geometric analytic h b phillips

8/12/2019 Geometric Analytic h b Phillips

1/218

'II

li mm

iiiiiili fiiilli 11IIIHi

jiiii^

i'h':--^;.tfi;:'-[M-'f: 'M i.T;f .TiLttMU- l


2/218

JItliaca, Hem ^otk

BOUGHT WITH THE INCOME OF THESAGE ENDOWMENT FUND

THE GIFT OFHENRY W. SAGE

1891


3/218

Cornell University Library

3 1924 031 220 795olin.anx


4/218

Cornell UniversityLibrary

The original of tliis book is intine Cornell University Library.

There are no known copyright restrictions inthe United States on the use of the text.

http://www.archive.org/details/cu31924031220795


5/218


6/218


7/218

ANALYTIC GEOMETEYBY

H. B. PHILLIPS, Ph.D.Assistant Professor of Mathematics in the MassachusettsInstitute of Technology.

FIRST EDITIONTOTAL ISSUE FIVE THOUSAND

NEW YOEKJOHN WILEY & SONS, Inc.London: CHAPMAN & HALL, Limited

19165


8/218

Copyright, 1916,BY

H. B. PHILLIPS

Stanbope iprcBSF. H. GILSOW COMPANYBOSTON, U.S.A.


9/218

PREFACEThe author of this text beUeves that the differential calculus

should be given to the student in college at the earliest possiblemoment, and that to accomplish this a short course ia analyticgeometry is essential. He has, therefore, written this text to supplya course that will equip the student for work in calculus and engi-neering without burdening him with a mass of detail useful only tothe student of mathematics for its own sake. The exercises areso numerous and varied that the teacher who desires to spend alonger time on analytic geometry can easily do so; and, indeed,if more than the briefest course is given, the best way to spend thetime is in working a large number of varied examples based upon thefew fundamental principles which occur constantly in practice.The author is indebted to Professor H. W. Tyler and Professor

E. B. Wilson for many suggestions and to Dr. Joseph Lipka for valu-able assistance both in the preparation of the manuscript and the re-vision of the proof.

H. B. PHILLIPSBoston, Mass.August, 1915.


10/218


11/218

CONTENTSCHAPTER 1

ALGEBRAIC PRINCIPLESArt. Page1. Constants and variables 12. Equations 33. Equations in one variable 54. Factors and roots ; 75. Approximate solution of equations 96. Inequalities 117. Simultaneous equations 128. Special cases 149. Undetermined coefficients 16

10. Functions 18CHAPTER 2

RECTANGULAR COORDINATES11. Definitions 2312. Segments 2613. Projection 2714. Distance between two points 2915. Vectors 3216. Multiple of a vector 3517. Addition and subtraction of vectors 3618. Slope of a line 3919. Graphs 4320. Equation of a locus 4721. Point on a locus 4922. Tangent curves 51

CHAPTER 3STRAIGHT LINE AND CIRCLE

23. Equation of a straight line 5424. First degree equation 5725. The expression Ax + By -\- C 60

V


12/218

vi ContentsAet. Page26. Distance from a point to a line 6127. Equation of a circle 6428. Circle determined by three conditions 66

CHAPTER 4SECOND DEGREE EQUATIONS

29. The ellipse 7030. The ellipse in other positions 7131. The parabola 7432. The hyperbola 7733. The rectangular hyperbola 8034. The second degree equation 8335. Locus problems 87

CHAPTER 5GRAPHS AND EMPIRICAL EQUATIONS

36. Intersections with the coordinate axes 9037. Real and imaginary coordinates 9238. Symmetry 9339. Infinite valjies 9540. Direction of the curve 9641. Sine curves 9942. Periodic functions 10243. Exponential and logarithmic curves 10444. Empirical equations 107

CHAPTER 6POLAR COORDINATES45. Definitions 11346. Change of coordinates 11647. Straight line and circle 11748. The conic 11749. Graphing equations 12150. Intersections of curves 12451. Locus problems 127

CHAPTER 7PARAMETRIC REPRESENTATION

52. Definition of parameter 13053. Locus of parametric equations 13254. Parametric from coordinate equations 13555. Locus problems 138


13/218

Contents viiCHAPTER 8

TRANSFORMATION OF COORDINATESAbt. Page56. Translation of the axes 14257. Rotation of the axes 14458. Invariants _. 14559. General equation of the second degree 146

CHAPTER 9COORDINATES OF A POINT IN SPACE

60. Rectangular coordinates 15061. Projection 15162. Distance between two points 15463. Vectors 15564. Direction of a line 15665. The angle between two directed lines 15866. CyUndrical and spherical coordinates 160

CHAPTER 10SURFACES67. Loci , 16368. Equation of a plane 16369. Equation of a sphere 16770. Equation of a cylindrical surface 16771. Surface of revolution 16872. Graph of an equation 170

CHAPTER 11LINES AND CURVES

73. The straight line 17674. Curves 17875. Parametric equations 180


14/218


15/218

ANALYTIC GEOMETRYCHAPTER 1

ALGEBRAIC PRINCIPLESArt. 1. Constants and Variables

In analytic geometry much use is made of algebra. Hence abrief review is here given of some algebraic principles and processesused in this book.

In a given investigation a quantity is constant if its value is thesame throughout that work, and variable if it may have differentvalues. It should be noted that a quantity that is constant in oneproblem may be variable in another. Thus, ia discussing a particu-lar circle the radius would be constant, but in a problem about acircular disk expanding under heat the radius would be variable.A quantity whose value is to be determined is often called anunknonm. Such a quantity may be either constant or variable.In some cases it is not even known in advance whether it is con-stant or variable.Real Numbers. The simplest constants are numbers. The proc-

ess of counting gives whole numbers. Division and subtraction givefractions and negative numbers. Whole numbers and fractions,whether positive or negative, are called rational numbers. Anumber, like V2, that can be expressed to any required degree ofaccuracy, but not exactly, by a fraction, is called irrational. Ra-tional and irrational numbers, whether positive or negative, arecalled real.The absolute value of a real number is the number without its

algebraic sign. The absolute value of is sometimes written \x\.Thus,

I-2] =,|-|-2l = 2.

1


16/218

2 Algebeaic Principles Chap. 1Graphical Representation. Real numbers can be represented

graphically by the points of a straight line. Upon any point of aline mark the number 0. Choose a unit of length.' On one side ofmark positive numbers, on the other negative numbers, making

the number at each point equal in absolute value to the distancefrom to the point. The result is a scale on the line. When theline is horizontal, as in Fig. 1, it is usual, but not necessary, to layoff the positive numbers on the right of 0, the negative numbers onthe left.

c A B-7 -6 -6 C-4 -3 -2 -1 01 3 30t 66 67 8

Fig. 1.

The point A representing the number a divides the scale into twoparts. On one side of A, called the positive, lie all numbers greaterthan a; on the other side, called the negative, he all numbers lessthan a. At a point B on the positive side of A is located a numberb greater than a, at a point C on the negative side of A is a number cless than a.The distance between two points of the scale is equal to the

difference of the numbers at those points. This is obvious if thenumbers are both positive. Thus

AB = OB-OA = h-a.It is still true if one or both are negative. Thus, since c is negative,CO = c and

C = CO + OB = -c + 6 = 6 - c.Imaginary Quantities. The extraction of roots sometimes leads to

expressions like V 1 or a + 6 ^/l, where a and 6 are real num-bers. These expressions are called imaginary. This means merelythat such expressions are not real numbers. It should not be in-ferred that imaginaries cannot be used or that they have no meaning.A quantity may have a meaning in one problem and not in another.For example, in determining the number of workmen needed in acertain undertaking the answer 3f would be absurd since 3f work-men cannot exist. In determining the area of a field the answer


17/218

Art. 2 Equations 3 10 acres would be meaningless since there is no negative area.In determining the ratio of two lengths the answer V2 is imagi-nary since the result must be a real number. But in still otherproblems, notably in work with alternating currents, an interpreta-tion can be given to the process of extracting the square root of anegative number and then such results are entirely real.

Art. 2. EquationsAn equation is the expression of equality between two quantities.

An identical equation is one in which the equality is true for allvalues of the variables. Thus, in

{x -yy + 4:xy= {x + yYthe two sides are equal whatever values be assigned to x and y.In many equations, however, the equality is true only for certain

values of the variables; thus x^ -|- x = 2 is an equation not true forall values of x, but only when x = +1 or 2.Two or more equations are called simultaneous if all are satisfiedat the same time. Equations often occur that are not simultaneous.Thus if x'' = 1, then x = l,orx = 1, but not both simultaneously.A solution of an equation is a set of values of the variables satis-fying the equation. Thus x = 3, ?/ = 4 is one solution of theequation x^ + y ^ = 2b. A solution of a set of simultaneous equa-tions is a set of values of the variables satisfying all of the equations.Eqmvalent Equations. Sets of equations are called equivalent if

they have the same solutions. Thus the pair of simultaneousequations

x^ + xy + y^ = 4, x'^ xy -\- y'^ = 2is equivalent to

x^ + y^ = 3, xy = 1(obtained by adding and subtracting the original equations) inthe sense that any values of x and y, satisfying both equationsof one pair, satisfy both equations of the other pair. Similarly,(^ + 2/) {x 2y) = is equivalent to the two equations

x + y = 0, X 2y =


18/218

4 Algebraic Principles Chap. 1in the sense that if x and y satisfy the equation {x + y) (x 2y) 0,then either x + y = Oorx 2y = Q; and, conversely, if x and ysatisfy either of the latter equations, they satisfy the former.The maia problem in handling equations is to replace an equation

or set of equations by a simpler or more convenient equivalent set.To solve an equation or set of equations is merely to find a particu-lar equivalent set of equations.Degree of Equation. The equations of algebra usually have the

form of polynomials equated to zero. By a polynomial is meantan expression, such as a;' + x 2 or x^y + 3 xy y-, containingonly positive integral powers and products of the variables.The degree of a term like x' or 3 xy is the sum of the exponents of

the variables in that term. Thus the degree of x' is three, thatof 3 xy is two. The degree of a polynomial is that of the highestterm in it. Thus, the polynomials given above are both of the thirddegree.

If a polynomial is equated to zero, or if two polynomials areequated to each other, the degree of the resulting equation is thatof the highest term in it. For example, x^ + y^ x = andxy = 1 are both equations of the second degree.

Exercises1. Determine which of the following equations are identities:

(a) x^s = aj^H- , (6) - + - = 2, (c) 1 + 1 = 5-+^'.^ ' y X X y xy2. Expand (x + y)' by the binomial theorem. Is the resulting

equation an identity?3. Show that x = V2 is a solution of the equation

x6 + 2x'-a;2 -8a; + 2 =0.4. Show that a;= 1, 2/ = 2isa solution of the simultaneous

equationsx^ + 6xy + y* = 5, x''+y' = 5.

5. Show that the pair of simultaneous equationsx' + y^ = 2, x + y = 1


19/218

Art. 3 Equations in One Variable 5is equivalent to the pair

x^ - xy -\- y'^ = 2, x+y = 1.6. Find a set of three equations equivalent to

(s^ - 1) (a;3 + 2) = 0.Explain in what sense the three are equivalent to the one.7. Is x^ 4 ^2/ + 3 1/^ = equivalent to the pair of simultaneous

equations x = y, x = Zyl8. The symbol V2 is generally used to represent the positive square

root of 2. Is a; = v^ equivalent to x^ = 2?9. Show that Vx + 1 + Vx 2 = 3 is equivalent to x = 3.10. The solution of the simultaneous equations

X -\-y = i, xy = Xcan be written

X = 1(3 VB), 2/ = i (3 =F VI).What do these mean? How many solutions are there?

11. What is the degree of the equation (x + z/)* = 3 xyl12. If X and y are the variables, what is the degree of ax^ = hxyl

Art. 3. Equations in One VariableQuadratic Equations. The quadratic equation

ax^ + 6x + c =can be solved by completing the square. Transposing c, dividingby a and adding 6^/4 c?- to both sides, the equation becomes

, & ,6^ 6' 4 ocx^+ -x-]r-T =a 4 a^ id?Extracting the square root and solving for x,

_ 6 -fc V62 _ 4 ac 2aIf the expression under the radical is positive, the square root canbe extracted and real values are obtained for x. If it is negative>no real square root exists and the values of x are imaginary.

Solution by Factoring. Another method for solving quadraticequations is factoring. Thus

x2 + 5a;- 6 =


20/218

6 Algebraic Principles Chap. 1is equivalent to

(a; - 1) (x + 6) = 0.Since a product can only be zero when one of its factors is zero, theabove equation is satisfied only when a; = 1 or a; = 6.

If the quadratic cannot be factored by inspection, it can still befactored by completing the square. Thus

Sx-' - 2x + 1 = 3 (x^ - ^x + I) = 3[ix - lY + U= 3 (a; - i - I V=2) (x - i + V^).

The solutions of the equation Sa; 2a; + l=0 are thena; = i (1 ^~2).

In this way any equation can be solved if the expression equatedto zero can be factored. For example, to solve the equation

a;' + x - 2 =write it in the form

a;3 - 1 + x2 - 1 = 0.Since x' 1 and x^ 1 both have x 1 as a factor, the equationis equivalent to

(x-1) (x2 + 2x + 2) =0.The solutions are consequently

X = 1 and X = 1 V 1.Exercises

Solve the following equations:1. 2x2 + 3x - 2 = 0.2. x^ + 4 X - 5 = 0.3. 3 x^ + 5 X + 1 = 0.4. x^ + X + 1 = 0.

Solve by factoring8. x^ - 3 X - 1 = 0.9. 2 x^ + X - 2 = 0.

10. x2 - X + 1 = 0.14. Solve the equation x* + 1 = by reducing it to the form

(x'' + 1)2 - 2 x^ = 0.

5.


21/218

Art. 4 Factors and Roots 715. Solve the equation a;' + x^ + 4 = by the method of the last

example.16. Factor 4x^ + ixy y^by completing the square of the first two

terms.

Art. 4. Factors and RootsIt has been shown above that the roots of an equation can be

found if the factors of the polynomial equated to zero are known.Conversely, if the roots are known the factors can be found. Thisis done by the use of the following theorem : If r is a root of a poly-nomial equation in one variable x, then x r is a factor of the poly-nomial. To prove this, let

P = ttx + bx -^ + +px + qbe a polynomial of the nth degree in which a,b, . . . , p, q are con-stants. If ? is a root of the equation given by equating this 'poly-nomial to zero,

ar + hr -^ + + pr + q = 0.Since subtractiag zero from a quantity does not change its value,P = ax -\- bx -^ + +px+ q {ar + br --^+ +pr+ q)= a (x - r ) + b ix -^ - r -') + + p (x -r).

Each term on the right side of this equation is divisible by a; r.Hence the polynomial, P, has a; r as a factor, which was to beproved.Number of Roots. It can be shown that any polynomial equation

in one unknown has a root, real or imaginary. Assuming this, itfollows that any polynomial of the rath degree in one variable is theproduct of n first degree factors. In fact, if n is a root of P = 0,then

P=ix- n) Q,Q being the quotient obtained by dividing P by a; ri. Similarly,if r2 is a root of Q = 0, Q= {x- ri) R.Hence

P = (a; - ri) (x - r^) R.


22/218

8 Algebraic Pkinciples Chap. 1In the same way R can be factored, etc. Now each time a factora; r is divided out the degree of the quotient is one less. Aftertaking out n factors, what is left will be of zero degree, that is, aconstant. If a is the constant

P = a{x -ri){x -n) . . . (a; ?).Hence P is the product of n first degree factors, a {x n), {x Ti),etc.

Since a product can only be zero when one of its factors is zero,it follows that the roots of P = are n, r^, . . . , r. It is thusshown that an equation of the nth degree has n roots. Some of theser's may be equal and so the equation may have less than n distinctroots.

Rational Roots. Though every polynomial equation in one un-known has a root, no very definite method can be given for findingit. If nothing in the particular equation suggests a better method,it is customary to try first to find a whole number or fraction that isa root of the equation. Such roots are found by trial. Somemethods that may be useful are shown in the following examples.Example 1. Solve the equation 4 x' + 4 a;^ a; 1 =0.Since a; is a factor of all the terms in this equation except the last, 1, it follows that any integral value of x must be a divisor of 1.

The only integral roots possible are then 1. By trial it is foundthat a; = 1 satisfies the equation. Hence a; + 1 is a factor of thepolynomial. Factoring, the equation becomes (a;+l) (4x'' 1) = 0.The roots are consequently 1, and |.

Ex. 2. Solve the equation 27 x^ + 9 a;^ - 12 a; - 4 = 0.Proceeding as in the last example it is found that the equation

has no integral root. Suppose a fraction -p/q (reduced to its lowestterms) satisfies the equation. Substituting and multiplying by g',

27 p3 + Qp\- 12pg2 _ 42' = 0.Since all the terms but the last are divisible by p, and p and q haveno common factqr, 4 must be divisible by p. For the same reason27 must be divisible by q. Any fractional root must then be equalto a divisor of 4 divided by a divisor of 27. It is found by trial that


23/218

Art. 6 Appeoxmate Solution of Equations 9f is a root. Hence a; f is a factor. Dividing and factoring thequotient, the equation is found to be

27 (x - f) (x + f) (x + i) = 0.The roots are consequently f and

Exercises

0.

Solve the following equations:1.2.3.4.5.6. x' + 4 a:^ -f- 4 X + 3 = 0.

x3-2x2-x + 2 = 0.3 x' - 7 x2 - 8 X + 204 x^ - 8 x - 35 X + 75 = 0.8x3 - 28x2 +30x - 9 = q.x3 - 4 x= - 2 X + 5 = 0.

7. 4x + 8x' + 3x'-2x-l=0.8. 6x


24/218

10 Algebraic Pkinciples Chap. 1above at A and below at B, it must cross the axis at some inter-mediate point C. At that point th'e value of the polynomial iszero.

Example 1. Find the roots of i' + 3 x^ 1 = accurate to onedecimal place.By substitution the following pairs of values are found:

X = -3, -2, -1, 0, +1,a;3 + 3x2- 1 = -1, +3, +1, -1, +3.

The polynomial changes sign between x = 3 and x .2, be-tween X = 1 and X = and between x = and x = 1. There isconsequently a root of the equation in each of these intervals. Tofind the root between and 1, make an enlarged table for this region.

X = 0, 0.5, 0.6, 1,x' + Sx''- 1 = -1, -.125, +.296, 3.

It is thus seen that the root is between 0.5 and 0.6. When x = 0.55the polynomial is positive. Hence the root hes between 0.5 and0.55. The value 0.5 is therefore correct to one decimal. In thesame way the value 2.9 is found for the root between 2 and 3,and 0.7 for the one between 1 and 0. Since the equation canhave only three roots this completes the list.Ex. 2. Solve the equation x' + x 3 = 0.Since x' + x increases with x it can equal 3 for only one realvalue of X. To two decimals this root is found to be 1.21. The

polynomial then has x 1.21 as an approximate factor. Dividingby this the quotient is

x^ + 1.21 X + 2.46.The solutions obtained by equating this to zero are

x= -.6 1.4 V^.Exercises

Fiiid to one decimal the roots of the following equations:1. z' - 3 x2 + 1 = 0. 4. a; - 3 x^ + 3 = 0.2. a;3 + 3 X - 7 = 0. 5. a^ + x - 1 = 0.3. x' + a;2 + X - 1 = 0. 6. x* _ 3 j; _ i = q.


25/218

Art. 6 Inequalities 11Art. 61 Inequalities

An inequality expresses that one quantity is greater than (>) orless than (x and (x - 1) (x + 2) or to < and < to >) when an inequality is multiplied or dividedby a negative quantity.The main problem in inequalities is to determine for what values

of the variable an inequality holds. How this is done is best shownby an example.Example. Find the values of x for which5x'^-x-3

x'' (2 - x)This is equivalent tobx'-x-Z

>I.

1 >0x (2 - x){x + 1) (x - 1) {x + 3) >0.x (2 - x)

The problem is to determine the values of x for which the expressionon the left is positive. Since x' is always positive, the sign of theexpression is determined by the signs of the other four factors. Thevalues of x making one of these factors zero are 3, 1, 1, 2. Mark

-++++++4- ++ -1 oFig. 6.

these values on a scale (Fig. 6). If a; < 3 the thi-ee factors in thenumerator are all negative, and (2 x) is positive. T^ wholeexpression, having an odd number of negative factors, is negative.


26/218

12 Algebraic Principles Chap. 1If X is between 3 and 1, there are two negative factors, a; + 1and X 1, and the whole expression is positive. If x is between 1and +1, the only negative factor is {x 1) and the expression isnegative. If x is between 1 and 2, all the factors are positive andthe whole expression is positive. If a; > 2 there is one negativefactor, (2 x), and the expression is negative.The expression is positive when x is between 3 and 1, or be-

tween 1 and 2. These conditions are expressed by the inequalities3 2.Art. 7. Simultaneous Equations

Simultaneous equations in more than one unknown are solved bya process called elimination. This is a name applied to any proc-ess by which equations are found equivalent to the given equationsbut some of which contain fewer unknowns. By a continuation ofthis process equations may eventually be obtained each containinga single unknown and these can be solved by the methods alreadygiven. In other cases it may not be possible to solve the equationscompletely but they may be reduced to a simpler form. If nothingin the equations indicates a simpler way, there are three generalmethods that may be useful:

(1) Multiply the equations by constants or variables and add orsubtract to get rid of an unknown or to obtain a simpler equation.


27/218

Art. 7 Simultaneous Equations 13(2) Solve one of the equations for one of the unknowns and sub-

stitute this value in each of the other equations.(3) Between one of the equations and each of the others eliminate

the same unknown. Proceed with the new equations in the sameway until finally (if possible) one of the unknowns is found. Thendetermine the other unknowns by substituting this value in theprevious equations.However the solutions be found they should be checked by sub-

stitution in each of the original equations.Example 1. Solve the simultaneous equationsx+ y + 2 = 2,2x- %j + Sz = 9,

3x + 2y- z= -1.Adding the second to the first and twice the second to the third,

3X + 4:Z= 11,7x + 5z = 17.

Subtracting 5 times the first from 4 times the second of these equa-tions, there is found 13 x = 13, whence x = 1. This value substi-tuted in either of the preceding equations gives z = 2. The valuesof X and 2 substituted in either of the original equations give y =1.The solution is x = 1, y = 1, z = 2. These values check whensubstituted in the original equations.

Ex. 2. Solve the equationsx'' + y^-2x + 4:y = 21,x'^ + y^ -\- x y = 12.

Subtraction gives 5y 3x = 9. Hence y = f (x -|- 3). Thisvalue substituted in the second equation gives

17x2 + 32x- 132 = 0.The roots of this are 2 and ff. The corresponding values of yare 3 and -fj. The solutions are x = 2, y = S and x = fr,y =-^7. These values check when substituted in the originalequations.


28/218

14 Algebraic Pkinciples Chap. 1Exercises

Solve the following simultaneous equations

:

1. 4:x-5y+ 6 = 0, 6. x^ + y^ + 2x=0,7x-9y + n=0. 2/ = 3x + 4.2. x + 2y-z + 3=0, 7. h' +k'' - 8h + 4k + 20 = f,2x- y -5=0, K' + k? + Qh->r2k-\-lQ = f\

x + 2z -8 = 0. h? + 8h + lQ = r\3. x + 2y + z = 0, 8. x'' + iy'' = 5,

X y z = 1, xy = 1.2x+ y-z^O. ., = 1 + 1i. x + 2y + 3z = 3, y^ zx-2y + 3z = l, =1-1^1x + 4:y + Qz = 6. '^ ~ z'^ x'5.- + - = l, z = --\X y X y1,1^2 10. a;2 + 2/2 + z2 = 6,

y z ' X +y + z =2,11^ 2x -y +3z =Q.z X

Art. 8. Special CasesInconsistent Equations. Sometimes equations are inconsistent,

that is, have no simultaneous solution. This is usually shown bythe equations requiring the same expression to have different values.For example, take the equations

x+ y+ z = l,2x + 32/ + 4z = 5,

a; + 22/ + 33 = 3.Elimination of x between the first and second and first and thirdgives

i/ + 2z = 3, y + 2z = 2.Any solution of the original equations must satisfy these. Sincethe expression y +2z cannot equal both 2 and 3, there is no solu-tion.Dependent Equations. Sometimes the solutions of part of the

equations all satisfy the remaining equations. These last give noadded information. Such equations are called dependent.


29/218

Art. 8 Special Cases 15For example, take the equations^

X + 2/ = 1, x - 2/2 + a; + 3 2/ = 2.Substituting 1 x icx y va. the second equation, it becomes 2 = 2.All the solutions of the first equation satisfy the second. The twoequations are equivalent to one equation x -\- y = 1. They havean infinite number of simultaneous solutions.Number of Solutions. In general, definite solutions are expected

if the number of equations is equal to the number of unknowns.Thus, two equations usually determine two imknowns, three equa-tions determine three unknowns, etc. This is, however, not alwaysthe case. The equations may be inconsistent and have no solutionor may be dependent and have an infinite mmiber of solutions. Ifthe equations determine definite solutions, the number of solutionsis expected to equal the product of the degrees of the equations.Special circumstances may, however, change this number. It canbe shown that unless the number of solutions is infinite it cannotexceed the product of the degrees of the equations.

If there are fewer equations than unknowns, the unknowns willnot be determined. In this case, if the equations are consistent,there wUl be an infinite number of solutions.

If there are more equations than unknowns, usually there wUlbe no solution. In particular cases, however, there may be solu-tions. To determine whether there is a solution, solve part of theequations and substitute the values found in the remaining equa-tions. If any of them satisfy all of the equations, there is a solu-tion, otherwise there is none.Homogeneous Equations. If all the terms of an equation have

the same degree the equation is called homogeneous. A set of homo-geneous equations can often be solved for the ratios of the variableswhen there are not enough equations to determine the exact values.For example, take the homogeneous equations

X y s = 0, 3x y 2g = 0.Solving for x and y

x = \z, y=-is.


30/218

16 Algebraic Principles Chap. 1Any value can be assigned to z and the values of % and y can thenbe determined from these equations. Let % = 2h. The solutionis then

x='k, y = k, z = 2k.Since h is arbitrary, x, y, z have any values proportional to 1, 1,2. This result can be written x:y:z = 1: 1:2.

ExercisesDetermine whether the following equations have no solutions,

definite solutions, or an infinite number of solutions:1. X + y + z = \, 6. x~y-\-2z = X,2x -2y + 5z = 7, ' Sx + y - 3 = 5,2x-3y+4z = 5. Sx + 2y-3z = 2.2. x + 2y+ z = 0, 7. 3x+y + l = 0,2x + y - 2 = 5, 2x + 3y-i = 0,5x + 7y + iz = 2. x + 2y-3=0.3. x + 2y + l = 0, 8. X - y =0,3x-y-i = 0, x^ - 2/2 = 1.2x + 3y + l=0. 9. {X- yy+{y - zy + {z-xy = 1,

4. s + 2/ - z = 4, x + 2/2 +z^ =2,x + 2y + 3z = 2, (_x+yy + (y+zy + {z + xy = 3.5x + 8y + 7z = li. 10. w + x+ y+ z = 1,

5. x + y ~ 5 = 0, w + 2x - 3y - 4:z = 2,3x + 2y -12 = Q, w - x-2y-3z = 3,2x + y - 6=0. w-bx+2y + 3z = 3.Find values to which the variables in the following equations are

proportional:11. X +y -2z = Q, 13. X +y - z = 0,

3 X - 2/ - 4 z = 0. x2 ^ 2/2 - 5 z2 = 0.12. X +2y +. z = 0, 14. x'' + y' = 2 z\

4 2/ + 3 z = 0. y^ = xz.Art. 9. Undetermined Coefficients

It is often necessary to reduce a given expression or equation toa required form. This form is indicated by an expression or equa-tion having letters for coefficients and the reduction is made bycalculating the values of these coefficient^.

In this work frequent use is made of the following theorem: Iftwo polynomials in one variable are equal for all values of the variable,


31/218

Art. 9. Undetermined Coefficients 17the coefficients of the same power of the variable in the two polynomialsare equal. To show this, suppose, for all values of x,Oo + aiCK + ajx^ + + ax = 60 + hx + fcax^ + + bnX -Then for all values of x

(do -bo) + {ai-bi)x+ + (a - &J x = 0.If the coefficients ia this equation are not zero, by Art. 4, it cannothave more than n distinct roots. Hence the coefficients must allbe zero and oo = 60, ai = 61, etc., which was to be proved.To reduce an expression to a given form, equate the expression tothe given form, clear of fractions or radicals, and determine the un-

known coefficients by the above theorem.Example 1. To find the coefficients a and 6 such that

X _ '^ 1 ^(x - 1) (s + 3) ~ a; - 1 ^ a; + 3

clear of fractions, getting,x = a{x + Z)+b{x-l) = {a + b)x + 2,a-b.

If this equation holds for all values of x,a + 6 = l, 3a-6 = 0.

Hence a = j, 6 = f . Conversely, if a and 6 have these values, theabove equations are identically satisfied. Therefore

X 1 + 3(x - 1) (x + 3) 4 (x - 1) ' 4 (x + 3)

In many cases the expression can be more easily changed to therequired form by simple algebraic processes. This is particularlythe case with second degree expressions where completing thesquare may give the required result.Example 2. To reduce the expression

l + 4x-2x2to the form a b {x cY, it can be written

l-2(x2-2x) =3-2(x-l)2,which is the result required.


32/218

18 Algebraic Principles Chap. 1In reducing equations to a required form it should be noted that

multiplying an equation by a number gives an equivalent equation.Thus, X + y = 1 and 2x + 2y = 2 are equivalent. Two equa-tions are then equivalent when corresponding coefficients are pro-portional.Example 3. To find k such that x + 8y 1 = and 3x y 3 + A; (a; + 3 2/ 1) = are equivalent, write the last equation

in the form(3 + k)x + {3k-l)y-{k + 3) -=0.

This is equivalent toai + Sj/ l = Oif corresponding coefficientsare proportional, that is, if3+k 3k-l _ k+3

1 ~ 8 1These equations are satisfied hy k = 5.

ExercisesReduce the following expressions to the forms indicated:1. 2 x2 + 3 a; + 4 = (ax + 6)2 + c.2. S + 2x-x' = b~(,x- ay.3. x'' -\- xy + y^ = a (x + myy + 6 (y mx)'.

X + 1 a , b4- r?::^sr = z +

:

X {x 2) X X 2ax + b , c

(x + 3) (x2 + 1) cc + 1 ' a; + 3Reduce the following equations to the forms indicated:6. 3a; 42/ = 5, y = mx + b.7. 2x + 3y = 4, ^ + |=1.8. dx' + 2y'-6x + iy = l, i^^' + l^^l^' = 1.9. x'-4y^-ix + 8y = 4, (x - ^' _ (y - fc)' ^ ^^

10. 3 a; - 2/ + 5 = 0, (x + y - 1) + k {x - y + 3) = 0.Art. 10. Functions

It is often desirable to state that one quantity is determined byanother. For this purpose the word function is used. A quantity


33/218

Art. 10 Functions 19y is called a function of x if values of x determine values of y. Thus,if 2/ = 1 x^, then t/ is a function of x, for a value of x determines avalue of y. Similarly, the area of a circle is a function of its radius;for, the length of radius being given, the area of the circle is deter-mined.

It is not necessary that a value of the variable determine a singlevalue of the function. It may be that a limited number of valuesare determined. Thus, y is a, function of x in the equation '

x^ 2 xy -\- y'^ + X = 1.To each value of x correspond two definite values of y obtained bysolving a quadratic equation.

If a single value of the function corresponds to each value of thevariable, the function is called single valued. If several values ofthe function correspond to the same value of the variable thefunction is called many valued.Kinds of Functions. Any expression containing a variable is a

function of that variable, for, a value of the variable being given, avalue of the expression is determined. Such a function is calledexplicit. Thus VxM-1 is an explicit function of x. Similarly, ify = Vx^ + 1, then y is an expUcit function of x.

If X and y are connected by an equation not solved for y, then yis called an implicit function of x. For example, 2/ is an implicitfunction of x in the equation

a;2 + 2/' + 2x + 2/ = l.Also X is an imphcit function of y.

ExpUcit and implicit do not denote properties of the functionbut merely of the way it is expressed. An implicit function isrendered expUcit by solving. For example, the above equation isequivalent to

2/ = i (- 1 V5-8x-4x2).A rational function is one representable by an algebraic expression


34/218

20 Algebraic Pbinciples Chap. 1containing no fractional powers of variable quantities. For example,

a; V5 + 3is a rational function of x.An irrational function is one represented by an algebraic expres-

sion whicli cannot be reduced to a rational form. For example,Vx + 1 is an irrational function of x.A function is called algebraic if it can be expressed explicitly orimplicitly by a finite number of algebraic operations (addition,subtraction, multiplication, division, raising to integral powers, andextraction of integral roots). All the functions previously men-tioned are algebraic.

Functions that are not algebraic are called transcendental. Forexample, x ' and 2^ are transcendental functions of x.The terms rational, irrational, algebraic, and transcendental de-

note properties of the function itself and do not depend on theway the function is expressed.Notation.A particular function of x is represented by the nota-

tion / (x), which should be read function of x, or/ of x, not/ times x.For example, / (z) = Vx^ + 1, means that / (a;) is the definitefunction VxM-1. Similarly, y f {x) means that y is a definite(though perhaps unknown) function of x.The / in the symbol of a function should be considered as repre-

senting an operation to be performed on the variable. Thus, if/ (x) = Vx^ + 1, /represents the operation of squaring the variable,adding 1, and extracting the square root of the result. If x is re-placed by any other quantity, the same operation is to be performedon that quantity. For example, / (2) is the result of performingthe operation / on 2. With the above value of /,

/(2) = V2M^ = V5.Similarly,

,

/(2/ + l) = V(2/ -f- 1)'' + 1 = V2/2-h22/-l-2.If it is necessary to consider several functions in the same dis-

cussion, they are distinguished by subscripts or accents or by the


35/218

Art. 10 Functions 21use of different letters. Thus, /i (x), f^ (x), f, {x), f (x), f (x),f (x) (read /-one of x, /-two of x, /-three of x, /-prime of x, /-secondof X, /-third of x), g (x) represent (presumably) dififerent functionsof X.

Functions of Several Variables. A quantity u is called a functionof several variables if values of u are determined by values of thosevariables. For example, the volume of a cone is a function of itsaltitude and the radius of its base; for the volume is determined bythe altitude and radius of base. This is indicated by the notation

v=f{h,r),which should be read, w is a function of h and r, orvisfoih and r.

Similarly, the volume of a rectangular parallelopiped is a functionof the lengths of its three edges. If a, b, and c are the lengths ofthe edges, this is expressed by the equation

v=fia, b,c),which should be read, w is a function of a, b, and c, or vis foi a, b, c.Independent and Dependent Variables. In most problems there

occur a nxunber of variable quantities connected by equations.Arbitrary values can be assigned to some of these quantities andthe others are then determined. Those taking arbitrary valuesare called independent variables; those determined are called de-pendent variables. Which are taken as independent and which asdependent variables is usually a matter of convenience. The num-ber of independent variables is, however, fixed by the equations.Example. The radius r, altitude h, volume v, and total surface S

of a cyhnder are connected by the equationsV = Tr%, S = 2nr'' + 2 irrh.

Any two of these four quantities can be taken as independent vari-ables and the other two calculated in terms of them. If, for ex-ample, V and r are taken as the independent variables, h and S havethe values

ht' r


36/218

22 Algebraic Pbinciplbs Chap. 1

Exercises1. If/ (a;) = a;2 - 3 a; + 2, show that/ (1) = / (2) = 0.2. lifix) =x + i, find/ (a; + 1). Also find/ (a;) + 1.3. If/ (x) = Vi^^n, find/ (2 a;). Also find 2/ (x).4. If/(x)=^3,find/(l). Alsofind^-l-^.6. If lA (a;) = a;< + 2 x^ + 3, show that i/- (-x) = v^ (x).6. If (^ (x) = X + -, show that [< > (x)]' = 4> (x^) + 2.7. If F (x) = 1^, show that F (a) F (-a) = 1.8. If/, (x) = 2\ f, (x) = xS find/i [/a (2/)]. Also find /a [/i (t/)].9. If / (x, y) = x^ + 2xy - 5, show that/ (1, 2) = 0.

10. If F (x, y) = x^ + xy+ y\ show that F {x, y) = F (y, x).11. If / (x, 2/) = x^ + 3 x^y + 2/^, show that / (x, wx) = x'f (1, u).12. If a, b, c are the sides of a right triangle how many of them can

be taken as independent variables?13. Express the radius and area of a sphere in terms of the volume

taken as independent variable.14. Given u = x^ + y^, v = x + y, determine x and y as functions

of the independent variables u and v.15. If X, y, z satisfy the equations

x + y + 2 = 6,x-2/ + 2z = 5,2x -\-y z = l,show that none of them can be independent variables.

16. The equations x+ y + z = 6,X- ^ + 2z= 5,2X + 42/+ z = 13are dependent. Show that any one of the quantities x, y, z can be takenas independent variable.

17. If u, V, X, y are connected by the equationsV? + uv y = 0, uv -\- X ~ y = 0,

show that u and x cannot both be independent variables.


37/218

CHAPTER 2RECTANGULAR COORDINATES

Art. 11. DefinitionsScale on a Line. In Art. 1 it has been shown that real numbers

can be attached to the points of a straight line in such a way thatthe distance between two points is equal to the difference (largerminus smaller) of the numbers located at those points.The line with its associated numbers is called a scale. Proceed-

ing along the scale in one direction (to the right in Fig. 11a) the-5-1-3-2-10133 15

Fig. 11a.numbers increase algebraically. Proceeding in the other directionthe numbers decrease. The direction in which the numbers in-crease is called positive, that in which they decrease is callednegative.

Coordinates of a Point. In a plane take two perpendicularscales X'X, Y'Y with their zero points coincident at (Fig. 116).It is customary to draw X'X, calledthe X-axis, horizontal with its positiveend on the right, and Y'Y, called they-axis, vertical with its positive endabove. The point is called theorigin. The axes divide the plane intofour sections called quadrants. Theseare numbered I, II, III, IV, as shownin Fig. 11&. Fig. 11&.From any point P in the plane drop

perpendicularsFM, PN to the axes. Let the number at Af in thescale X'X be x and that at N in the scale Y'Y be y. These num-

23


38/218

24 Rbctangulab Coordinates Chap. 2bers X and y are called the rectangular coordinates of P with respectto the axes X'X, Y'Y. The number x is called the abscissa, thenumber y is called the ordinate of P.

If the axes are drawn as in Fig. 116, the abscissa x is equal inmagnitude to the distance PN from P to the y-a,xis, is positive forpoints on the right and negative for points on the left of the j/-axis.The ordinate y is equal in magnitude to the distance PM from P tothe X-axis, is positive for points above and negative for points belowthe X-axis. For example, the point P in Fig. 116 has coordinatesx = 3, 2/ = 2, whUe Q has the coordinates x = 3, y = 2.

Notation. The point whose coordinates are x and y is representedby the symbol (x, y). To signify that P has the coordinates x and ythe notation P (x, y) is used. For example, (1, 2) is the pointX = 1, 2/ = 2. Similarly, A (2, 3) signifies that the abscissa ofA is 2, and its ordinate 3.

Plotting. The process of locating a point whose coordinates aregiven is called plotting. It is convenient for this purpose to usecoordinate paper, that is, paper ruled with two sets of lines as inFig. lie. Two of these perpendicular lines are taken as axes. Acertain number of divisions (one in Figs, lie, d and e) are taken asrepresenting a unit of length. This unit should be long enough tomake the diagram of reasonable size but not so long that any pointsto be plotted fail to lie on the paper. By counting the rulings from

the axes it is easy to locate the point having givencoordinates (approximately if it does not fall at anintersection).Example 1. Show graphically that the pointsA (-1, -3), B (0, -1), C (1, 1) and D (2, 3) lie

on a straight line.The points are plotted in Fig. lie. By apply-ing a ruler to the figure it is found that a straight

line can be drawn through the points.Ex. 2. Plot the points Pi (-2, -1), P^ (3, 4)

and find the distance between them.The points are plotted in Fig. lid.Let the horizontal line through Pi cut the vertical line through

AFig. lie.


39/218

Art. 11 Definitions 25P2 in R. From the figure it is seen that PiR = 5, RP2 = 5.Consequently,

P1P2 = Vp,E^ + BP22 = VSO.


40/218

26 Rectangular Coordinates Chap. 2through A meet the vertical line through B in P. What are the co-ordinates of PI What are the lengths of AP and BPl Calculate thedistance AB.

9. Find the distance between A (2, 3) and B (3, 4).10. Find the area of the triangle formed by the points (3, 4), (5, 3)and (2, 0).

Art. 12. SegmentsIn this book the term line (meaning straight line) is used only

when referring to the infinite line extending indefinitely in bothdirections. The part of a line between two points will be called asegment.

In many cases a segment is regarded as having a definite direction.The symbol AB is used for the segment beginning at A and endingat B. The segment beginning at B and ending at A is written BA.The value of a segment may be ainy one of three things that should

be carefully distinguished. Whenever the symbol AB occurs in anequation it must be understood which of these things is meant.

(1) In many cases the value AB means the length of the segment.In this case AB = 3 means that AB has a length of three units andAB = CD means that AB and CD have equal lengths.

(2) In other cases the symbol AB represents the length togetherwith a sign positive or negative according as the segment is directedone way or the other along the line. For example, the a;-co6rdinateof a point is equal to the segment NP (Fig. 116) considered positivewhen drawn to the right, negative when drawn to the left. Thevalue of the segment is in this case a number with a positive ornegative sign.

(3) Certain physical quantities, such as velocities, include in theirdescription the direction of the hnes along which they occur as wellas their magnitudes and directions one way or the other along thoselines. Two segments representing such quantities are equal onlywhen they have the same length and direction. The value of sucha segment (called a vector) is not a number but a number anddirection.Segments Parallel to a Coordinate Axis. In most cases the value

of a segment parallel to a coordinate axis will be the number equal


41/218

Art. 13 Projection 27in magnitude to the length of the segment and positive or negativeaccording as the segment is drawn in the positive or negative direc-


42/218

28 Rbctangulah Coordinates Chap. 2For example, in Fig. 13a or 136, in both magnitude and sign,

AB: BC = A'B': B'C.

a'M P^ C B' NFig. 13&.

In using this formula it should be noted that if AB and BC aredistances, their projections must be distances and if AB and BChave algebraic signs their projections must have algebraic signs.For example, if the segments are projected on the coordinate axesand the values of the projections determined by equation (12), ABand BC must be considered opposite in sign when they have oppositedirections.Example 1. Find the middle point of the segment joining

Pi {xi, yi) and P2 (x^, 1/2).Let P (x, y) be the middle point. Project on the axes (Fig. 13c).

Since P is the middle point ofP1P2, in both length and directionPiP =PP2. Hence

MiM = MM2, NiN-=NN2,and soX xi = Xi X, y yi = yi y.Consequently,

Fig. 13c. x = i{xi + X2), 2/ = ^J/i + 2/2).The sum of several quantities divided by their number is called theaverage. Hence each coordinate of the middle point is the averageof the corresponding coordinates of the ends.

Ex. 2. Given Pi (-1, 1), P2 (3, 4), find the point P {x, y), onP1P2 produced, which is twice as far from Pi as from P2 (Fig. 13d).

T


43/218

Art. 14 Distance between Two Points 29Since PiP is twice as long as PP2 and they have opposite direc-

tions,

PiP= -2PP2.Consequently,MiM= -2 MMi, NiN = - 2 NN2,

and soa;+l=-2(3-x), j/-l = - 2(4 - j/). -PifHence x = 7, y 7.

Y


44/218


45/218

Art. 14 Distance between Two Points 31Let P {x, y) be the point required (Fig. 14d). Since PA = PB

V{x - 9)2 + 2/2 = VCa; + 6)^ + {y - Z)\Y Squaring and cancelling,

5 a; 2/ = 6.Similarly, sincePA = PC,

2x-3y = 5.S 1 V i n g simultaneouslygives X = l,y = 1. Therequired point is therefore

Fig. Ud. (1, -1).

c

f

Exercises1

.

Find the perimeter of the triangle whose vertices are (2, 3), ( 3, 3)and (1, 1).2. Show that the points (1, 2), (4, 2) and ( 3, 5) are the verticesof an isosceles triangle.

3. Show that the points (0, 0), (3, 1), (1, -1) and (2, 2) are thevertices of a parallelogram.

4. Given A (2, 0), B (1, 1), C (0, 2) show that the distances AB, BCand AC satisfy the equation AB + BC = AC. What do you concludeabout the points?

5. Show that (6, 2), ( 2, 4), (5, 5), ( 1, 3) are on a circle whosecenter is (2, 1).6. It can be shown that four points form a quadrilateral inscribed

in a circle if the product of the diagonals is equal to the sum of the prod-ucts of the opposite sides. Assuming this, show that ( 2, 2), (3, 3),(1, 1) and (2, 0) he on a circle.

7. Find the coordinates of two points whose distances from (2, 3)are 4 and whose ordinates are equal to 5.

8. Find a point on the a;-axis which is equidistant from (0, 4) and(-3, -3).9. Find the center of the circle passing through (0, 0), (3, 3) and

(5, 4).10. Given A (0, 0), B (1, 1), C (-1, 1), D (1, -2), find the point

in which the perpendicular bisector of AB cuts the perpendicular bisec-tor of CD.

11. Find the foot of the perpendicular from (1, 2) to the line joining(2, 1) and (-1, -5).


46/218

32 Rectangular Coordinates Chap. 2

Fig. 15o.

Art. 15. Vectors

A vector is a segment of definite length and direction. ThepoiQt Pi is the beginning, the point P2 the end of the vector P1P2.The direction of the vector along itsline is often indicated by an arrow asshown in Fig. 15a.Two vectors are called equal if they

have the same length and direction. Forexample, in Fig. 15a, P1P2 = P3P4- Iftwo vectprs have the same length butopposite directions, either is calledthe negative of the other. For ex-

ample, P1P2 = -P4P3 = -P2P1.Let the projections of P1P2 on the coordinate axes be M1M2 and

NiNi (Fig. 156). The x-component of the vector P1P2 is definedas the length of M1M2 or the negative ofthat length, according as MiMi has thepositive or negative direction along thea;-axis. Similarly, the y-component is thedistance iViA''2 or the negative of thatdistance, according as N1N2 is drawn inthe positive or negative direction alongthe 2/-axis. For example, in Fig. 156, thecomponents of P1P2 are 3 and 4, whilethose of P2P3 are 7 and 3.

Let the points be Pi {xi, 2/1), P2 {xi, 2/2). In Art. 12 it has beenshown that in magnitude and sign MiMi and iViA^2 are representedby X2 Xi and 2/2 2/1. Hence x^ Xi is the x-component and2/2 2/1 is the 2/-component of P1P2. That is, the componentsof a vector are obtained by subtracting the coordinates of thebeginning from the corresponding coordinates of the end of thevector.

If two vectors are equal their components are equal. For letP1P2 = P3P4 (Fig. 15c). Then, since by definition of equaUtyPxPi and P3P4 have the same length and direction, the triangles

p


47/218

Art. 15 Vectors 33


48/218


Fig. 15/.

Ex. 3. Find the area of the triangle PP1P2, givenPPi = [ai, 61], PP2 = [02, 62].In Fig. 15/ the sides of the

^2 ^i shaded triangles have their lengthsmarked on them. The area of thetriangle PP1P2 is equal to the areaof the whole rectangle less the sumof the areas of the shaded triangles.Hence

PP1P2 = 0162 J ai6i \ aj)i i (i ^2) (62 &i) = 2 (o^i^z 2&i)-This result is shown for a particular figure. By drawing otherfigures it will be found that the result is always correct if the anglefrom PPi to PP2 is positive (that is, drawn in the counter-clockwisedirection). If that angle is negative the formula gives the negativeof the area.

Exercises1. If the vectors AB and CD are equal show that AC and BD are

equal.2. If AB = AiBi, BC = BiCi, show that AC = A,Ci.3. The components of a vector are a, 6. Show that the components

of its negative are a, b.4. Show that the vector [a, b] is equal to the vector from the origin

to the point (o, 6).5. Construct vectors equal to [2, 3], [-2, 3], [-2, -3], and [2, -3].6. Show that the points P (-1, 2), Q (1, -2), R (3, 4), S (5, 0) are

the vertices of a parallelogram.7. A vector equal to [ 3, 4] begins at the point (1, 2). What are

the coordinates of its end?8. The points (1, 2), (2, 1), (3, 2) are the vertices of a triangle.

Find the vectors from the vertices to the middle points of the opposite9. Given A (2, 3), B (-4, 5), C (-2, 3), fmd D such that AB = CD.

10. The middle point of a certain segment is (1, 2) and one end is( 3, 5). Find the coordinates of the other end.

11. By showing that the area of the triangle ABC is zero show thatthe points A (0, -2), B (1, 1) and C (3, 7) he on a hne.

12. Find the area of the quadrilateral whose vertices are the points(-2, -3), (1, -2), (3, 4), (-1, 5).


49/218

Art. 16 Multiple of a Vector 3513. Show that the vectors [oi, h], [ch, bi] are parallel if and only if

01/02 = bi/62. In this way show that the hne through ( 1, 2) and(3, 4) is parallel to that through (1, 1) and (9, 3).

Art. 16. Multiple of a VectorIf tt is a vector and r a real number, the symbol ru is used to rep-

resent a vector r times as long as u and having the same directionif r is positive, but the opposite di-rection if r is negative.

In Fig. 16, let u = [a, b] andV = ru. Since u and v are paralleltheir components have lengths pro-portional to the lengths of u and v. Also corresponding componentshave the same or opposite signs according as u and v have thesame or opposite directions. Hence the components of v are raand rb. That is,

V = [ra, rb\.Hence to multiply a vector by a number, multiply each of its com-ponents by that number.Example 1. Given Pi ( 2, 3), P2 (1, 4), find a vector having

the same direction as PiPz but 3 times as long.The vector required is3PiP2 = 3[3, -7] = [9, -21].

Ex. 2. Find the point on the line joining Pi (2, 3), P2 (1, 2)and one-third of the way from Pi to P2.By hypothesis PiP = \ P1P2, that is,

[x - 2, 2/ - 3] = H-1, -5] = [-i -|].Hence x 2 = \, y 3 = , and consequently a; = j,v = i-

Ex. 3. Show that the segment joining the points A (1, 1),B (5, 7) is parallel to and twice as long as that joining C (4, 3),D (2, -1).


50/218

36 Rbctangulak Coordinates Chap. 2The vectors are

A5 = [4,8], CD = [-2, -4].Hence AB = 2 CD. Consequently the segments are paralleland the first is twice as long as the second.

Art. 17. Addition and Subtraction of VectorsSum of Two Vectors. Draw a vector equal to v, beginning at the

end of u. The vector from the beginning of u to the end of v iscalled the simi of u and v.

Let u = [oi, 6i], V = [02, 62]. From the diagram it is seen thatthe components oi u + v are ai + ^2 and 61 + 62- This is true not

Y


51/218

Art. 17 Addition and Subtraction of VectobsFig. 17d let u = AB,v = BC,w = CD. Then

(.u + v)+w = AC + CD = AD,u + (v + w) = AB + BD = AD.

The two sums are consequently equal.

37

r


52/218

38 Rectangular Coordinates Chap. 2point P satisfying the equation

wiPPi + W2PP2 + WiPPz + etc. = 0.Find the center of gravity of three weights of 2, 3 and 5 poundsplaced at the points Pi (-2, 1), Pi (1, -3), P3 (4, 5) respectively.The center of gravity P satisfies the equation

2PPi + 3PP2 + 5PPs = 0,that is,

2[- 2 - x,l - y] + Zll - X, - 3 - y] + 514: - X, 5 - y]= [19-lOx, 18-10?/] = 0.

Consequently the coordinates of the center of gravity are a; = 1.9and y 1.8.

Exercises1. Given P (1, -3), Q (7, 1), R (-1, 1), S (2, 3), show that PQ haa

the same direction as RS and is twice as long.2. Given Pi (2, -3), P2 (-1, 2), find the point on P1P2 which istwice as far from Pi as from P2. Also find the point on P1P2 produced

which is twice as far from Pi as from P2.3. Find the points P and Q on the line throughPi (2, -1), P2 (-4, 5)

if PiP = -f PP2, PiQ = -f P1P2..4. One end of a segment is (2, 5) and a point one-fourth of the dis-

tance to the other end is ( 1, 4). Find the coordinates of the other end.5. Given the three points A ( 3, 3), B (3, 1), C (6, 0) on a line, find

the fourth point D on the line such that AD: DC = AB: BC.6. Show that the line through ( 4, 5), ( 2, 8) is parallel to thatthrough (3, -1), (9, 8).

7. If the vectors AC and AB satisfy the equation AC = rAB, wherer is a real number, show that A,B,C lie on a line.. In this way show that(2, 3), (-4, -7) and (5, 8) lie on a Une.

8. Given A (1, 1), B (2, 3), C (0, 4), find the point D such thatBD = 2 DC. Show that AD = i [AB + 2 AC].9. In Ex. 8 find the point P such that PA + PB + PC = 0.10. If V is any vector let v^ mean the square of the length of v. Given

vi = [2, 5], Vi = [10, 4], show that (vi + ws)^ = Vi^ + vi and conse-quently that Wi and v^ are perpendicular to each other.

11. Show that the vectors from the vertices of a triangle to themiddle points of the opposite sides have a sum equal to zero.

12. Find the center of gravity of two weights of 1 and 4 poundsplaced at the points (3, 4) and (2, 7) respectively.


53/218

Art. 18 Slope of a Line 3913. Show that the center of gravity of four equal weights placed at

the vertices of a quadrilateral is the middle point of the segment joiningthe middle points of two opposite sides of the quadrilateral.

Art. 18. Slope of a LineLet Pi (xi, 2/0, Pa {x2, yi) be two points of the line MN. The

ratio2/2-m = ViX2 Xi (18a)

is called the slope of the line MN. The same value of the slopeis obtained whatever pair of points on the line is used. For, if

J^Fig. 18a. Fig. 186.

Pi, P2 and Pi, Pi are pairs of points on the same line, the trianglesP1RP2 and P3SP4 are similar andVi-yi ^ RP2 ^ SP4 ^ Vi-yz

X2 xi PiR P3S Xi X3Let

. (18b)Hence the slope of a line is equal to the tangent of the angle fromthe positive direction of the x-axis to the line.

Since the x-axis is horizontal, the steeper the line the nearer is the


54/218

40 Rbctangtjlak Coordinates Chap. 2angle to 90 and consequently the greater the slope. The slope isthus a measure of the steepness of the line.

If the liae extends upward to the right, as in Fig. 18a, the com-ponents PiR and RP^ have the same sign and the slope is positive.If the hne extends upward to the left, as in Fig. 186, the componentshave opposite signs and the slope is negative.

Parallel Lines. If two lines are parallel, they have the same slope;for the angles i^i and 02 (Fig. 18c) are then equal and their slopes,tan i and tan ^2, are equal. Conversely, if the slopes are equal,the angles are equal and the lines are parallel.

Y


55/218

Art. 18 Slope of a Line 41L2 (positive when measured ia the counter-clockwise direction),then

Hencetan /3 = tan (02


56/218

42 Rectangular Coordinates Chap. 2these is interior, the other exterior to the triangle. From the

figure the positive interior angle A is seento extend from AB to AC. Consequently,

3


57/218

Art. 19 Graphs 4314. By showing that the angles CAD and CBD are equal show that

the points A (6, 11), B (-11, 4), C (-4, -13), D (1, -14) lie on acircle.

15. A point is 7 units distant from the origin and the slope of theHne joining it to (3, 4) is |. Find its coordinates.16. A point is equidistant from (2, 1) and ( 4, 3) and the slope of

the line joining it to (1, 1) is |. Find its coordinates.

Art. 19. Graphs

In many cases corresponding values of two related quantities areknown. Each pair of values can be taken as the coordinates of apoint. The totality, or locus, of such points is called a graph.This graph exhibits pictorially the relation of the quantities repre-sented. There are three cases differing in the accuracy with whichintermediate values are known.

Statistical Graphs. Sometimes definite pairs of values are givenbut there is no information by which intermediate values can beeven approximately inferred. In such cases the values are plottedand consecutive points connected by straight lines to show theorder in which the values are given.Example 1. The temperature at 6 a.m. on ten consecutive days

at a certain place is given in the following table:Jan. 12 3 419 27 11 40 5 6 7 8 9 1034 28 36 18 42 38These values are represented by points in Fig. 19ffl. Points repre-senting temperatures on consecutive days are connected by straightlines. At times between thosemarked no estimate of the temper-ature can be made.

Physical Graphs. In other casesit is known that in the intervalsbetween the values given the vari-ables change nearly proportionally.In such cases the points are plotted

I 50B 30. 20


58/218

44 Rectanqotab Coordinates Chap. 2

2 1 G 8 10 12 li IC 18 20Time

Fia. 19&.

of this curve between those plotted are assumed to representapproximately corresponding values of the variables.

Ex. 2. The observed temperatures 5 of a vessel of cooling waterat times t, in minutes, from the beginning of observation were< = 1 2 3 5 7 10 15 20e = 92 85.3 79.5 74.5 67 60.5 53.5 45 39.5These values are plotted in Fig. 196. Since the temperature de-creases gradually and during the intervals given at nearly constant

rates a smooth curve drawn throughthese points will represent approxi-mately the relation of temperatureand time throughout the experi-ment.The Graph of an Equation. In

other cases the equation connect-ing the variables is known. Thegraph then consists of all points

whose coordinates satisfy the equation. Arbitrary values are as-signed to either of the variables, thecorresponding values of the other vari-able calculated, and the resulting pointsplotted. When the points have beenplotted so closely that between consecu-tive points the curve is nearly straight,a smooth curve is drawn through them.Ex. 3. Plot the graph of the equa-

tion y = x'-.In the foUowiujg table values are as-

signed to X and the correspondingvalues of y calculated:a; = -5 -4 -3 -2 -1 1 2 3 4 5y= 25 16 9 4 1 1 4 9 16 25The points are plotted in Fig. 19c. Thepart of the curve shown extends from x h\a x=-\-h. The wholecurve extends to an indefinite distance. Horizontal lines cut the


59/218

Art. 19 Graphs 45

Fig. 19d.

the curve at equal distances right and left of the y-axis. Thisis expressed by saying that the curve is symmetrical with respectto the 2/-axis.Ex. 4. The force of gravitation between two bodies varies in-

versely as the square of their distance apart, that is, F = k/d^, kbeing constant. Assuming k = 10, plot the curverepresenting the relation of force and distance.

In the following table values are assigned tod and the corresponding values of F calculatedd = 0.5 1 1.5 2 3 5 10F=oo 40 10 4.4 2.5 1.1 0.4 0.1

The distance d cannot be negative. For verylarge values of d, F is very small. Hence when dis large the curve very nearly coincides with thehorizontal axis. When d is very small F is verylarge and the curve very nearly coincides withthe vertical axis (Fig. 19d).

Units. Before a graph can be plotted a length must be chosen oneach axis to represent a unit measure of the quantity representedby that coordinate. If the quantities represented by x and y areof different kinds, for example temperature and time, these lengthscan be chosen independently. If, however, these quantities areof the same kind it is usually more convenient to have the sameunit of length along both axes. This is also the case in graphingmathematical equations where no physical interpretation is givento X and y. Other things being equal, a large curve is better thana small one. Such units of length should then be chosen as tomake the curve spread both vertically and horizontally verynearly over the region available for it. If several graphs are tobe compared, the same unit lengths should be used for all.

Exercises1. The price of steel rails each year from 1892 to 1907 was as follows,

in dollars per gross ton:30 28 24 24 28 19 18 28 32 27 28 28 28 28 28 28Make a graph showing the relation of year and price. (Let x be the


60/218

46 Rectangular Coordinates Chap. 2number of years beyond 1892 and y the amount that the price exceeds18 dollars.)

2. The amplitudes of successive vibrations of a pendulum set inmotion and left free wereNumber of vibration 12 3 4 5 6 7

Amplitude 69 48 33.5 23.5 16.5 11.5 8Plot points representing these pairs of values and draw a smooth curvethrough them. Do points of this curve between those plotted have anyphysical meaning?

3. The atomic weight, W, and specific heat, S, of several chemicalelements are shown in the following table:TF = 7 9.1 11 12 23 28 39 55 56 108 196S = 0.94 0.41 0.25 0.147 0.29 0.177 0.166 0.122 0.112 0.057 0.032Make a graph showing the relation of specific heat and atomic weight.

4. The table below gives the maximum length of spark betweenneedle points of an alternating current under standard conditions ofneedles, temperature and barometric pressure.L = length in millimeters. V = electromotive force in kilovolts.F = 10 15 20 25 30 35 40 45 50 60 70 80L = 12 18 25 33 41 51 62 75 90 118 149 180

Make a curve showing the relation of voltage and length of spark.5. In the table below are given the maximum vapor pressures of

water at various temperatures, yhere T = temperature in degrees Centi-grade and P = pressure in centimeters of mercury:T = 10


61/218

Art. 20 Equation of a JLocus 4711. On the same diagram plot the curves y = ar* and y = {x + ly.How are they related? Show that they both pass through the point

(-4, tV)-12. On the same diagram plot the curves a; = l/y^andx = l/{y 2y.

How are they related? Show that both pass through (1, 1).13. On the same diagram plot the graphs of x + y = 2 and2x 3y = 1. What are they? Find their point of intersection. (It

must have coordinates satisfying both equations.)

Art. 20. Equation of a LocusIf a locus and an equation are such that (1) every point on the

locv,s has coordinates that satisfy the equation and (2) every pointwhose coordinates satisfy the equation lies on the locus, then the equa-tion is said to represent the locus and the locus to represent the equation.

Usually a locus is defined by a property possessed by each of itspoints. Thus a circle is the locus of points at a constant distancefrom a fixed point. To find the equa-tion of a locus express this property bymeans of an equation connecting thecoordinates of each locus point. Thegraph is constructed by using the defini-tion or by plotting from the equation.Example 1. Find the equation of a

circle with center C ( 2, 1) and radiusequal to 4.

Let P {x, y) (Fig. 2Qa) be any pointon the circle. By the definition of a circle, CPequivalent to

V{x + 2y +{y- ly = 4,

P (.x,y)

Fig. 20a.

4, and this is

which is the equation required. By squaring this can be reducedto the form(x + 2y + {y- ly = 16.

Ex. 2. A curve is described by a point P {x, y) moving in such away that its distance from the a-axis equals its distance from thepoint Q (1, 1). Construct the curve and find its equation.


62/218


'P (x,y)

Fig. 20b.

The curve is constructed by locating points such that MP = QP.Since no point below the a;-axis can be equally distant from Q and

the a;-axis, the curve lies entirely abovethe .r-axis. Hence y is positive andequal to the distance MP. AlsoQP = V{x - ly + (y - ly. Con-sequently, if P {x, y) is any point onthe curve,

2/ = V(a; - 1)2 + (2/ - V,\Conversely, if this equation is satis-

fied, the point P is equidistant from Q and the a;-axis. and so heson the curve. It is therefore the equation of the curve. Since ycannot be negative the equation is equivalent to

2/2 = (a; - 1)2 + (2/ - 1)2,and consequently to

a;2-2a;-22/ + 2 = 0.Exercises

1. What loci are represented by the equations, (a) a; = 3, (b) y = 2,(c) 2/ = 2 X, (d) x^+y^ = 1?

2. The point P (x, y) is equidistant from (1, 2) and (3, 4). Whatis the locus of PI Find its equation.

3. The point P {x, y) is twice as far from the s-axis as from the 2/-axis.What is the locus of P? (Two parts.) Find its equation.4. The slope of the Une joining P (x, y) to (2, 3) is equal to 3.What is the locus of P? Find its equation.5. Given A (-3, 1), B (2, 0), P {x, y). If the slopes of AB and BP

are equal what is the locus of P? Find its equation.6. The point P (x, y) is equidistant from the 2/-axis and the point

(3, 4). Construct the locus of P. Find its equation.7. Given A (2, 3), P (-1, 1), C (2, -3), P {x, y). If P moves alongthe line through A perpendicular to BC, show that

PC^ - PP2 = AC2 - AB'.Find the equation of the line described by P.

8. Find the equation of a circle with center ( 1, 2) and radius 5.9. Find the equation of the circle whose diameter is the segment

joining (2, 3) and ( 1, 4).


63/218


64/218

60 Rectangular Cocjrdinates Chap. 2

Fig. 21a.

intersection, solve tlie equations simultaneously. All solutionsshould be checked by substitution.Example 1. Plot the curves represented by the equations y^ = 2x

and x^ = 2y and find their points ofintersection.The curves are shown in Fig. 21a.

From the figure it is seen that thereare two real points of intersection.Squaring the first equation and substi-tuting the value of x'^ from the second,

Consequently, y {y^ 8) = 0. Thetwo real solutions are y = 0, 2. The

correspondiag values of x are a; = 1 2/^ = 0, 2. The points ofintersection are then (0, 0) and (2, 2). Substitution shows thatthe coordinates of thesepoints satisfy both equa-tions.Ex. 2. Plot the curves

xy = 2, x^ + y^ = A: x, andfind their points of intersec-tion.The curves are, shown in

Fig. 216. It is seen thatthere are two points of in-tersection. Elimination ofy gives

x*-4: x^ 4-4 = 0.The equation can only be solved approximately. The figure indi-cates that the solutions are near a; = 1.2 and a; = 4. By substitu-tion the following values are found:

a;= 1.1 1.2 3.9 4a;4_4a;3 + 4= 0.14 -0.84 -1.9 4

There is a root between 1.1 and 1.2 and another between 3.9 and 4.When X = 1.15 the expression a;* 4 a;' -F 4 is found to be negative.


65/218

Art. 22 Tangent Curves 51

N


66/218


67/218

Art. 22 Tangent Curves 5311. 2a;2+ /2 = 6, 13. 2/ = 2s',X y = 1. y = 3 x' 5.12 xh/ + 3y = 4, 14. x^ + 4 ?/ = 0,X +2y = 3. x'^ + y^ + 6x + 7 = 0.15. Show that the following curves have a common point:

a?+y^ = lQ, 2/3= a; + 4, 2/ = ^^.


68/218

CHAPTER 3STRAIGHT LINE AND CIRCLEArt. 23. Equation of a Straight Line

Let Pi (xi, ?/i) be a fixed point and P {x, y) a variable point on theline MN. If the line is not perpendicular to the x-axis (Fig. 23a),let its slope be m. Since the line passes through Pi and P, by thedefinition of slope,

m = X X\This is an equation satisfied by the coordinates, x and y, of anypoint P on the line. Conversely, if the coordinates of any point

Y


69/218

Art. 23 Equation of a Straight Line 55numerically equal to the distance along the ?/-axis to the Une, ispositive when the line is above the origin and negative when it isbelow. Equation (236) represents the line with slope m and inter-cept 6 on the 2/-axis.

If the line is perpendicular to the x-axis (Fig. 236) its slope isinfinite and equations (23a) and (236) cannot be used. In this casethe figure shows that

X = Xx. (23c)Conversely, if the abscissa of a point is Xi, it lies on the line. There-fore (23c) is the equation of a line through (xi, 2/1) perpendicular tothe X-axis.Example 1. Find the equation of a line through (1,2), the angle

from the s-axis to the line being 30.The slope of the line is

m = tan (30) = i V^.The equation of the line is then

2/ - 2 = i V3 (x + 1).Ex. 2. Find the equation of the line through the points (2, 0)

and (1, 3).The slope of the line is

=^^ = 4-3.1-2 ^Since the line passes through (2, 0) and has a slope equal to 3, itsequation is 1/ = 3 (x 2), whence

2/ = 3x 6.Ex. 3. Find the equation of the line through (1, 1) perpen-

dicular to the line through (2, 3) and (3, 2).The slope of the line through (2, 3) and (3, -2) is -5. The

slope of a perpendicular line is -l/(-5) = i. The equation ofthe line with this slope passing through (1, 1) is

2/ + l = i(x-l),which is the equation required.


70/218

56 Straight Line and Circle Chap. 3Ex. i. Show that the equation 2x 3y = 5 represents a

Straight line. Find its slope.Solving for y,

2/ = fa;-|.Comparing this with the equation y mx + b, it is seen that thetwo are equivalent if m = f , & = ^. Therefore the given equa-tion represents a straight line with slope f and intercept | onthe 2/-axis.

Exercises1. The angle from the a;-axis to a line is 60. The line passes through

( 1, 3). Find its equation.2. Find the equation of the line through (2, 1) and (3, 2).3. Find the equation of the line through (2, 3) and (2, 4).4. Find the equation of the line through (1, 2) parallel to the a;-axis.5. Find the equation of the perpendicular bisector of the segment

joining ( 3, 5) and ( 4, 1).6. An equilateral triangle has its base in the a;-axis and its vertex at(3, 5). Find the equations of its sides.7. A line is perpendicular to the segment between ( 4, 2) and

(2, 6) at the point one-third of the way from the first to the secondpoint. Find its equation.

8. Find the equation of the line through (3, 5) parallel to thatthrough (2, 5) and (-5, -2).

9. One diagonal of a parallelogram joins the points (4, 2) and(4, 4). One end of the other diagonal is (1, 2). Find its equationand length.

10. A diagonal of a square joins the points (1, 2), (2, 5). Find theequations of the sides of the square.

11. The base of an isosceles triangle is the segment joining ( 2, 3)and (3, 1). Its vertex is on the y-axis. Find the equations of itssides.

12. Show that the equation 2x y = S represents a straight line.Find its slope and construct the line.

13. Show that the equations2x + Sy = 5, 3x -2y = 7

represent two perpendicular straight lines.14. Perpendiculars are dropped from the point (5, 0) upon the sides

of the triangle whose vertices are the points (4, 3), (4, 3), (0, 5).Show that the feet of the perpendiculars lie on a line.


71/218

Art. 24 First Degree Equation 67Art. 24. First Degree Equation

Any straight line is represented by an equation of the first degreein rectangular coordinates. In fact, if the line is not perpendicularto the X-axis, its equation has been shown to be

V -yi = m{x- xi),xi, 2/1 and m being constants. If it is perpendicular to the x-axisits equation is

X = Xi.Since both of these equations are of the first degree in x and y, itfollows that any straight Line has an equation of the first degree inrectangular coordinates.

Conversely, any equation of the first degree in rectangular coordi-nates represents a straight line. For any equation of the first de-gree in X and y has the form

Ax + By + C = 0, (24)A,B,C being constant. If B is not zero, this equation is equivalentto A Cy=-^x~-,which represents a line with slope A/B and intercept C/B onthe y-ajds. If B is zero, the equation is equivalent to

C

which represents a line perpendicular to the x-axis passing throughthe point {C/A, 0). Hence in any case an equation of the firstdegree represents a straight fine.Graph of First Degree Equation. Since a first degree equation

represents a straight line, its graph can be constructed by findingtwo points and drawing the straight line through them. The bestpoints for this purpose are usually the intersections of the line andcoordinate axes. The intersection A (Fig. 24a) with the x-axis isfound by letting y = and solving the equation of the line for the


72/218

58 Straight Line and Circle Chap. 3corresponding value of x. Similarly, the intersection B with the2/-axis is found by letting x = and solving for y. The abscissa of

YS,B_

o s X

Y

o y

Fig. 24a. Fig. 246.

A and the ordinate of B are called the intercepts of the line on thecoordinate axes.

If the line passes through the origin (Fig. 246) the points A and Bcoincide at the origin and it is necessary to find another point on theline. This is done by assigning any value to one of the coordinatesand calculating the resulting value of the other coordinate.Example 1. Plot the line 2 a; + 3 ?/ = 6 and find its intercepts onthe axes.Substitutiag y = Q gives a; = 3 and substituting x Q gives

y = 2. Hence the liae passes through the points A (3, 0) andB (0, 2). Its intercept on the a-axis is 3 and on the 2/-axis 2.Ex. 2. Construct the line whose equation is 2a; 3?/ = 0.When X is zero y is zero. The line then passes through the

origin. When y = \, x = %. Hence the line OPi (Fig. 246)through the origin and the point (|, 1) is the one required.Slope of a Line. If the line is perpendicular to the a;-axis, its slope

is infinite. If it is not perpendicular to the a;-axis, its equation isy = mx + 6.

The slope is the coefficient of x in this equation. Consequently, ifthe equation of a line is solved for y, its slope is the coefficient of x.Example 1. Find the slope of the Une 3 a; 5 2/ = 7.

Solving for y y=^%x-i.The slope of the line is therefore f

.

Ex. 2. Find the angle from the line x + y = Z to the line2/ = 2 a; + 5.


73/218

Art. 24 First Degree Equation 59The slope of the first line is 1, that of the second 2. The angle

/3 between the lines is determined by the equation

The negative sign signifies that the angle is negative or obtuse.Ex. 3. Find the equation of the line through (3, 1) perpendicular

to the line 2 a; + 4 2/ = 5.The given equation can be written 2/ = | a; + |. Its slope is

consequently J. The slope of a perpendicular line is 2. Theequation of the liae through (3, 1) with slope 2 isy-l=2ix-3),which is the equation required.

ExercisesPlot the straight lines represented by the following equations, find

their slopes and intercepts:1. 2j/-3 = 0. 5. 3a;-6y + 7 = 0.2. 5x + 7 = 0. 6. 2x + 5y + 8 = Q.3. x + y = 2. 7. ix + 3y = 0.4. 2x + dy-5 = 0. 8. 3x-42/ = 6.9. Show that the equation

{2x + Zy-l){x-7y + 2) =0represents a pair of Hnes.

10. Show that (x + iyy = 9 represents two parallel lines.11. Show that x^ = (j/ 1)2 represents two lines perpendicular toeach other.

12. Show that the hnes 3x + 4:y -7 = 0, 9a; + 12j/-8 =are parallel.

. 13. Show that the lines x + 2y + 5=0, ix - 2y - 7 = a,Teperpendicular to each other.

14. Find the interior angles of the triangle formed by the HnesX = 0, x.-y + 2 = 0, 21 + 32/- 21 =0.

15. Find the equation of the line whose intercepts on the x and yaxes are 2 and 3 respectively.

16. Find the equation of the line whose slope is 5 and intercept onthe 2/-axis 4.

17. Find the equation of the Une through (3, 1) parallel to the lineX y = S.


74/218

60 Straight Line and Circle Chap. 318. Find the equation of the line through (2, 1) perpendicular to

the hne 9x-8y + 6-0.19. Find the projection of the point (2, 3) on the line a; 4 ?/ = 5.20. Find the equation of the line perpendicular to 3a; 5^ = 9and bisecting the segment joining ( 1, 2) and (4, 5).21. Find the lengths of the sides of the triangle formed by the lines

4:x + 3y-l=0, 3x-y-4 = 0, x + 4:y - 10 = 0.22. A line passes through (2, 2). Find its equation if the angle from

it to 3 X - 2 i/ = is 45.23. Find the equation of the line through (4, |) and the intersection

of the lines 3x - iy - 2 = 0, 9a;-ll2/-6=0.24. Find the equation of the line through the intersection of the lines

2a; ^ + 5=0, x + y + 1 = 0, and the intersection of the linesX - y + 7 =0, 2x + y-5 = Q.

25. Find the locus of a point if the tangents from it to two fixedcircles are of equal length.

26. What angle is made with the axis of j/ by a straight line whoseequation is ly + i x = 11

Art. 25. The Expression Ax + By + CAt each pointP of the plane a first degree expression Ax+By + C

has a definite value obtained by putting for x and y the coordinatesof P. Thus at the point (1, 2) the expression has the valueA + 2 B + C. Points where the expression is zero constitute aline whose equation is Ax + By + C = 0. If the point P movesslowly the value of the expression changes continuously. A numberchanging continuously can only change sign by passing through zero.

If the point P does not crossthe line the expression cannotbecome zero and so cannotchange sign. Therefore at allpoints on one side of the lineAx + By + C = the expres-sion Ax + By + Chas the samesign.

Fig. 25a. Example 1. Determine theregion in which x + y 1>0.The equation x + y 1 = represents the line LK (Fig. 25a).


75/218

Art. 26 Distance from a Point to a Line 61At all points on one side of LK the expression then has the samesign. At (1, 1)

a; + 2/ - 1 = 1 + 1 _ 1 = 1which is positive. It is seen from the figure that (1, 1) is above LK.Hence, at all points above LK, x + y l\s positive. At the originx+y-l =0+0-1 = -1which 'is negative. The origin is below the line. Hence at allpoints belowLK the expression x+ 2/ 1 is negative. The region inwhich a; + 2/ l>Ois thereforethe part of the plane above the line.Ex.2. Determine the region in

which x + y>Q, x + 2y-2 Qabove (1), x + 2y 2


76/218

62 Straight Line and Circle Chap. 3Since Q is on the line LK, its coordinates, x\ and MQ, must satisfythe equation of LK. Therefore

and consequently

(c)

Axi + B MQ + C = 0,Axi + CMQ B

id)

The slope of LK is tan = A/B, whenceBcos =

Substituting the values from (6), (c), (d) in (a),Axi + Bj/i + CDPt

=VA2 + B^ (26)Equation (26) gfiVes i/ie distance from the point (xi, j/i) to i/ie line

whose equation is Ax + By + C = 0. The distance being positive,such a sign must be used in the denominator that the result ispositive.Example 1. Find the distance from the point (1, 2) to the line

2x -By = 6.The distance from any point

(xi, 2/i) to the line is by (26)2xi-3y,-Q

-\/l3The distance from (1, 2) is then

2 (1) - 3 (2) - 6 10Fig. 26b. =Vl3 Via

Ex. 2. The lines (1) y - x - 1 = 0, (2) x + y - 2 = 0, (3)x + 2y + 2 = determine a triangle ABC. Find the bisector ofthe angle A between the lines (1) and (2) (Fig. 266).The bisector is a locus of points equidistant from the lines (1)


77/218

Art. 26 Distance from a Point to a Line 63and (2). If {x, y) is any point of the bisector, x and y must thensatisfy the equation

y X 1 _x + y 2V'2 V'2

The signs must be so chosen that these expressions are positive atpoints inside the triangle. At the origin these expressions become 1/(V2), 2/(V2). Hence the negative sign must be usedin both denominators. The bisector required is therefore

y X 1 X -\-y 2

5.


78/218

64 Straight Line and Circle Chap. 316. The sum of the distances from the point P (x, y) to the lines

2/ = a;, a; + 2/ = 4, y + 2 = 0, is 4. Find an equation satisfied by thecoordinates of P. Do aU points whose coordinates satisfy this equationhave the required property?

Art. 27. Equation of a CircleA circle is the locus of a point at constant distance from a fixed

point. The fixed point is the center ofthe circle and the constant distance is itsradius.

Let C {h, k), Fig. 27, be the center ofthe circle and r its radius. If P (x, y) isany point on the circle

r = CF = V{x-hy+(y-ky,or (x - hy +{y- ky = r\ (27a)

Y

P ix,v)

Fig. 27.This is an equation satisfied by the coordinates of any point on thecircle. Conversely, if the coordinates x, y satisfy this equation,the point P is at the distance r from the center and consequentlylies on the circle. Therefore it is the equation of the circle.Example 1. Find the equation of the circle with center (2, 1)

and radius 3. In this case, h = 2, k = 1, r = 3.By (27a) the equation of the circle is then

(x + 2)2 +(2/ -1)2 = 9.Ex. 2. Find the equation of the circle with center (1, 1) which

passes through, the point (3, 4).The radius is the distance from the center to the point (3, 4).

Consequentlyr = V(3-l)2+(-4-l)2 = Vig.

The equation of the circle is therefore(x - ly +iy- ly = 29.

Form of the Equation. Expanding (27a),xi + tf-2hx- 2ky+h'^ + k^-r^ = 0.


79/218

Art. 27 Equation of a Circle 65This is an equation of the form

A iz' + y'')+Bx + Cy + D = 0, (276)in which A, B, C, D are constant.Conversely, if A is not zero, an equation of the form (276) repre-sents a circle, if it represents any curve at all. To show this divideby A and complete the squares of the terms containing x aiid thosecontaining y separately. The result is

E' + C^-iADh^lj+h^ljIf the right side of this equation is positive, it represents a circlewith center (-B/2 A, -C/2 A) and radius Vb^ + C - iAD/2 A.If the right side is zero, the radius is zero and the circle shrinks to apoint. If the right side is negative, since the sum of squares of realnumbers is positive, there is no real locus.Example 1. Find the center and radius of the circle

2a;'' + 22/2-3a; + 42/ = 1.Dividing by 2 and completing the squares,

{x - lY + (2/ + 1)^ = \%.The center of the circle is (|, 1) and its radius is \ V33.

Ex. 2. Determine the locus ofa;2-|-2/'' + 4a;-62/-|-13 = 0.Completing the squares,

{x + 2Y + (2/ - 3)== = 0.The sum of two squares can only be zero when both are zero. Theonly real point on this locus is then (2, 3). The circle shrinksto a point.Ex. 3. Discuss the locus of

x2 + 2/2 + 22/ + 3 = 0.Completing the squares,

x^ + (2/ + 1)^ = -2.There are no real values for which this is true. The locus isimaginary.


80/218

66 Straight Line and Circle Chap. 3

Art. 28. Circle Determined by Three ConditionsWhen a circle is given, h, k, r, in equation (27a), have definite

values. Conversely,

geometric analytic h b phillips

Documents