points form an isoscelestriangle.

EXAMPLE 2.1.3 Prove that the points are vertices of a righttriangle.

EXAMPLE 2.1.4 Show that the points are collinear.


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The midpointof a line segment is the point on the segment that is equidistant from

both endpoints. Given the coordinates of two endpoints of a line segment, it is not difficult to

find the coordinates of the midpoint of the segment.

THEOREM 2.1.5 MIDPOINT FORMULA

The midpoint of the line segment with endpoints and is , where and

Proof: See Figure 2.3. Let be the midpoint, the midpoint with coordinates , thepoint with coordinates , and the point with coordinates . Since line isparallel to line and is the midpoint of the line segment , is the midpoint ofthe line segment. Thus, we have . Similarly, can be computed.

Figure 2.3

One can also prove Theorem 2.1.5 by noting that, by ASA Congruence Theorem,

.

EXAMPLE 2.1.6One endpoint of a line segment is and its midpoint is . Find theother endpoint.

EXAMPLE 2.1.7 Find the coordinates of the points that divide the line segment joining the

points and into four equal parts.

EXAMPLE 2.1.8The line segment is extended an equal length in both directions so thatthe length of the resulting segment is three times that of the original length. Find thecoordinates of the endpoints and of the extended line segment if and .

EXAMPLE 2.1.9If the point divides the line segment from to such that||||

, find the coordinates of point .


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EXAMPLE 2.1.10Prove that the diagonals of a parallelogram bisect each other.


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EXERCISES 2.1

In exercises 1 to 2, determine in which quadrant or on what coordinates axis each point lies.

Describes the distance of each point from the and the .

1.

2.

In exercises 3 to 14, find the distance between and and find the midpoint of the linesegment fromto .3.and 4. and 5.and 6. and

7. and

8. and

9. and 10. and 11. and 12. and 13. and 14.and

15. Find the lengths of the medians of the triangle having vertices and.

16. Find the length of the medians of the triangle having the vertices and.

17. Determine whether the following points are vertices of a right triangle: and

18. Determine whether the following points are vertices of a right triangle: and

19. Determine whether is on the line segment joiningand .20. Determine whether is on the line segment joining and21. Determine whether the points and are collinear.22. Determine whether the points and are collinear.23. If the point is equidistant from and , find .24. Find the point on the equidistant from and .25. Find the point on the equidistant from and .26. If the point is equidistant from and (, find .27. One endpoint of a line segment is and its midpoint is . Find the other point.


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28. One endpoint of a line segment is and its midpoint is . Find the other

endpoint.

29. Show that the triangle with vertices and is isosceles. Find itsarea.

30. Show that the triangle with points and are verticesof a square. Find its area.

31. Without using the distance formula, find the points and that divide into congruent segments if the points and are and .

32. Find the coordinates of a point which is on the line segment joining the points and and is three times as far from as it is from .

33. Find the point(s) on the line through and such that .34. Let be a point on the line passing through and but not on the segment

. Find the coordinates of if it is twice as far from as from.35. Given and , find the point of the way fromto .

36. Givenand , find the point of the way fromto .

37. If the point divides the line segment fromto such that is thepoint

of the way fromto , find the coordinates of the point

38. If the point divides the line segment from to such that | |

| |

,

find the coordinates of point .


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Lesson 2.2 Circles

An application of the Distance Formula (Theorem 2.1.1) leads to one of the most

familiar shapes in geometry. Acircleis the set of all points in a plane that lie a fixed distance

from a fixed point. The fixed point is called the center, and the fixed distance is called the

radius.

Figure 2.5

Let be a point on a circle with radius and center at . See figure 2.5. Then,by the Distance Formula, we have .Squaring both sides gives the following theorem.

THEOREM 2.2.1

The equation of a circle of radius with center at is

In particular, a circle of radius with center at the origin has equation

The first equation in Theorem 2.2.1 is the centerradius formor the standard formof

the equation of a circle.

A diameter of a circle is a line segment joining two points of the circle and passing

through the center.

EXAMPLE 2.2.2 Fine an equation of a circle whose diameter has endpoints at and.Solution: The center of the circle is the midpoint of the given diameter. If

is the

center of the circle, then

and

.


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Thus, the center is at . The radius of the circle is equal to ||(or ||or

||).

||

Using Theorem 2.2.1, an equation of the circle is

()Or, simply,

The equation in Example 2.2.2 can also be written in another form by expanding the

squared expressions:

The last equation is the so called general form of the equation of a circle. Every

equation of a circle can be written in the general form Where are constants, . Note that we can assume that

EXAMPLE 2.2.3 Determine the center and the radius of the circle whose equation is

EXAMPLE 2.2.4 Find all values of such that the circle whose equation is is tangent to the .

Every circle has an equation in general form. However, the graph of an equation of theform

Where are constants, is not necessarily a circle. We can explain this statement bycompleting the squares, and writing the above equation in the form

Here, the value of will either be positive, zero, or negative.

(i) The graph is a circle if and only if (ii) The graph is a single point if and only if (iii) The graph is an empty set if and only if

EXAMPLE 2.2.5Determine whether the graph of the following equation is a circle, a point, or an

empty set:

EXAMPLE 2.2.6Find the values of so that the graph of the equation

Is a circle, a point, or an empty set.


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EXERCISES 2.2

In exercises 16, find the center and the radius of the circle. Then sketch a graph of the circle.

1. 2.

3.

4.

5. 6.

In exercises 7 12, find the equation of the circle with center at and radius . Write theequation in both centerradius and general forms.

7. 8.

9. ( )

10. 11. 12.

In exercises 1318, find the center and the radius of the circle.

13. 14. 15.

16. 17. 18.

In exercises 19 23, determine whether the graph is a circle, a point, or the empty set. It is a

circle; find its center and radius.

19.

20. 21.

22.

23.

In exercises 2432, find the equation of the circle(s) satisfying the given conditions.

24. center at and passes through the point .25. center at and passes through the point .26. having a diameter whose endpoints are .27. having a diameter whose endpoints are .28. center is at the midpoint of the segment joining the center and one of the points

on the circle is

29. passing through with being the pointof the way from the center

.


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30. through the points 31. through the points 32. radius is and passing through the points 33. The circle passes through the point . Find the value of , the

radius, and the center of the circle.

34. For what values of and will be the equation of circle havingcenter at ? Find the radius of this circle.

35. Determine the value(s) of so that the graph of

(a) is a circle.

(b) is a point.

(c) is an empty set.


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Lesson 2.3 Lines

Two points determine a line. This familiar axiom from high school geometry tells us

that all we need in order to draw a line are two distinct points.

A line also can be determined by a point on the line and some measure of the

steepness of the line. This method can determine a line because, from the given point, themeasure of steepness can determine another point on the line. One way to get a measure of

the steepness of a line is to compare the vertical change in the line (the rise) to the horizontal

change (the run) while moving along the line from one fixed point to another.

Figure 2.7

Suppose that and are two different points on a line . Let be thepoint with coordinates . See Figure 2.7. Then, going along from to , the valuechanges from to , the value of changes from to by the amount . Let

We show that the value of is not affected by the choice of points on . Let ( )and( )be two other (different) points on . the point with coordinates ( ), andcompute

Since , we have

||||

||

||

Thus, the value of is the same number no matter what two points on are selected. Thisconstant is called the slopeof the line.


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DEFINITION 2.3.1

Consider a line that is not parallel to the . Let and be twodistinct points on the line. The slopeof the line is given by

The slope of a line parallel to the (that is, a vertical line) is not defined.

Figure 2.8

From Definition 2.3.1, a line with a positive slope goes up (rises) from left to right, while

a line with negative slope goes down (falls) from left to right. A horizontal line has zero slope.

Figure 2.8 shows lines of positive, zero, negative, and undefined slope.

EXAMPLE 2.3.2

The slope of a line segment is and one endpoint is . If the other endpoint is on

the what are its coordinates?EXAMPLE 2.3.3

Find the value(s) of so that the points are collinear.

By an equation of a linewe mean an equation that is satisfied by those, and only those, points

on the line.

To find an equation of the non vertical line that passes through a certain point

and has a certain slope , consider a point on the line, . The slopedetermined by and must be the same as the prescribed slope ; that is

Clearing the denominator provides an equation in and , which is called the point slopeform of an equation of the line. The line through of slope is described by theequation


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If two points on a line are known, it is possible to find an equation of the line. First, find

the slope using the slope formula, and then use the slope and one of the given points in the

pointslope form.

EXAMPLE 2.3.4

Find an equation of the line passing through the points and EXAMPLE 2.3.5

Find an equation of the line passing through the midpoint of the line segment joining

the pointsand , and whose slope is half the slope of the line through and .(Express your answer in the form , where are integers with .)

If, in the point slope form, we choose the particular point (that is, the pointwhere the line intersects the ) for the point , we have

or .The number , the coordinate of the point where the line intersects the , is the of the line. Consequently, the equation is called theslope intercept formof an equation of the line. This form is especially useful because

(1) it enables us to find the slope of a line from its equation, and

(2) it expresses the - coordinate of a point on the line explicitly in terms of itscoordinate.

Every equation of a nonvertical line has a unique (one and only one) slopeintercept form.

EXAMPLE 2.3.6

Find the slope of the line having the equation .We similarly define the of a (nonhorizontal) line as the - coordinate

of the point where the line intersects the .

EXAMPLE 2.3.7

The - intercept of a line is three times its intercept. With the coordinate axes, theline forms a triangle of area two in the first quadrant. Find an equation of the line in slope

intercept form.

EXAMPLE 2.3.8

Determine the value(s) of in the equation so that this line will form aright triangle with the coordinate axes whose area is square units.

The horizontal line that intersects the - axis at the point ahs slope .Therefore, from the slopeintercept form, an equation of this line is . However, becausethe slope of a vertical line is not defined, we cannot apply the point slope form to obtain its

equation. The vertical line that intersects the - axis at the point contains those and only


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those points having the same coordinate. Thus, is any point on this vertical line ifand only if .

If a line is not vertical, then it has an equation of the form , which can bewritten as . If a line is vertical, then it has an equation of the form , whichcan also be written as . Each of the above equations is a special case of an equationof the form , where are constants, and are not bothzero. It implies that every line has an equation of the form . The followingtheorem tells us that the converse of this fact is also true.

THEOREM 2.3.9

The equation where are constants, and are notboth zero, is an equation of a line.

Proof: We consider two cases: and .CASE 1: Suppose

, then

. The equation becomes

, which

gives

,

an equation of a vertical line with - intercept .CASE 2: Suppose , then the equation can be written into

which is an equation of a line in slopeintercept form.

Therefore, in either cases, the equation gives an equation of a line.

The equation given in Theorem 2.3.9 is the general formof an equation of a line.

We now discuss two applications of the concept of slope in geometry: parallel lines and

perpendicular lines.

In Euclidean geometry, we say that two lines are parallel if they never intersect. Two

distinct vertical lines are always parallel to each other. The following theorem tells us when two

distinct nonvertical lines are parallel.

THEOREM 2.3.10

Let be two distinct non vertical lines with slopes , respectively.Then is parallel to if and only if .

Proof: Let the respective equations of and be and . Let and . Refer to Figure 2.9.


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Since and are non vertical, they intersect with vertical line . Let the line intersect at and at .

The two lines are parallel if and only if

or, simply

Thus, are parallel if and only if .

Figure 2.9

In Euclidean geometry, two lines are perpendicular if they meet at a right angle. A

vertical line and horizontal line are perpendicular to each other. The following theorem tells us

when two distinct nonvertical lines are perpendicular.

THEOREM 2.3.11

Let be two distinct non vertical lines with slopes , respectively.Then is perpendicular to if and only if

Figure 2.10


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Proof: The two lines must intersect; otherwise, they are parallel and .See Figure 2.10.

We can choose the coordinate axes so that intersect at the origin . Then has an equation of the form , and has .

Since are non vertical, they intersect with the vertical line . Let the intersect at and at .

The lines are perpendicular if and only if is rightangled at Thatis, by the Pythagorean Theorem, are perpendicular if and only if

|| || ||.Applying the Distance Formula (Theorem 2.1.1), we get

||

||

And

||

Therefore, are perpendicular if and only if

Or, simply, .

Because is equivalent to

and

,

Theorem 2.3.11 states that two non vertical lines are perpendicular if and only if the slope of

one of them is the negative reciprocal of the slope of the other.

EXAMPLE 2.3.12Given the line having the equation , determine the general form of

an equation of the line through and(a) parallel to , and(b) perpendicular to .


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EXAMPLE 2.3.13

Find the point in the third quadrant so that the points andare vertices of a parallelogram.

A tangent lineto a circle is a line that intersects the circle at exactly one point, called the

point of tangency. The radius drawn to the point of tangency is perpendicular to the tangentline. If a line is tangent to a circle, we sometimes say that the circle is tangent to the line.

EXAMPLE 2.3.14

Find the equations (in general form) of the tangent lines to the circle

at the points of the circle on the

EXAMPLE 2.3.15

Find the equation (in general form) of the circle tangent to the line atand whose center in on the .

Before we proceed to the succeeding examples, we need to recall some concepts and

results about circles from high school geometry.

1. A chord of a circle is a line segment joining any two points of the circle. Thus, a

diameter is an example of a chord.

2. Aperpendicular bisectorof a chord is the line perpendicular to the chord and passing

through the midpoint of the chord. The perpendicular bisector of a chord passes

through the center of the circle.

3. The line through the midpoint of a chord and the center of the circle is perpendicular

to the chord.

4. The line perpendicular to a chord and passing through the center bisects the chord.

5. Three non collinear points determine exactly one circle.

EXAMPLE 2.3.16

Find an equation of the circle containing the point and tangent to the line at .

EXAMPLE 2.3.17

Write an equation of the circle (in general form) containing the points

and .


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EXERCISES 2.3

In exercises 1 to 14, find the slope of the line passing through the given points.

1. 2. 3. 4. 5. 6. 7. 8.

9. ()

10.

11.

12.

13. 14.

In exercises 15 to 53, find the equation(s) of the lines satisfying the given conditions.

15. slope is and the through the point .

16. slope is and the through the point .

17. slope isand the through the point

.

18. slope is and through the point .19. slope is and through the point .

20. slope is and through the point

.

21. slope is undefined and through the point ().

22. slope is undefined through the point .23. passing through the points and .24. passing through the points and .

25. passing through the points and

.

26. passing through the points

and

.

27. passing through the point and parallel to the .28. passing through the point and parallel to the .29. passing through the point and perpendicular to the .30. passing through the point and perpendicular to the 31. passing through the point and parallel to the line


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32. passing through the point and parallel to the line .33. passing through the point and perpendicular to the line .34. passing through the point and perpendicular to the line 35. passing through the point

and perpendicular to the line containing

and

36. passing through the point and perpendicular to the line containing and

.37. passing through the point and parallel to the line containing and 38. passing through the point and parallel to the line containing and 39. perpendicular to and containing the midpoint of the segment joining and .40. perpendicular to and containing the midpoint of the segment joining and

41. perpendicular to the segment joining and , and containing the point ofthe way from .

42. all points on the line are equidistant from and .43. if is any point on the line, then the distance from to is equal to the distance

from to .44. having twice the and half the of the line .45. having the same as and twice the slope of .46. through the points of intersection of and 47. through the points of intersection of and .48. containing segment where the of the point is and the distance

from to is 49. containing segment where the of the point is and the distance

from to is .50. tangent to the circle at the point .51. tangent to the circle at the point .52. tangent to the circle .

52. tangent to the circle at the point .53. tangent to the circle at the point .


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In exercises 54 to 68, find the equation(s) of the circle(s) satisfying the given conditions. Write

your answers in centerradius form.

54. center at and tangent to the line .

55. center at and tangent to the line

.

56. center at and tangent to the line .57. center at and tangent to the line .58. passes through the points and , and whose center is on the line 59. passes through the points and and whose center is on the line .60. tangent to the and axes, and the center is units away from the origin and lies on

the 4th

quadrant.

61. tangent to the and axes, and the center is units away from the origin and likeson the 1

st

quadrant.62. tangent to the line at and passing through .63. tangent to the line at and passing through .64. containing the point and tangent to the line at the point .65. tangent to the line at the point , and the center is on the line

.66. tangent to the line at the point , and the center is on the line

.

67. tangent to the line at the point and also tangent to the line at the point .68. tangent to the line at the point and also tangent to the line

at the point .


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Review for Chapters 1 and 2

1. Solve the following

(a)

(b) | | | | (c) | |

2. Find an equation of the circle that passes through the points and , and has itscenter on the line .

3. Find an equation of the line that contains a diameter of the circle ,and which is parallel to the line .

4. Prove analytically (i. e. using coordinate geometry) that the diagonals of a rhombus and

perpendicular.

5. Let

and

be positive real numbers. Prove that

math iv circles and lines

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