Coordinate Geometry Loney SL

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(5) —. —. BY THE SAME AUTHOR. A. Treatise on Elementary Dynamics. Third Edition, Revised and Enlarged.. Crown. 7s.. 8vo.. 6 c?.. Solutions of the Examples in the Elementary Dynamics. Crown 8vo. 75. Qd.. The Elements. of Statics and Dynamics. Fcap. 8vo. Elements of Statics. Fourth Edition, 4s. 6d. II. Elements of Dynamics. Fourth Ed., 3s. Qd. two Parts bound in one Volume. The 7s. Qd.. Part Part. I.. "Mr Loney shows that he knows how to combine perspicuity with brevity in a remarkable degree. One feature of both books is that the author points out the portions that are adapted for a first reading, and also those that are required for particular examinations." GUdTd'tQjTt. PART. ELEMENTS OF. I.. STATICS.. "Students reading for the different examinations at Cambridge, for the London University Matriculation, and intermediate Science, for the Woolwich Entrance Examinations, will fi^nd a statement of the part of the book to be read on this subject. Mr Loney deserves much praise for the uniformly high standard he has shown in this his Glasgow Herald. latest effort.". and. Solutions of the Examples Statics. and Dynamics.. in. the Elements of. Ex. Fcap. Svo.. 7s. Qd.. Mechanics and Hydrostatics for Beginners. Fcap. 8vo.. Third Edition.. Ex.. 3s. Qd.. Parts. Crown 8vo. Second and including the Solution of. Plane Trigonometry. In two Edition. Part Triangles. 5s.. Part. II.. tions.. De. I.. up. to. Moivre's Theorem and the higher por-. 3s. 6d.. The two Parts bound. in one. Yolume.. 7s. Qd.. NATURE. says ; " ...It would be difficult to find a better introduction to Plane Trigonometry.". SCHOOL GUARDIAN. says; " ...It is a model of method, and the simplicity of its explanations, its numerous illustrative diagrams, and its large and varied collection of exercises and examples combine to render it especially suitable for the use of young pupils and private students.... The typography of the volume is in every respect admirable....". The. clearness,. and accuracy. ;.

(6) ". ". The SCHOOLMASTER says; " ...Mr Loney, using the soundest judgment in his choice of matter, presents it with inimitable brevity and clearness ; the publishers, too, have vied with the author in their efforts to excel, and, as a result, the book is in every way worthy of commendation." The SCOTSMAN says " ...It expounds the subject with a skill and fulness which give evidence of a peculiar experience in teaching and of a special appreciation of the needs of students. ;. The. PRACTICAL TEACHER. competitors. it. cannot. says; "...Among its numerous a deservedly high and popular. fail to attain. position.". The NATIONAL TEACHER (Ireland) says '• ...Seldom do such books come under the notice of the reviewer. For conciseness and clearness it has few compeers. It is deep without dullness ; comprehensive without wearisomeness. It comes before us with a newness and freshness almost amounting to novelty...." ;. The SPEAKER says ; "Mr Loney as a writer of elementary mathematical treatises maintains a high standard. His Elementary Dynamics is marked by its brevity and clearness and deserves its success.... The Cambridge Press has every reason to be proud of its achievement.". The made a. EDUCATIONAL REVIEW. says;. "...The author has not. special point of developing any one particular branch of the subject, but his work is of uniform character throughout, and, we may add, of uniformly good quality.... The arrangement of sections is excellent ; the attention is clearly directed to important points, and the style in which the book is produced may be fairly described as. luxurious.". GLASGOW HERALD. says ; " ...Mr Loney's text-book is sure The to be widely used in the several mathematical classes where sound work The publisher's part of the work is also most admirably is expected.. done.. CORRESPONDENT. says; "...It is well The UNIVERSITY written and the chapters relating to the changes of sign and magnitude of trigonometrical ratios in different quadrants are especially deserving of praise....". SCIENCE AND ART. says; "...In the analytical part of the subject, considerable attention has been given to complex quantities, and the author is to be congratulated on the lucid way in which he has treated them.... The book may be strongly recommended as a first-rate text-book.". aontion:. C.. J.. CLAY and SONS,. CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AVE MARIA LANE. (ffilagfloia:. 263,. ARGTLE STREET..

(7) THE ELEMENTS OP. COOEDINATE GEOMETRY..

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(9) THE ELEMENTS OF. COOEDINATE aEOMETRY. BY. S.. L.. LONEY,. M.A.,. LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE, PROFESSOR AT THE ROYAL HOLLOWAY COLLEGE.. ^^^l^^mU. MASS. MATH, DEPTi. MACMILLAN AND AND NEW YOEK. 1895 [All Bights reserved.']. CO..

(10) CTambrtlJse:. PRINTED BY. J.. &. C.. F.. CLAY,. AT THE UNIVEBSITY PRESS.. 150553.

(11) PKEFACE. "TN. the following work I have tried to present the. elements of Coordinate Geometry in a manner suitable. present. Junior Students.. The. book only deals with Cartesian and. Polar. Beginners and. for. Within these. Coordinates.. that the book. is fairly. limits I venture to. complete, and that no proposi-. tions of very great importance have. The Straight Line and more. fully. since. it. is. hope. been omitted.. Circle have been treated. than the other portions of the subject, generally in the. elementary conceptions. that beginners find great difficulties.. There are a large number of Examples, over 1100 in. all,. and they. character.. are,. in. The examples. general, of an are especially. the earlier parts of the book.. elementary. numerous in.

(12) PREFACE.. vi. for. reading. portions of the proof sheets, but especially to. Mr W.. I. J.. am much. indebted to several friends. Dobbs, M.A. who has kindly read the whole of the. book and made many valuable suggestions. For any shall. be. criticisms,. suggestions, or. grateful. S.. EoTAIi. corrections, I. HOLLOWAY COLLEGE,. Egham, Surbey. July. 4, 1895.. L.. LONEY..

(13) CONTENTS. CHAP. I.. II.. Introduction.. .... Algebraic Kesults. Coordinates. Lengths of Straight Lines and Areas of Triangles Polar Coordinates. Equation to a Locus. Locus.. IV.. The Straight Line. Eect angular Coordinates. .... .. .. .. .. .. .. Length of a perpendicular. .. .. .. Line.. 51. .. .. .. .. .. .. Equations representing two or more Straight Lines lines given. by one equation. Greneral equation of the second degree. VII.. 66. .. loci. Angle between two. Transformation of Coordinates Invariants. 42. Polar Equations and. Equations involving an arbitrary constant. VI.. 39. 58. Oblique Coordinates Examples of. 31. .44. .. .. ". .. Bisectors of angles. The Straight. 8. 24. Straight line through two points Angle between two given straight lines Conditions that they may be parallel and per-. V. 1. 19. III.. pendicular. PAGE. .. .. .. .. .. 73 80. 88 90 94 109 115.

(14) .. CONTENTS.. Vlii. PAGE. CHAP.. VIII.. The Circle. 118. Equation to a tangent. 126 137. Pole and polar. Equation to a circle in polar coordinates Equation referred to oblique axes Equations in terms of one variable .. .150. .. 160. Systems of Circles Orthogonal circles. X.. 148. .. .. .. IX.. .145. .. .. .. .. ,. .. .160. .. Kadical axis. 161. Coaxal circles. 166. The Parabola. Conic Sections.. 174. .. 180. Equation to a tangent. Some. properties of the parabola. .. 187. .. .. 190. Pole and polar. 195. Diameters Equations in terms of one variable XI.. The Parabola. {continued'). .. .198. .. ..... Loci connected with the parabola Three normals passing through a given point Parabola referred to two tangents as axes .. .. .. 211. The Ellipse. 225. Auxiliary circle and eccentric angle. Equation to a tangent. Some. Conjugate diameters. 237. ....... 249. .. Pour normals through any point Examples of loci XIII.. .231. .. .. ..... .. properties of the ellipse. Pole and polar. 206. .217. .. XII.. 206. .. .. .. .. .. .. 254 265. 266. The Hyperbola Asymptotes Equation referred to the asymptotes as axes One variable. Examples. 242. 271. 284 .. 296 299.

(15) .. CONTENTS. CHAP.. XIV.. Polar Equation to, a Conic. IX. ..... Polar equation to a tangent, polar, and normal. XV.. General Equation.. Tracing of Curves. Particular cases of conic sections. PAGE. 306 313. ,. 322. .... .. 322. Transformation of equation to centre as origin Equation to asymptotes Tracing a parabola. ...... ...... ....... Tracing a central conic. .. .. .. .. .. .. Eccentricity and foci of general conic. XVI.. General Equation Tangent Conjugate diameters. 326. .. 329 332 338 342. 349 349 352. Conies through the intersections of two conies. 356. The equation S=Xuv. 358. ...... ........ General equation to the. j)air. of tangents. drawn. from any point. The director The foci The axes. circle. 367 369. Lengths of straight lines drawn in given directions to meet the conic Conies passing through four 23oints .. .. .. Conies touching four lines. The. LM=B?. conic. XVII. Miscellaneous Propositions. On. 364 365. ..... 370 378 380 382 385. the four normals from any point to a central conic. Confocal conies. ........ Circles of curvature. and contact of the third order. 385. .. Envelopes. Answers. 392 398 407. .. i. —. xiii.

(16) ERKATA. Page „ ,,. ,,. 87,. Ex.. 27, line 4.. 235, Ex. 18, line 3.. „. ,,. 282, Ex.. ,,. 3.. line 5.. For "JR" read " S.". For "odd" read "even." Dele. "and Page. 37,. Ex. 15.". For "transverse" read "conjugate.".

(17) CHAPTER. I.. INTRODUCTION.. SOME ALGEBRAIC RESULTS. 1.. Quadratic Equations.. The. roots of the quad-. ratic equation a'3^. may. easily. + 6x + c =. be shewn to be. -. &. +. •JlP'. — 4ac. 1. -b- s/b^ — 4:aG. '^"'^. •. 2^. 2i.. They are. and unequal,. equal, or imaginary, according as the quantity b^—iac is positive, zero, or negative,. therefore real. i.e.. 2.. and. according as. b^. =. 4:ac.. Relations between the roots of any algebraic equation of the terms of the equation.. the coejicients. If any equation be written so that the coefficient of the highest term is unity, it is shewn in any treatise on Algebra that. the sum of the roots is equal to the coefficient of (1) the second term with its sign changed, (2). the. at a time,. is. sum of the products of the roots, taken two equal to the coefficient of the third term,. the sum of their products, taken three at a time, (3) equal to the coefficient of the fourth term with its sign changed, is. and so L.. on.. e. 1.

(18) ,. .. COORDINATE GEOMETRY. Ex.. If. 1.. a and. /3. ax'^. 2.. If a,. b. + bx + c = 0,. i.e.. x^. c. + - x + ~ = 0, a. — b. a + p=. we have. Ex.. be the roots of the equation. a. c. ^. and a^ = -.. and 7 be the roots of the cubic equation. j8,. ax^ + bx^ + cx + d=0, i.e.. x^+-x^. of. a. we have. +-x + - = 0, a. a. + p + y:. a. ^y + ya + a^=:and. o-Pl-. 3.. shewn that the solution. It can easily be. of the. equations. +. a^x. h^y. +. G^z. = 0,. a^ + h^y + c^z = 0,. and. X IS. y. ~ ^2^1. ^1^2. ^1^2 ~ ^2^1. '^1^2. ~ ^2^1. Determinant Notation. 4.. The. is called. quantity-. a determinant of the. second order and stands for the quantity. Exs.. (1). d-yf. d^. ^1,. h. a-})^. — aj)^,. so that. = Ob^^ — 6»2&i. \%^\ = 2x5-4x3 = 10-12=-2; ;'. |. !4, 5i. 3, (ii). -7,. -4|. -6 = 3. X. (. -. 6). -{-. 7). X. (. - 4) = 18 - 28 = -. 10..

(19) DETERMINANTS. «!,. 5.. The quantity.

(20) COORDINATE GEOMETRY.. 8.. The quantity j. (h.1. ^2>. %J. ^4. 61,. &2). hi. h. ^11. ^25. ^1) ^2. ^3> 5. ^3) ^4. called a determinant of the fourth order and stands for the quantity. is. K «i X. ^2». <^2>. h, ^3. ^4. J. — Clo. ^3> ^4. X. i^lJ. ^35. C-,. ^3}. \. 1. 1. &i,. + 6^3. h 4. 33. 5. 62J. X. ^4! C^. 1. ?. 2. 5. 4. &1, cCj_. X.

(21) .. ELIMINATION.. 5. Elimination. Suppose. 11.. we have. the two equations. + a^y =. aj^x. (1),. \x +b^y ^0 two unknown quantities x and. between the be some relation holding between the four ôr, from (1), we have bi, and 63. (2),. y.. There must. coefficients. 6*i, ctaj. •. and, from. (2),. we have. y~. %'. -=— y. K. Equating these two values. i.e.. a-J)^. =-^. X. of -. we have. — ajb^ =. (3).. The result (3) is the condition that both the equations and (2) should be true for the same values of x and y. The process of finding this condition is called the eliminating of X and y from the equations (1) and (2), and the result (3) is often called the eliminant of (1) and (2). Using the notation of Art. 4, the result (3) may be (1). written in the form. 1. ). '^. 0.. is obtained from (1) and (2) by taking the x and y in the order in which they occur in the equations, placing them in this order to form a determinant, and equating it to zero.. This result. coefficients of. 12.. we have a-^x + a^y + a^^ = \x+ h^y^ h^z = G^x + G^y + C3S = unknown quantities. Suppose, again, that. and between the three. the three equations (1),. (2),. (3), x, y,. and. z..

(22) COORDINATE GEOMETRY.. 6. By. dividing each equation by z. we have. three equations. X. between the two unknown quantities — and y. z. %,. z. Two. of.

(23) ELIMINATION. 14.. and. If again. we have. the four equations. a-^x. +. dil/. +. cf'zZ. + aû =. 0,. h^x. +. h^y. +. b^z. +. bû. =. 0,. Ci«;. +. cî/. +. G^z. +. cû. =. 0,. djX + d^y. +. d.^z. +. d^ — 0,. could be shewn that the result of eliminating the four quantities cc, y, z^ and u is the determinant. it. «1J.

(24) CHAPTER COORDINATES.. II.. LENGTHS OF STRAIGHT LINES AND AREAS OF TRIANGLES.. OX. 07. and 15. Coordinates. Let be two fixed The line is straight lines in the plane of the paper. the axis of y, whilst the called the axis of cc, the line two together are called the axes of coordinates.. OX. OY. The point. is. called the origin of coordinates or,. more. shortly, the origin.. F. From any point in the plane draw a straight line. OF to meet OX M. The distance OM is called the Abscissa, and the distance. parallel to. in. MP the Ordinate of the point P, whilst the abscissa and the ordinate together are called its Coordinates.. OX. Distances measured parallel to or without a suffix, {e.g.Xj, x.-^... x\ measured parallel to OY are called suffix, (e.g.. 2/i, 2/2,---. If the distances. the coordinates of. 2/'.. y",---)-. OM and MP. P are,. are called a?, with x",...), and distances y, with or without a. be respectively x and ?/, by the symbol. for brevity, denoted. {x, y).. Conversely, when we are given that the coordinates of For from we a point are (x, y) we know its position. {—x) along and have only to measure a distance. P. OM. OX.

(25) COORDINATES. then from 21 measure a distance. 9. MP. OY. {=y) parallel to arrive at the position of the point P. For example be equal to the unit of length and in the figure, if. and we. MP= WM,. OM. P is the point (1, 2). Produce XO backwards to form then. the line OX' and backwards to become OY'. In Analytical Geometry we have the same rule as to signs that the student has already met with in Trigonometry. Lines measured parallel to OX are positive whilst those measured parallel to OX' are negative ; lines measured parallel to OY are positive and those parallel to OY' are. 16.. YO. negative.. quadrant YOX' and P^M^, drawn y, meet OX' in M^^ and if the numerical values of the quantities OM^ and J/aPg be a and h, the coordinates of P are {-a and h) and the position of Pg is given by the symbol (—a, h). If. P2 b® i^. *li®. parallel to the axis of. Similarly, if P3 be in the third quadrant X'OY', both of coordinates are negative, and, if the numerical lengths of Oi/3 and J/3P3 be c and d, then P3 is denoted by the. its. symbol (—. c,. Finally,. positive 17.. and Ex. (i). —. d).. in the fourth quadrant its abscissa is its ordinate is negative.. if. P4. lie. Lay down on "paper (2,. -1),. (ii). (-3,. the position of the points 2),. and. (iii). (-2, -3).. To get the first point we measure a distance 2 along OX and then a distance 1 parallel to OF'; we thus arrive at the required point. To get the second point, we measure a distance 3 along OX', and then 2 parallel to OY. To get the third point, we measure 2 along OX' and then 3 parallel to OT. These three points are respectively the points P4 P., and Pg in ,. ,. the figure of Art. 15.. 18. When the axes of coordinates are as in the figure of Art. 15, not at right angles, they are said to be Oblique Axes, and the angle between their two positive directions and 07, i.e. the angle XOY, is generally denoted by. OX. the Greek letter. w..

(26) COORDINATE GEOMETRY.. 10 In general,. it is. however found. to. OX. be more convenient to They are then. and OZat right angles. take the axes said to be Rectangular Axes.. It may always be assumed throughout this book that the axes are rectangular unless it is otherwise stated.. The system. spoken of in the last System of CoordiIt is so called because this system was first intronates. duced by the philosopher Des Cartes. There are other systems of coordinates in use, but the Cartesian system is by far the most important. 19.. few. articles is. To find. 20.. of coordinates. known. as the Cartesian. the distance between. two points whose co-. ordinates are given.. Let Pi and P^ be the two given points, and let their coordinates be respectively {x^ y^) ,. and. (a^sj 2/2)-. Draw rallel. J/j. to. to. Pji/i and P^M^ paOY, to meet OX in. and M^.. Draw P^R parallel. OX to meet M-^P^ in R.. '. q. M. jvT. Then. P^R = M^Mt^ = OM^ - OMc^ = oi^-X2, RP, = M,P,-M,P, = y,~y,, and. z Pî^Pi. = z6>ifiPa-l 80° -PiJfiX^l 80° -<o.. We therefore have P^P^^. [Trigonometry, Art. 164]. = P^R^ + RP^^ - 2P^R PPi cos P^RP^ .. - (^1 - x^Y + (2/1 - 2/2)' - 2. (aî. - x^). (2/1. - 2/2) cos. (180°. - (o). = (Xi-X2)2 + (yj_y2)2+2(Xi-X2)(yi-y2)COSCO...(l). If the axes be, as is generally the case, at right angles,. we have. <o ==. 90°. The formula. and hence cos to = (1). 0.. then becomes. P^P^ -. (x,. - x^Y +. (2/1. - y^Y^.

(27) DISTANCE BETWEEN TWO POINTS.. 11. SO that in rectangular coordinates the distance between the two points (x^j y^ and (a-g, 2/2) is. V(Xi. - x^)^ +. (Yi. - y^)^. (2).. (x^, y-^ from the origin Cor. The rectangular. being the axes This follows from + 2/1^, Jx^ equal to zero. both and by making x^ (2) y^. distance of the point. is. The formula of the previous article has been proved for the when the coordinates of both the points are all positive.! Due regard being had to the signs of the coordinates, the formula 21.. case. will be found to be true for all points.. As a numerical example, let Pj be the point (5, 6) and Pg be the point (-7, -4), so that we have and Then. y2= -^.. P^ = 31^0 + OM^ = 7 + 5 and RPt^ = EM-^ + l/iPj = 4 + 6. = The. Yi. -2/2 +. 2/1.. rest of the proof is as in the last article.. Similarly any other case could be considered.. 22. To find tJie coordinates of the point which divides in a given ratio (ni^ m^ the line joining two given jyoints :. (a?!, 2/1). and. (x^, y^).. O. M,. Let Pi be the point. M {x^, y^),. the required point, so that. M,. X. Po the point. we have. (x^, y^),. and. P.

(28) COORDINATE GEOMETRY.. 12. Let P be the point (sc, y) so that if P^M^, PM, and P^M^ be drawn parallel to the axis of y to meet the axis of £C in i/i, Mj and M^, we have Oi/i. MP = y,. = £Ci, M^P^ = y^, OM=x,. and. =. i/^zâ. QM^^x^,. 2/2-. Draw PiEi and P-Sg, parallel to OX, to meet J/P and M^P^ in Pi and Pg respectively. Then PjPi = M^^M^^ OM- OM^ = x-x^, PR^ = MM^ = OJ/2 - 0M= x,^ - X, R,P^MP-M,P, = y-y,, P2P2 = M^P^ -. and .. From. PiPjP and PR^P^ we have X — Xt^ PiRi. the similar triangles. PjP PP^. m^ m^. ,. x=. t.e.. Again -. MP = y^-y.. *. PR^. ifv-tt/Uey *T". x^. i/VoOO-i. ^^. .. P,P R,P PP2 P2P2 mi (3/2 - 3/) = 7^2. y-y.. mi m^. so that. y=. and hence. -^^ Wi +. — x'. 2/2-2/'. {y. - 3/1),. ^-^ .. 7?22. The coordinates of the point which divides PiP^ internally in the given ratio rrii tyi^ are therefore :. nil If the point. mi + nig. + ^2 Q. divide the line P1P2 externally in the :: mj mî its coordinates. same ratio, i.e. so that P^Q QP^ would be found to be :. nil. The proof. "". *. ^^2. :. 'î. " ^^2. of this statement is similar to that of the. preceding article and. is left. as. an exercise. for the student..

(29) .. LINES DIVIDED IN A GIVEN RATIO.. Cor. joining. Ex.. D is B. Take. the. ABC 'prove. 5C. D is. that. AB^ + AC^ = 2 {AD^ + DG^), middle point of BG. and a. line through. B i>er-. BC. BG=a,. Then. In any triangle. 1.. as the axis of x, as origin, as the axis of y.. pendicular to. Let. of the line. {x^,. 23. lohere. The coordinates of the middle point y^ to {x^, y^ are. 13. so that. G. the point. is. the point (|>. C>. ^D2=ra;i. Hence. 2 (^D^ + DC^). and. let. A. be the point. j. -^Y + i/i^. Hence. (a, 0),. ::=. DG^=f~y.. and. 2 ["x^^ + y^^ - ax^ + ^~|. = 2xi2 + 2yi2_2o.x.^ + a2. Also. ^C'2.= (a;i-a)2. and. + ?j^2^. AB^=^x^-\-y^.. Therefore. AB'^ + ^(72 = 'Ix^ +. This. is. 2?/i2. _. + a^.. 2aa;i. ^52 + ^(72^2(^2)2 + 2)(72)_. Hence. the well-known theorem of Ptolemy.. Ex. 2. ABG is a triangle and D, E, and F are the middle points of the sides BG, GA, and ; prove that the point lohich divides 1 also divides the lines BE and GF in internally in the ratio 2. AD. AB :. the same ratio.. Hence prove that. the medians of a triangle meet in a point.. Let the coordinates of the vertices A,. and. (ajg,. The. coordinates of. Let. G. and. let its. By. J5,. and G be. D are therefore. -^. and. ^. be the point that divides internally coordinates be x and y.. AD. the last article. 2Xiy "T Xq. ^_ So. (x-^, y-^), (arg, 2/2),. y^) respectively.. 2 ;. 2+1. ^ ^1 + 3^2 + ^3 3. ^^2/rt^±^3. 3. -. ^^ .. in the ratio 2. :. 1,.

(30) ;. COORDINATE GEOMETRY.. 14. In the same manner we could shew that these are th^ coordinates and CF in the ratio 2 : 1. of the points that divide Since the point whose coordinates are. BE. x-^. + x^ + x^ and ^L±^2+l_3 3. 3 lies. on each of the lines meet in a point.. AD, BE, and CF,. it. follows that these three. lines. This point. is called. the Centroid of the triangle.. EXAMPLES.. I.. Find the distances between the following pairs of. 3_. and (5, 7). _ _ 3, 2) and ( - 6, (. 4.. (a, o). 6.. {a cos a, a sin a). 1.. 2.. (2, 3). {am^^,. 7.. and. -7) and (-1,. (4,. 5).. the axes being inclined at 60°.. 7),. (o, 6).. 5.. and. points.. {a cos. |S,. {b. + c, c + a) and. a sin. {c. + a,. a. + b).. /3).. 2ami) and (am^^ 2am^.. in a figure the positions of the points (1, - 3) and that the distance between them is 5. prove and ( 2,1), Find the value of x^ if the distance between the points [x^^, 2) 9. and (3, 4) be 8. 8.. Lay down. - 3) 10. A line is of length 10 and one end is at the point (2, the abscissa of the other end be 10, prove that its ordinate must be 3 or - 9.. if. 11. Prove that the points (2a, 4a), (2a, 6a), and (2a + s/3a, oa) are the vertices of an equilateral triangle whose side is 2a.. 12.. Prove that the points (-2, -1),. (1, 0), (4, 3),. and. (1, 2). are. at the vertices of a parallelogram.. 13.. Prove that the points. (2,. -2),. (8, 4),. (5, 7),. and (-1,. 1) are. at the angular points of a rectangle.. 14. Prove that the point ( - xV. f I) is the centre of the circle circumscribing the triangle whose angular points are (1, 1), (2, 3),. and ( -. 2, 2).. Find the coordinates 15. ratio 3. 16.. of the point. which. divides the line joining the points :. (1,. 3). and. (2,. 7) in. the. 4.. divides the. same. line in the ratio 3. 17. divides, internally to (4, - 5) in the ratio 2. and :. 3.. :. -. 4.. externally, the line joining. (. -. 1,. 2).

(31) .. [EXS.. - 8,. (. 15. divides, internally and externally, the line joining the ratio 7 : 5.. 18. to. EXAMPLES.. I.]. ;. (. -. - 4). 3,. 7) in. 19. The line joining the points (1, - 2) and find the coordinates of the points of trisection.. (. - 3,. 4) is trisected. 20. The line joining the points ( - 6, 8) and (8, - 6) is divided into four equal parts ; find the coordinates of the points of section.. rind the coordinates of the points which divide, internally externally, the line joining the point {a + b, a-h) to the point (a-&, a + 6) in the ratio a h. 21.. and. :. The coordinates of the vertices of a triangle are [x-^, y^ and (xg, y^. The line joining the first two is divided in. 22. {x^,. 2/i)». the. and the. line joining this point of division to the opposite angular point is then divided in the ratio : + Z. Find the coordinates of the latter point of section. ratio. I. :. h,. m. Jfc. 23. Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation. x-y + l=:0. If. 24.. G. be the centroid of a triangle. ABC and O. be any other. point, prove that. ^{GA^-^GB'^+GCr-) = BG^+GA^ + AB\. OA^ + OB^-\-OG'^=GA^ + GB'^+GG^- + ^Ga\. and. 25. Prove that the lines joining the middle points of opposite sides of a quadrilateral and the Une joining the middle points of its diagonals meet in a point and bisect one another.. B, G, D... are n points in a plane whose coordinates are 2/2)' (^3> 2/3)j---* -^-S is bisected in the point G-^; G^G is (î» 2/i)> divided at G^ in the ratio 1:2; G^D is divided at G^ in the ratio 1:3; GgE at G^ in the ratio 1 4, and so on until all the points are exhausted. Shew that the coordinates of the final point so obtained are 26.. -4,. (^2'. :. ^1 + ^2. + 3^3+ •••+^n. ^j^^ yi. + y^ + Vz+'-'-^Vn. n [This point. is. n. called the. Centre of Mean Position of the n given. points.]. 27. Prove that a point can be found which distance from each of the four points. (am,,. ^). ,. (a»„. ^). ,. {am,,. ^J. ,. and. is. at. {am^m^,. the. same. ^^). To prove that the area of a trapeziitm, i. e. a quadhaving two sides parallel, is one half the sum of the two parallel sides multiplied by the perpendicular distance 24.. rilateral. between them..

(32) COOEDINATE GEOMETRY.. 16 Let. BC. ABGD. be the trapezium having the sides. AD. and. parallel.. AC and draw AL perpento BG and ON perpendicular. Join. dicular to AD^ produced if necessary. Since the area of a triangle is one B half the product of any side and the perpendicular drawn from the opposite angle,. area. we have. ABGD = ÂBG ¥ ÂGD = l.BG ,AL + l,AD.GN =î{BG + AD). AL.. X. 25. To find the area of the triangle^ the coordinates of whose angular 'points are given^ the axes being rectangular. Let ABG be the triangle and let the coordinates of its angular points A, B and G be {x^, 2/1), (a?2, 2/2), and {x^, y^). Draw AL, BM, and Ci\^ perpendicular to the axis of x, and let. A. denote the required area.. Then. A == trapezium A LNG + trapezium GNMB — trapezium A LMB = \LN {LA + NG) + \NM {NG + MB) - \LM (LA + MB), by the. last article,. = i [(^3 - ^1) (2/1 + ys) + {002 - ^z) (2/2 + 2/3) - (^2 - aî) {Vi + 2/2)]On simplifying we easily have ^ ^ I (^172 - XaYi + y^zYz - ^372 + XgYi - x^yg), or the equivalent form. ^=J. [^1. (2/2. - Vz) +. ^2. (2/3. - 2/1) + ^3. (2/1. we. use the determinant notation this (as in Art. 5) If. ,. Cor.. The area of the and the points. origin (0, 0). ^1). 2/1). ^2>. 2/2?. ^3j. 2/35. - 2/2)]. may be. written. ^ '-. ^. triangle. whose vertices are the. {x^, y-^, {x^,. 2/2). is. J. (^12/2. — ^'22/1)*.

(33) AREA OF A QUADRILATERAL,. 17. 26. In the preceding article, if the axes be oblique, the perpendiculars AL, BM, and CN, are not equal to the ordinates y^ y^ and 2/3, but are equal respectively to yi sin w, 2/2 sin w, and y^ sin w. ,. ,. The area. of the triangle in this case becomes. isin. w. {x^y^. xu fsm. I.e.. yi, 1. wx. 27. In order that the expression for the area in Art. 25 may be a positive quantity (as all areas necessarily are) the points A, B, and G must be taken in the order in which they would be met by a person starting from A and walking round the triangle in such a manner that the area of the triangle is always on his left hand. Otherwise the expressions of Art. 25 would be found to be negative.. 28. To find the area of a quadrilateral the coordinates of whose angular foints are given.. Let the angular points of the quadrilateral, taken in be A^ B, C, and D, and let their coordinates be. order,. respectively {x^, y^\. Draw ALy BM,. (x.^,,. CJV,. y^), (x^,. and. DH. y^\ and. {x^, y^).. perpendicular to the axis. of X.. Then the area. of the quadrilateral. = trapezium ALRD + trapezium BRNO + trapezium GNMB — trapezium ALMB. = ILR {LA + RD) + IRN (RD + NG) + ^NM {JVC + MB) -lLM{LA-\-MB) _ 1 {(^4 - ^1) + + + (^2 - ^^) (2/3 + 2/2) (^3 ^4) (2/3 + 2/4) (2/1 2/4) - (a?2 - x^) (2/1 + 2/2)} _ 1 {(^12/2 - «^22/i) + fez^s - ^zvi) + {xsy4. - ^42/3) + (^42/i - ^y^)}' L.. 2.

(34) COORDINATE GEOMETRY.. 18. 29. The above formula may also be obtained by drawing the lines OA, OB, OC and OD. For the quadrilateral. ABCn = AOBG+ AOCD- aOBA- AOAD.. But the coordinates of the vertices of the triangle OBG are (0, 0), (ajg, 2/2) ^^^ (^35 2/3) ^ hence, by Art. 25, its area is ^ i^^y-^ — ^zV^)So for the other triangles.. The required area therefore. ^ h [(«^22/3 - a^32/2) + {^zVa, - ^m) - (^Wi - ^12/2) - {î2/4 - ^'42/l)l = i [{^^2 - ^22/1) + (^22/3 - ^32/2) + {^3y4 - ^m) + (^42/l - «^l2/4)]In a similar manner it may be shewn that the area of a polygon of n sides the coordinates of points, taken in order, are (ÎJ. 2/1/5. V^2). (^35. 2/2/3. 2/3)? •••(.'^)i3 2/ra/. i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) +. is. whose angular. • • •. EXAMPLES.. +. (^n2/i. - a'l^/ri)]-. II.. Find the areas of the triangles the coordinates of whose angular points are respectively 1.. (1, 3),. (. -. and. 7, 6). (5,. 3.. 4.. {a,. 5.. {a,c. 6.. {a cos. 7.. (a7?ij2^ 2a7?i;^),. 8.. {awiimg, a(??ii + ??i2)},. 9.. l>. -. 1).. 2.. (0, 4), (3, 6). and. (. -. 8,. - 2).. (-9, -3) and (-3, -5). + c), (a, h-c) and {-a, c).. (5,2),. + a), (pi,. lam-,.. —. {a, c). <p-^),. {a cos ^^, b sin ^g). [arn^^. lam^ and [am^,. b sin. I. ,. and {-a, c-a).. \am^,. {aWaWg, a. —. y. and. (7?i2. and. (a cos ^3, b sin ^g).. 2avi^.. + %)}. -Jawo,. —. [. ^-^cl. .. Prove (by shewing that the area of the triangle formed by them a straight line :. zero) that the following sets of three points are in. 10.. (1,4), (3, -2),. 11.. (-i,. 12.. («.. and (-3,16).. (-5,6), and (-8,8). b + c), (6, c + a), and (c, a + b). 3),. is.

(35) .. [EXS.. POLAR COORDINATES.. II.]. 19. Find the areas of the quadrilaterals the coordinates of whose angular points, taken in order, are -2), and. 13.. (1,1),. 14.. (-1,. 6),. 15.. If. be the origin, and. (3,4),. (5,. (-3, -9),. .. .. cos. (3, 9).. the coordinates of any two points and [x^, y.^, prove that. if. Pj and Pg ^6 respectively (%, y^. OP^ OP2. -7).. (4,. -8), and. (5,. P1OP2 = x-^Xc^ + y-^y^. •. 30. Polar Coordinates. There is another method, which is often used, for determining the position of a point in a plane.. Suppose to be a fixed point, called the origin or pole, and OX a fixed line, called the initial line.. Take any other point P in the plane of the paper and The position of P is clearly known when the XOP and the length OP are given.. join OP. angle. [For giving the angle. XOP. drawn, and giving the distance. OP. shews the direction in which. OP. tells. the distance of. P. is. along this. direction.]. The angle. XOP. which would be traced out by the line is called the in revolving from the initial line vectorial angle of and the length OP is called its radius vector. The two taken together are called the polar coordinates of P.. OX. OP. P. If the vectorial angle be. position of. P is. and the radius vector be. denoted by the symbol. The radius vector. is. positive. if it. r,. the. (r, 0).. be measured from the. along the line bounding the vectorial angle; measured in the opposite direction it is negative. origin. 31. (ii). Ex.. Construct. 150°), (iii) 330°), (v) (3,. (3,. (-3, (-3, -30°).. the. positions. (-2, 45°), -210°) and. 'points. (i). (2,. 80°),. (iv). (vi). To construct the first point, the radius vector revolve from OX through an angle of 30°, and then mark off along it a distance equal to two units of length. We thus obtain the point P^. (i). of the. if. /^. ff\. ^\^ ,->^^ ^^^^^^^^^"^^ '/^^^. let. >/. y. yC ""-., ''-•.,. 'p. M". ^. (ii) For the second point, the radius vector revolves from OX through 150° and is then in the position OP^ ; measuring a distance 3 along it we arrive at Pg. 2—2.

(36) COORDINATE GEOMETKY.. 20. (iii) For the third point, let the radius vector revolve from OX through 45° into the position OL. We have now to measure along OL a distance - 2, i.e. we have to measure a distance 2 not along OL but in the opposite direction. Producing iO to Pg, so that OP3 is 2 units of length, we have the required point P3.. To. get the fourth point, we let the radius vector rotate from 330° into the position and measure on it a distance -3, i.e. 3 in the direction produced. thus have the point P^y which is the same as the point given by (ii). (iv). OX through. OM. MO. We. (v) If the radius vector rotate through - 210°, position OP2, and the point required is Pg.. it. be in the. will. For the sixth. point, the radius vector, after rotating through in the position OM: then measure - 3 along it, i.e. 3 in the direction produced, and once more arrive at the point Pg. (vi)^. - 30°,. We. is. MO. 32. It will be observed that in the previous example the same point P^ is denoted by each of the four sets of polar coordinates (3, 150°),. -210°) and (-3, -30°). be found that the same point is given. (-3, 330°),. (3,. In general it vîll by each of the polar coordinates (r, 0), (- r, 180° + 6), {r, - (360° or,. and {- r, - (180° expressing the angles in radians, by each of the 6)]. 6% co-. ordinates (r,. e\ {-r,7r +. 6), {r,. - (27r - 0)} and. {- r,. -. (tt. -. $)}.. It is also clear that adding 360° (or any multiple of 360°) to the vectorial angle does not alter the final position of the revolving line, so that {r, 6) is always the same point as (r, ^ + ?i 360°), where n is an integer. .. So, adding 180° or any odd multiple of 180° to the vectorial angle and changing the sign of the radius vector. gives the. same point as. before.. [-r, ^ is. the same point as [—. r,. Thus the point. + (2n + 1)180°] 6. +. 180°],. i.e. is. the point. [r,. 6\. 33. To find the length of the straight line joining two points whose polar coordinates are given. Let. A and B. coordinates be. (r^,. be the two points and let their polar 6y) and (r^, 6^ respectively, so that. OA^r^, OB = r^, lXOAÔ^, and. lX0B = 6^,.

(37) POLAR COORDINATES.. 21. Then (Trigonometry, Art. 164) AB" - OA'' + OB'' -20 A. OB cos = r-^ + r^ - 2r-^r^ cos {0^ - 6^.. AOB. 34. To find the area of a triangle the coordinates of whose angular points are given. Let ABC be the triangle and be the polar coordinates of angular points. We have. let. (r-^,. 0^),. (r^,. 62),. and. (rg, ^3). its. AABO=AOBC+aOCA -AOBA. (1).. Now. A0BC = i0B,0C sin BOC [Trigonometry, Art. 198] '. So. and. = ^r^r^ sin (^3 - $^). A OCA = \0G OA sin CO A = ^r^r, sin (6, - 6,), AOAB^ÔA. OB sin AOB = ^r^r^ sin {6^ - 6.^ = - Jn^2 sin (^2 - ^1). .. Hence. (1) gives. A ABC = J. \r<^r^. sin (^3 -. 6^ +. r^r^ (sin. 0-^. -. 0^). + r^r^ sin. To change from Cartesian Coordinates. 35.. Coordinates,. Let. {Oo. and. to. 0^)].. Polar. conversely.. P be any point whose Cartesian coordinates, referred. to rectangular axes, are x and y, and whose polar coordinates, reas as pole and ferred to initial line, are (r, 6). Draw Pit/'perpendicular to so that we have. OX. OX. OM=x,. MP = y, LMOP =. e,. OP = r.. and. From. the triangle. X'. MOP. O:. we. have. x=. OM=OPcosMOP = rcosO. (1),. y=. MP= OPsinMOP ^r smO. (2),. r=OP= sJOM^ + MP^^ s/x" + y'. (3),.

(38) COORDINATE GEOMETRY,. 22. and. Equations (1) and (2) express the Cartesian coordinates in terms of the polar coordinates. Equations. (3). and. (4) express. the polar in terms of the. Cartesian coordinates.. P. be in anyThe same relations will be found to hold if other of the quadrants into which the plane is divided by. XOX'. YOT.. and. Ex.. Change. to. Cartesian coordinates the equations. {!) r. and (2)r=a^cos-. a. Multiplying the equation by. (1) i.e.. = asind,. by equations. (2). and. (3), x^-\-y^. Squaring the equation. (2). 2^2. (2), it --. becomes (1. {2x^ +. i.e.. = ar + ar cos 6,. 2y^-ax)^ = a^{x^ + y^).. EXAMPLES. Lay down (3,45°).. 5.. (-1, -180°).. 9.. (2a.-^).. 2.. (-2, -60°). 6.. (1,. -210°).. (. 30°). «>. I. ). and. 3.. 7.. (4,135°). (5,. (4, 120°).. and («"'!). (2,330°).. 8.. («>. |). •. lines joining the pairs of points. 13. •. -675°).. 4.. n.(-2a.4').. 10.{-a,l).. Find the lengths of the straight whose polar coordinates are (2,. III.. the positions of the points whose polar coordinates are. L. 14.. + cos^),. 2{x^ + y^) = a sjx^ + y^-\- ax,. i.e.. 12.. Q,. = ay.. r=acos2- = i. e.. r, it. becomes r^rrar sin. (-3, 45°) and. (7, 105°)..

(39) 23. EXAMPLES.. [EXS. III.]. 15. Prove that the points. (0, 0),. (3,. |). ,. and. ( 3,. ^. form an equiJ. lateral triangle.. Find the areas of the triangles the coordinates of whose angular points are (1, 30°),. 16.. (2, 60°),. 17. (_3, -30°),. (-^. i)'. 18.. and. (5, 150°),. (3, 90°).. and. (7,. 210°).. («'i)'^^*l(-2«.. -y)-. Find the polar coordinates (drawing the. figure in each case) of the. points 19. x = J3, y = l.. 20.. x=-^B,. y = l.. 21.. x=-l,. y = l.. Find the Cartesian coordinates (drawing a figure in each case) of the points whose polar coordinates are 22.. (5,. Change 25.. 23.. I).. 24.(5,-^).. (-6, I).. to polar coordinates the equations. 26. y = xtana.. x^+y^=a\. 28. x^-y^=2ay.. 29. x^=y^{2a-x).. 27. x^ + y^ = 2ax. 30.. {x^. + y^)^ = a^{x^-y^).. Transform to Cartesian coordinates the equations 31.. r=a.. 32. ^ = tan-i??i.. = a2cos2^.. 34. r = asin2^.. 35.. 37. r'^G0s2d = a^.. 38. r^cos- = a^, 2 2i. 40.. 'T. (cos 3^ + sin 3^). ?-2. 33. r = acos^. o'^. 39^. r^=ai sin-.. e. = 5h sin 6 cos. d. sin 2d =2a^.. 36.. a.

(40) CHAPTER. III.. EQUATION TO A LOCUS.. LOCUS.. 36. When a point moves so as always to satisfy a given condition, or conditions, the path it traces out is. Locus under these conditions. For example, suppose to be a given point in the plane of the paper and that a point F is to move on the paper so shall be constant and equal to a. that its distance from. called its. It. is. clear that all the positions of the. moving point must. and on the circumference of a circle whose centre is whose radius is a. The circumference of this circle is therefore the " Locus" of P when it moves subject to the shall be equal to the condition that its distance from lie. constant distance. a.. Again, suppose A and B to be two fixed points in the plane of the paper and that a point P is to move in the plane of the paper so that its distances from A and B If we bisect AB in G and through are to be always equal. it draw a straight line (of infinite length in both directions) perpendicular to AB, then any point on this straight line Also there is no is at equal distances from A and B. point, whose distances from A and B are the same, which This straight line is does not lie on this straight line. therefore the "Locus" of P subject to the assumed con-. 37.. dition.. 38. Again, suppose A and and that the point P is to move. APB AB. so that the angle describe a circle on. is. B. to be. two. fixed points. in the plane of the paper always a right angle. If we. as diameter then. P may. be any.

(41) :. EQUATION TO A LOCUS.. 25. point on the circumference of this circle, since the angle in a semi-circle is a right angle; also it could easily be shewn that APB is not a right angle except when P lies under the The "Locus" of on this circumference. assumed condition is therefore a circle on -4^ as diameter.. P. 39. One single equation between two unknown quanx and y, e.g. x+y=l (1), cannot completely determine the values of x and y. tities. *\P6. \Q vPs. ^1?. M. OM. \1. Vs \P Such an equation has an infinite number of Amongst them are the followins: a:. = 0,1. solutions..

(42) :. COORDINATE GEOMETRY.. 26. Similarly the point Pg,. the equation. (2,. — 1), and. P4,. (3,. — 2),. satisfy. (1).. Again, the coordinates (—1, 2) of Pg and the coordinates (—2, 3) of Pg satisfy equation (1). the measurements carefully we should find the points we obtain lie on the line P^P^ (produced both ways).. On making. that. all. Again, if we took any point Q, lying on P^P^, and draw to OX, we should find on measurement a perpendicular that the sum of its x and y (each taken with its proper sign) would be equal to unity, so that the coordinates of Q. QM. would. (1),. satisfy (1).. Also we should find no point, whose coordinates satisfy which does not lie on P1P2.. All the points, lying on the straight line P1P2, and no others are therefore such that their coordinates satisfy the equation (1). This result is expressed in the language of Analytical Geometry by saying that (1) is the Equation to the Straight Line P^P^.. 40.. Consider again the equation £c2. Amongst an. + 2/2 = 4. number. infinite. (1).. of solutions of this equa-. tion are the following. x = 2A. x= V3| J?>\. x = J'I\. 2/=2r. 2/=x/3r. 2/=V2. x = -'2,\ 2/^0. x^-.^?>,\ /' 2/^--i. x=. = 0, 2/ = -2j. x=l,. r. 53. \ '. y. x^l x=î. J\. \. V3/'. r. -J2A. 3/=-v2r. \. y=i x = -l,. r \. y=-si^)'. x=J2, }L and x=J3) -./2/'-'"% = -l/2/ =.

(43) EQUATION TO A LOCUS.. 1. 27. •.

(44) COORDINATE GEOMETRY.. 28. a large number of values of x and the corresponding values of ?/, the points thus obtained would be found all to lie on the curve in the figure. If. we took. Both. of its branches. would be found. to stretch. away. to. infinity towards the right of the figure.. took any point on this curve and measured accuracy its x and y the values thus obtained would be found to satisfy equation (1). Also we should not be able to find any point, not lying on the curve, whose coordinates would satisfy (1). In the language of Analytical Geometry the equation This curve is called (1) is the equation to the above curve. a Parabola and will be fully discussed in Chapter X. Also,. with. we. if. sufficient. If a point move so as to satisfy any given condition describe some definite curve, or locus, and there can always be found an equation between the x and y of any. 42.. it will. point on the path.. This equation. is. called the equation to the locus or. Hence. curve.. Def.. Equation to a curve.. The equation. to. a. curve is the relation which exists between the coordinates of any foint on the curve^ and which holds for no other points except those lying on the curve.. 43. will be. Conversely to every equation between x and y it is, in general, a definite geometrical. found that there. locus.. Thus in Art. 39 the equation. is. x + y=\, and the P^P^ (produced. definite path, or locus, is the straight line. indefinitely. both ways).. In Art. 40 the equation path, or locus,. is. the dotted. is. x'^. +. y'^^ 4,. and the. definite. circle.. Again the equation 2/ = 1 states that the moving point is such that its ordinate is always unity, i.e. that it is always at a distance 1 from the axis of x. The definite path, or locus, is therefore a straight line parallel to and at a distance unity from it.. OX.

(45) EQUATION TO A LOCUS.. 29. In the next chapter it will be found that if the equation be of the first degree {i.e. if it contain no products, squares, or higher powers of x and y) the locus. 44.. corresponding is always a straight line. If the equation be of the second or higher degree, the corresponding locus is, in general, a curved line.. We append. 45.. a few simple examples of the forma-. tion of the equation to a locus.. Ex. 1. A point moves so that from tioo given perpendicular axes find the equation. the algebraic is. equal. to. sum of. its. distances. a constant quantity a;. to its locus.. Take the two straight lines as the axes of coordinates. Let {x, y) be any point satisfying the given condition. We then ha,wex + y = a. This being the relation connecting the coordinates of any point on the locus is the equation to the locus. It will be found in the next chapter that this equation represents a straight. line.. Ex. 2. The sum of the squares of the distances of a moving point from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant quantity 2c^. Find the equation to its locus. Let (a;, y) be any position of the moving point. Then, by Art. 20, the condition of the question gives {. [x. - af + /} +. { (a;. + af + if] = 2c\. x^ + y'^ = c^- a~.. i.Ci. This being the relation between the coordinates of any, and every, point that satisfies the given condition is, by Art. 42, the equation to the required locus. This equation tells us that the square of the distance of the point {x, y) from the origin is constant and equal to c^ - a^, and therefore the locus of the point is a circle whose centre is the origin.. is. Ex. 3. A point moves so that its distance from the point (-1,0) always three times its distance from the point (0, 2).. Let {x, y) be any point which then have. satisfies. the given condition.. We. J{x + iy' + {y-0)^=Bj{x - 0)2+ {y - 2)2, so that,. on squaring, x'^. i.e.. + 2x + l + y'^=9{x'^ + y'^-4:y + 4), 8(a;2 + y2)_2a;-36?/ + 35 = 0.. This being the relation between the coordinates of each, and that satisfies the given relation is, by Art. 42, the. every, point. required equation. It will be found, in a later chapter, that this equation represents a circle..

(46) COOEDINATE GEOMETRY.. 30. EXAMPLES.. IV.. By taking a number of solutions, as in Arts. 39 the loci of the following equations. —41,. sketch. :. 1.. 2x + dy = l0.. 4.. a;2-4aa; + ?/2 + 3a2. 2.. ^x-y = l.. = 0.. 5.. x'^-2ax-Vy'^ = Q.. 3. y'^. = x.. 6.. ^x = y^-^.. ^' + ^'=1. '4^9. 7. A and B being the fixed points (a, 0) and ( obtain the equations giving the locus of P, when 8.. is. - P52 _ a constant quantity = 2fc2.. PA"^. 10.. PA = nPB, n being constant. P^+PjB = c, a constant quantity.. 11.. PB^ + PC^=2PA^, C. 9.. a, 0) respectively,. being the point. (c, 0).. 12. Find the locus of a point whose distance from the point equal to its distance from the axis of y.. which. Find the equation to the locus of a point distant from the points whose coordinates are 13.. (1, 0). 15.. {a. +. b,. and. (0,. -2).. 14.. (2, 3). is. and. (1, 2). always equi(4, 5).. a-h) and {a-b, a + b).. Find the equation to the locus of a point which moves so that 16. its distance from the axis of x the axis of y.. 17. its distance from the point tance from the axis of y. 18. the. sum. is. three times. (a, 0). is. its. distance from. always four times. of the squares of its distances. from the axes. its dis-. is. equal. to 3.. 19. the square of. 20.. from 21.. its. its. distance from the point. distance from the point. (3, 0) is. (0, 2) is. equal to. 4.. three times its distance. (0, 2).. its. from the. distance from the axis of x. is. always one half. its. distance. origin.. A. fixed point is at a perpendicular distance a from a fixed 22. straight line and a point moves so that its distance from the fixed point is always equal to its distance from the fixed line. Find the equation to its locus, the axes of coordinates being drawn through the fixed point and being parallel and perpendicular to the given line.. 23. In the previous question if the first distance be (1), always half, (2), always twice, the second distance, find the equations to the. and. respective loci..

(47) CHAPTER THE STRAIGHT. LINE.. IV.. RECTANGULAR COORDINATES.. 46. To find the equation to a straight line which is parallel to one of the coordinate axes. Let CL be any line parallel to the axis of y and passing through a point C on the axis of x such that OG = c. Let F be any point on X and y.. Then the always. c,. this line. abscissa of the point. whose coordinates are. F. is. so that. x=c. (1).. This being true for every point on the line CL (produced indefinitely both ways), and for no other point, is, by Art. 42, the equation to the line.. X. It will be noted that the equation does not contain the. coordinate. y.. Similarly the equation to a straight line parallel to the axis oi X is y — d.. Cor. The equation to the axis of a? is The equation to the axis oi y is x — 0.. 2/. = 0.. 47. To find the equation to a st7'aight line which cuts off a given intercept on the axis of y and is inclined at a given angle to the axis of x. Let the given intercept be. c. and. let. the given angle be a..

(48) COORDINATE GEOMETRY.. 32. Let C be a point on the axis Through C draw a straight line Z(7Z' inclined at an angle a (= tan~^ m) to the axis of x^ so that tan a. OC. of y such that. is. c.. — m.. ^^^^ O. The straight line LCL' is therefore the straight line required, and we have to -'l find the relation between the lying on coordinates of any point. P. Draw PM perpendicular G parallel to OX.. to. MX. it.. OX. to. meet in. â. line. through. Let the coordinates = y. and Then MP = NP +. of. P. be. cu. and. ?/». so that. OM=x. MP. MN =C]Srtîia + 00 = m.x +. c,. y = mx+c.. i.e.. This relation being true for any point on the given straight line is, by Art. 42, the equation to the straight line.. [In this, and other similar cases, it could be shewn, is only true for points lying. conversely, that the equation on the given straight line.]. to any straight line passing through which cuts off a zero intercept from the axis found by putting c — O and hence is 3/ = mx.. The equation. Cor.. the origin, of. 2/,. is. i.e.. 48. The angle a which is used in the previous article is the angle through which a straight line, originally parallel to OZ, would have to turn in order to coincide with the given direction, the rotation being always in the positive direction. Also m is always the tangent In the case of such a straight line as AB, in the figure of this angle. is equal to the tangent of the angle PAX (not of the of Art. 50, angle PAO). In this case therefore wi, being the tangent of an obtuse angle, is a negative quantity.. m. The student should verify the truth of the equation on the straight line LCL', and also. of the last for straight. for such a straight line as. A^B^ in the. article for all points. Hnes in other positions, e.g. figure of Art. 59. In this. latter case. both. m. and. c are negative. quantities.. A. careful consideration of all the possible cases of a few propositions will soon satisfy him that this verification is not always necessary, but that it is sufficient to consider the standard figure..

(49) THE STRAIGHT. 33. LINE.. 49. Ex. The equation to the straight line cutting off an intercept 3 from the negative direction of the axis of y, and inclined at 120° to the axis of a;, is. = a;tanl20° + (-3), y= -x^S-S, y + x^S + S = 0. ?/. i.e. i.e.. 50.. 1^0. off given. the equation to the straight line. find. a and. i7itercepts. OX. Let A and B be on such that OA = a and OB =. AB. Join. and produce. definitely both ways.. which cuts. h from the axes.. and. OY. respectively,. and be. h.. it in-. P. Let. be. any point (x, y) on this straight perpendicular line, and draw. PM. to. OX.. We require the. relation that. always holds between x and long as P lies on AB.. By. Euc. YI.. 3/,. so. we have. 4,. OM_PB. MP _AP ^"""^. OA~AB'. 'OB~AB. OM MP PB + AP = + ^.e.. This. AB. OB. OA. X. y. a. D. 1,. ^. therefore the required equation ; for it is the between the coordinates of any point the given straight line. is. relation that holds. lying on 51.. The equation. in the preceding article. may. he also obtained. by expressing the fact that the sum of the areas of the triangles and OPB is equal to OAB, so that. OP A. \axy + \hy.x = \ax'b, and hence. a. 52. Ex. 1. Find the equation to through the -point (3, - 4) and cutting opposite signs,. from. straight. line passing equal but of. the tioo axes.. Let the intercepts cut. — a.. the. off intercepts,. off. from the two axes be of lengths a and.

(50) COORDINATE GEOMETRY.. 34 The equation. to the straight line is then. -a. a. x-y = a. i.e.. (3,. (1).. Since, in addition, the straight line is to go through the point -4), these coordinates must satisfy (1), so that. 3-(-4) = a, and therefore The required equation. a = l. is. therefore. x-y = 7. Ex. 2. Find the equation to the straight line lohich passes through the point (-5, 4) and is such that the portion of it between the axes is divided by the point in the ratio ofl 2. :. + t = 1. b. Let the required straight line be a in the points whose coordinates are {a,. and. 0). The coordinates. This meets the axes (0, &).. the. of the point dividing points in the ratio 1 2, are (Art. 22). joining these. line. :. 2.a+1.0 If this. be the point. ,. (. -. 2.0 + 1.&. 2(1. .. i.e,-^. b. ,. and -.. we have. 5, 4). 2a „ b -5:=and 4=-, so that. The. ,. ,. a=--Y- and. b. required straight line. is. therefore. X y -i^^l2 oy. I.e.. 53.. To find. = 12.. 8a;. '. = 60. a straight. the equation to. line in tenns. of. perpendicular let fall upon it from the origin and the angle that this perpendicular makes with the axis of x. the. Let be jo.. OR. be the perpendicular from. Let a be the angle that with OX.. Let. P. OR. makes. be any point, whose. co-. AB. ordinates are x and y, lying on draw the ordinate PM, and also perpendicular to OR and perpendicular to ML.. PN. ',. ML. and. let its. length.

(51) 85. THE STRAIGHT LINE.. OL = OMcoâ. Then. (1),. LR = NP = MF&inNMP.. and. lNMP^W - lNMO= iMOLâ.. But. LR = MP&m.a Hence, adding Oil/ cos a. (2).. and (2), we have a=OL + LR=OR ifPsin + (1). =79,. X COS a + y sin a = p.. i.e.. This. is. the required equation.. —. 54. In Arts. 47 53 we have found that the corresponding equations are only of the first degree in x and y. shall now prove that. Any. We. equation of the first degree. i7i. x and y always repre-. a straight line. For the most general form of such an equation is Ax + By^C = ^ (1), which do where A^ B, and C are constants, i.e. quantities not contain x and y and which remain the same for all points on the locus. Let (cCi, 2/1), (a?2) 2/2)) îicl (rt's, 2/3) be any three points on sents. the locus of the equation. (1).. Since the point {x-^, y^) lies on the locus, its coordinates when substituted for x and y in (1) must satisfy it.. Hence. Ax^ +. Ry^+C-=0. (2).. CÔ. (3),. Ax^ + Ry^ +. So and. Axs +. £ys+C =. (4).. Since these three equations hold between the three quantities A, B, and C, we can, as in Art. 12, eliminate them.. The. result is. = ^35. 2/35. (5).. -•-. But, by Art. 25, the relation (5) states that the area of the triangle whose vertices are (x^, y^), (x^, 3/2)5 ^^^ (^3> 2/3) is zero.. Also these are any three points on the. locus.. 3—2.

(52) ,. COORDINATE GEOMETRY.. 36. must therefore be a straight line ; for a curved not be such that the triangle obtained by joining any three points on it should be zero.. The. locus. line could. The proposition. 55.. of the preceding article â;. may be and. may also. be deduced. For the equation. from Art. 47.. % + (7=0. A C y=- — x-^,. written. this is the. +. same. as the straight line. y = mx + c,. A. C. ^. ?3i=-— and. if. c. =-—. .. is. x>. But in Art. 47 it was shewn that y = mx + c was the equation to a straight line cutting off an intercept c from the axis of y and inclined at an angle tan~^m to the axis of x.. Ax + By + C=0. The equation. therefore represents a straight line cutting off. C. an intercept - — from x>. the axis of y and inclined at an angle tan~^. We. 56.. (. -. —. |. to the axis of x.. can reduce the general equation of the. Ax + By + C =. degree. first. (1). to the form of Art. 53. For, if p be the perpendicular from the origin on (1) and a the angle it makes with the axis, the equation to the straight line must be. X cos a 4- 2/ sin a - /» = This equation must therefore be the same. (2).. as. ABC. cos a. Hence p. cos a. sin a. C. -A. -B. (1).. —p. sin a. \/cos^. a + sin^ a. Ja^ + B'. 1. sJa^'. + B^. Hence cos a -. -A s/A^. -B. .. -. sm a =. + B^'. ,. \fA-'. ,. C. and^ p =. + B''. The equation (1) may by dividing it by JA^ + B^ and arranging constant term. is. negative.. + B^ form (2). sfA^. therefore be reduced to the it. so that the.

(53) THE STRAIGHT Ex.. 57.. Reduce. to tlie. 37. LINE.. perpendicular form the equation. ^ + 2/\/3 + 7 = + JA'' B^= ^TTs = sJ4:=2.. Here Dividing. by. (1). we have. 2,. i.e.. ^(-i)+y(--^)-i=o,. i.e.. X cos 240° + y sin 240° - 1 = 0.. To. 58.. (1).. trace the straight line given hy. an equation of. the first degree.. Let the equation be. Ax + By + G =. (1).. This can be written in the form. (a). A Comparing. this. B. with the result of Art. 50,. we. see that it. —. represents a straight. Hne which cuts. ——. Its position. is. equation. reduces to the form. from the axes.. off intercepts. (J -^. and. therefore known.. jO. If. G. be. zero, the. (1). A and thus (by Art.. 47, Cor.). represents a straight. passing through the origin inchned at an angle tan~^ to the axis of (^). The. x.. Its position is therefore. may. straight line. If. we put y —. in (1). therefore lies. on. G. we have x — —-r.. it.. by. firnding. it.. JL. i-'i-'). ~ r). known.. also be traced. the coordinates of any two points on. I. hne. The point.

(54) COORDINATE GEOMETRY.. 38 If. we put G^. (»-.). oj. on. lies. = 0, we have. G. 2/. =— ^. so that the point. ,. it.. Hence, as before, we have the position of the straight line.. Ex.. 69.. Trace the straight (1). 3a;-4i/. (3). %y = x',. +. lines. 7 = 0; (4). (2). x = î. 7a;. + 8y + 9 = 0j. (5). Putting 2/ = 0, we have rc= -|, (1) and putting x = Q, we have y = ^. Measuring 0A-^{= -^) along the axis. 2/= -2.. of. x we have one point on. the Hne.. Measuring OB^ (=t) along the axis of y we have another point. A-^B^ produced both ways, is the required line,. Hence. ,. Putting in succession y and x equal to zero, we have the (2) intercepts on the axes equal to - f and - f. If then 0-42= -f and 0^2= - |, we have A^B^, the required line. (3). The point. (0, 0) satisfies. the equation so that the origin. is. on. the line.. Also the point therefore OC3. (4). (3,. The line ic = 2. 1),. is,. by. i.e.. C.^,. lies. on. it.. The required. Art. 46, parallel to the axis of y. line is. and passes. through the point A^ on the axis of x such that 0A^ = 2.. The line y= - 2 is parallel to the axis of x and passes through (5) the point B^ on the axis of y, such that 0B^= - 2.. 60.. Straight Line at Infinity. We have seen Ax + By + (7 = represents a straight line. that the equation.

(55) STRAIGHT LINE JOINING TWO POINTS. which cuts. oiF intercepts. c. c. Ji.. Jj. — - and — — from. 39. the axes of. coordinates. If. X. of. A. vanish, but not. B. or C, the intercept on the axis. The equation of the straight line the form y = constant, and hence, as in. is infinitely great.. then reduces to Art. 46, represents a straight line parallel to Ox.. B. vanish, but not A or C, the straight line meets So if the axis of y at an infinite distance and is therefore parallel to. it.. B. If A and both vanish, but not C, these two intercepts are both infinite and therefore the straight line Q .x + .y + C = is altogether at infinity.. The multiplication of an equation by a constant Thus the equations it. and 10a;- 152/+ 25 2a;-32/+5 = represent the same straight line. Conversely, if two equations of the first degree represent the same straight line, one equation must be equal to the other multiplied by a constant quantity, so that the ratios of the corresponding coefficients must be the same. For example, if the equations and A-^^x + B^y + Cj = a^x + \y + Ci = we must have 61.. does not alter. \. «!. CjL. 62. To jind the equation to the straight line which passes through the two given points {x\ y') and (x", y").. By. Art. 47, the equation to y--. By. any. straight line. is. mx -VG. (1).. m. and properly determining the quantities (1) represent any straight line we please.. c. we can. make. If (1) pass through the point 2/'. Substituting for. c. from. (a;',. y')^. we have. = mas' + c. (2),. (2).. the equation (1) becomes. y-y' = m(x-x'). (3)..

(56) X. COOKDINATE GEOMETRY.. 40. This is the equation to the line going through (x\ y') making an angle tan~^ with OX. If in addition (3) passes through the point {x", y"), then. m. —y=m{x. y *. -y. X'. -. *. Substituting this value in equation. 63.. Ex.. Find. the. r. we. (3),. get as the required. V" — v' X" — x^. *^. through the points (-1,. — x),. y. ti. equation 3). and. to. (4,. '. the straight. line. which passes. -2).. Let the required equation be. y=mx + c Since. (1). goes through the. 3=-m + Hence. first c,. point,. (1).. we have. so that c =. m + S.. becomes. (1). y = mx + m + S If in addition the line. -2 = 47?i + m + 3, Hence. (2). (2).. goes through the second point, so that. m=. we have. -1.. becomes. y=-x + 2,. i.e.. x + y = 2.. Or, again, using the result of the last article the equation is. y-B = ^-^^^{x + l)=-x-l, y + x-=2.. i.e.. 64.. To. fix definitely. the position of a straight line. we. must have always two quantities given. Thus one point on the straight line and the direction of the straight line will determine it; or again two points lying on the straight line will determine. it.. Analytically, the general equation to a straight line two arbitrary constants, which will have to be determined so that the general equation may represent any particular straight line. will contain. m. Thus, in Art. 47, the quantities and c which remain the same, so long as we are considering the same straigld line, are the two constants for the straight line..

(57) 41. EXAMPLES.. Similarly, in Art. 50, the quantities a and h are the constants for the straight line.. 65. In any equation to a locus the quantities x and y, which are the coordinates of any point on the locus, are called Current Coordinates. ;. traced out by a point which. the curve. may. be conceived as. " runs " along the locus.. EXAMPLES.. V.. Find the equation to the straight line 1. cutting off an intercept unity from the positive direction of the axis of y and inclined at 45° to the axis of x. 2. cutting off an intercept - 5 from the axis of y and being equally inclined to the axes. 3. cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to OX.. 4.. cutting off an intercept - 3 from the axis of y to the axis of x.. and inclined. at. an angle tan~i f. Find the equation. to the straight line. 5.. cutting off intercepts 3 and 2 from the axes.. 6.. cutting off intercepts. - 5 and. 6. from the axes.. Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes equal in magnitude and both positive, (1) equal in magnitude but opposite in sign. (2) 7.. 8.. Find the equations to the straight lines which pass through (1, - 2) and cut off equal distances from the two axes.. the point. 9. Find the equation to the straight line which passes through the given point {x\ y') and is such that the given point bisects the part intercepted between the axes.. 10. Find the equation to the straight line which passes through the point ( - 4, 3) and is such that the portion of it between the axes is divided by the point in the ratio 5 3. :. Trace the straight lines whose equations are. + 2?/+3 = 0. + 7r/ = 0.. 11.. a;. 13.. 3a;. 12.. 5a--7//-9. 14.. 2a;-3?/. Find the equations to the straight lines passing following pairs of points. 15.. (0, 0). 17.. (-1,. and 3). (2,. and. -2).. (6,. -7).. 16.. (3, 4). 18.. (0,. = 0.. + 4 = 0.. and. through the (5, 6).. -a) and. (&, 0)..

(58) COORDINATE GEOMETRY.. 42 and. {a. + h, a-h).. 19.. (a, &). 20.. {at^, 2at-^). 22.. (« cos 01. 23.. (acos0jLJ & sin 0j). 24.. (* sec 01, 6. ,. and. a sin. (at^^ 2at;).. <pi). and. (a cos. 21. (p^,. 26.. (a«„^)and(a«„^j.. a sin. tan 0i) and (a sec 02, 6 tan. (1,4), (2,-3), (0,1), (2,0),. â)-. and (acos02> ^sin^g)*. Find the equations to the sides of the whose angular points are respectively 25.. [Exs. v.]. 02).. triangles the coordinates of. and (-1,-2).. and (-1, -2).. 27. Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a\ y = b, and y = b\. 28. Find the equation to the straight line which bisects the distance between the points {a, b) and {a', b') and also bisects the distance between the points ( - a, b) and (a', - b'). 29. Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3a; + 2/ = 12 which is intercepted between the axes of coordinates.. Angles between straight. lines.. To find the angle between two given straight lines. Let the two straight lines be AL^ and AL^j meeting the axis of X in L^ and L^, 66.. I.. Let their equations be. y — m^x^-G-^ and y ~ in.j,x ^r c.^ By Art. 47 we therefore have tan^ZjA'^mi, and td^Vi. AL.^X^Wj.^,.. Now. — L AL^X — L AL.2.X. tan LÂL^ — tan \AL^X — AL^X\ L. L-ÂL^^. AL^X— tan AL^X 1 + tan AL^X. tan AL^X ta,n. rn^ 1. — n^. +mi«i2. (1)..

(59) 43. ANGLES BETWEEN STRAIGHT LINES. Hence the required angle — lLÂL. = tan-i. "'^•""'^. l. [In any numerical example, if the quantity tity it is the tangent of the acute angle it is the tangent of the obtuse angle.]. II.. and. (2).. + mim2 (2). Let the equations of the straight î£c + î2/ + Ci = 0,. + G^Ô. By dividing the equations by B^ and. written. and. be a positive quan-. between the lines. lines. ;. if. negative,. be. A^^x^- B^^y. B^, they. may be.

(60) COORDINATE GEOMETRY.. 44. To find. 67.. the condition that. two straight lines. may. he parallel.. Two. straight lines are parallel when the angle between therefore the tangent of this angle is zero.. them is zero and The equation. (2) of the last article. then gives. Two. straight lines whose equations are given in the "m" form are therefore parallel when their "7?i's" are the same, or, in other words, if their equations differ only in the constant term.. The straight line Ax + By + G' = parallel to the straight line Ax + By two equations are the same.. is. any straight. + C = 0. For. the. line. which. "m's". is. of the. Again the equation A {x-x')+B {y-y') = clearly represents the straight line which passes through the point {x', y') and is parallel to. Ax + By + C=0.. The. result (3) of the last article gives, as the condition. for parallel lines,. Ex.. 68.. Find. through the point. (4,. the equation to the straight line, which passes - 5), and which is parallel to the straight line 3:c. Any. straight line. which. + 4r/ + 5--=0. (1).. is parallel to (1). has. its. equation of the. form 3a;. [For the. "w". of both (1). and. + 4^/ + (7=0 (2) is. (2).. the same.]. This straight line will pass through the point. (4,. - 5). if. 3x4 + 4x(-5) + C = 0, (7=20-12 = 8.. i.e. if. The equation. (2). then becomes 3a;+42/. 69.. To find. the condition that. equations are given,. Let the straight. and. + 8 = 0.. may lines. two. st^'aight linesj. he 'perpendicular.. be. y — m^x y — m.^x. -i-Ci, -\- G.2_.. whose.

(61) CONDITIONS OF PERPENDICULARITY.. 45. be the angle between them we have, by Art. 66,. If. tan^^ r^""^^ 1 +mim2. (1).. If the lines be perpendicular, then ^. = 00. tan. be. =. 90°,. and therefore. .. The right-hand member of equation (1) must therefore and this can only happen when its denominator. infinite,. is zero.. The condition 1. The to. +. of perpendicularity is therefore that m^TTi^. straight line y. y = »...H-.c.,. — O, —. Tn-^Tn2. i.e.. tu^x. +. c.^. is. =—. I.'. therefore perpendicular. «, = -!.. if. y/c'-t. It follows that the straight lines. A^x +B^y + C^ = which m^ = —. for. ^. and. AA a). m^^ — ^ /. ,. are at right angles. if. A,,. _ V AÂ^+B^B^ = 0.. a. i.e.. and A^x + B^y + 0^ = 0,. From. 70.. A. the preceding article. it. follows that the. two. straight lines. and are at right angles. ;. A^x + B,y + Ci = Q. (1),. B,x-A,y+C^ =. (2),. for the product of their m's. derived from (1) by interchanging the coefficients y, changing the sign of one of them, and changing the constant into any other constant.. Also of. a;. (2) is. and. Ex. where. The. straight line through. B^x'. (x', y'). perpendicular to. - A-^y' + 62= 0, so that Cg = A^y'- B^x'.. This straight line. is. therefore. B,{x-x')-A^{y-y') = 0.. (1) is (2).

(62) COORDINATE GEOMETRY.. 46. 71. Ex. 1. Find the equation to the straight line which passes through the point (4, —5) and is perpendicular to the straight line. Sx + 4ij + 5 =. Any. First Method.. (1).. straight line perpendicular to (1) is by the. last article. 4:X-Sij. + C=0. (2).. [We should expect an arbitrary constant in (2) because an infinite number of straight lines perpendicular to (1).] The straight line (2) passes through the point (4, - 5) if i.e.. there are. 4x4-3x(-5) + C = 0, (7= -16-15= -31.. a. The required equation. is. therefore. 4:X-Sy = 31.. Any. Second Method. point is. straight line passing through. the given. y -{-5)=m{x~4:). This straight line m's. is. -. perpendicular to. is. 1,. mX. i.e. if. (. - 1) = -. (1) if. the product of their. 1,. m=|.. i.e. if. The required equation. is. therefore. y + 5=i{x-4), 4:X-'6y. i.e.. Any. Third Method. the point It is. (4,. - 5),. = Sl.. straight line is. y=mx + c.. It. passes through. if. -5 = 4m + c. perpendicular to. (3).. (1) if. mx{-i)=-l. (4).. Hence m = f and then (3) gives c = —V. The required equation is therefore y = '^x-^-^,. 4x-By = Sl.. i.e.. [In the first method, we start with any straight line which is perpendicular to the given straight line and pick out that particular straight line which goes through the given point. In the second method, we start with any straight line passing through the given point and pick out that particular one which is perpendicular to the given straight hne. In the third method, we start with any straight line whatever and determine its constants, so that it may satisfy the two given conditions.. The student should. illustrate. by. figures. ]. Ex. 2. Find the equation to the straight line which passes through the point (x', y') and is perpendicular to the given straight line yy' = 2a {x. + x')..

(63) THE STRAIGHT The given. straight line is. yy'. Any. 47. LINE.. - 2ax - 2ax' = 0.. straight line perpendicular to it is (Art. 70). 2ay + xy'+G=0 This will pass through the point straight line required. the coordinates x'. if. 2ai/ +. i.eAt. xY+C = 0,. G=-2ay' -x'y'. G the required equation. i.e. if. Substituting in. (1).. and therefore will be the and y' satisfy it,. (x', y'). (1) for. 2a{y-y'). is. therefore. + y'{x-x') = 0.. 72. To find the equations to the straight lines which pass through a given point (x', t/') and make a given angle a with the given straight line y — nix + c.. Let. P be. the given point and let the given straight line. be LMJSf, making an angle with the axis of x such that. = m.. tan. (i.e. except when a right angle or zero) there. In general a. is. two straight lines PMR and making an angle a with. are. FNS. the given. line.. Let these. lines. the axis of. of x in R and S and let with the positive direction of. meet the axis. them make angles ^ and. <^'. x.. The equations. two required straight. to the. lines are. therefore (by Art. 62). y -y' = tan ^ x (x — x) X (x — x') y ~y' ~ t^^. and. (1), (2).. <fi'. Now. cf,. and. <^'^L. = L LMR + L RLM = a +. 6*,. LNS+ L SLN= (180° â) + e.. Hence <^. = tan (a + ^) -. ——. tan a + tan ^. ^.. ,. tan. .j. i. and. tan „. = tan (^ — .. .. a). </>'. = tan tan d. = ^— 1. tan a tan (. 1. 80°. +. ^. ^. i. + tan. —=. tan a. —m tan a. - a). m— tan a. — tan a -p,. tana + m. =. ff. .,. 1. +7n tan a. .. ,.

(64) COORDINATE GEOMETRY.. 48. On substituting these values in (1) and the required equations + tan a ^ ^ ^ 1-mtana^ ,. m. ,. m — tan a. ,. as. ,.. ,. 1. EXAMPLES.. we have. ,,. 1 = y-y \^ - ^ ^ ^ 1 + m tan a. and. (2),. )•. VI.. Find the angles between the pairs of straight lines 1. x-ijsj^ — ^ and ^/3a;+2/ = 7. 3. 2/ = 3a3 + 7 and 3?/-a; = 8. 2. ic-4?/ = 3 and ^x-y = ll.. = (2-V3)a: + 5 and 2/= (2 + ^/3) a;- 7. = (inn^-n^)x + n^ and (?n7H- m^) y = (??i?i - w'^) a; + m^ {m'^-mn)y 5. 6. Find the tangent of the angle between the lines whose intercepts on the axes are respectively a, - 6 and 6, - a. 7. Prove that the points (2, - 1), (0, 2), (2, 3), and (4, 0) are the 4.. 2/. coordinates of the angular points of a parallelogram and find the angle between its diagonals.. Find the equation. to the straight line. 8.. passing through the point. 9.. passing through the point. 3). (2,. and perpendicular. to. the. straight line 4a;-3i/ = 10.. straight line. 7aj. (. -. 6,. 10. passing through the point straight line joining the points (5,. (2,. 7). and perpendicular. to the. -3) and perpendicular ( - 6, 3).. Find the equation. the straight line. — a. to the. and. 11. passing through the point (-4, straight line joining (1, 3) and (2, 7). 12.. 10). + 8?/ = 5.. -3) and perpendicular. to the straight line. drawn. v = 1 through the point where. to the. at right angles to it. meets the axis. b. of X.. 13. Find the equation to the straight line which bisects, and is perpendicular to, the straight line joining the points (a, b) and (a',. &')•. 14. Prove that the equation to the straight line which passes through the point {a cos^ 6, a sin^ 6) and is perpendicular to the straight line xsecd + y cosec d = ais x cos d-y sin d = a cos 26. 15. Find the equations to the straight lines passing through {x', y'). and. respectively perpendicular to the straight lines. xxf-\-yy'=a\.

(65) EXAMPLES.. [Exs. VI.]. XX. yy. a-. 62. and and and. = 1,. + xy' = a-.. x'y. Find the equations. to the straight lines externally, the line joining (-3,7) to (5, which are perpendicular to this line.. 16.. 49. which divide, internally. - 4). in the ratio of 4. :. 7. 17. Through the point (3, 4) are drawn two straight lines each inclined at 45° to the straight line x-y = 2. Find their equations and find also the area included by the three lines.. 18.. Shew. the point. (3,. that the equations to the straight lines passing through 2) and incHned at 60° to the line. -. iJ3x + y = l are y. +2=. and y. -J3x + 2 + 3^S = 0.. 19. Find the equations to the straight lines which pass through the origin and are inclined at 75° to the straight line. x + y + ^S{y-x) = a. 20. Find the equations to the straight lines which pass through the point {h, k) and are inclined at an angle tan~'^m to the straight. y = mx + c. Find the angle between the two straight lines 3a; = 4?/ + 7 and 5y = 12x + 6 and also the equations to the two straight lines which pass through the point (4, 5) and make equal angles with the two. line. 21.. given lines.. 73. To sheiv that the point (x', y') is on one side or the other of the straight line Ax + By +(7 = according as the quantity Ax + By' + C is positive or negative.. Let. LM. be the given straight line and. P. any point. ix\ y).. P. Through the axis of straight line. ordinates of. Since. Q. draw P^,. parallel to. meet the given in Q^ and let the co-. 3/,. Q. lies. to. be. (.'«',. y").. on the given. line,. we. i>x:x. have ^£c'. so that. +. %" + C=:0, Ax + C. y. B. .(1).. It is clear from the figure that PQ is drawn parallel to the positive or negative direction of the axis of y according as is on one side, or the other, of the straight line LM^ i.e. according as y" is > or < y\ i.e. according as y" — y is positive or negative.. P. L.. 4.

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