CO-ORDINATE GEOMETRY

Transcription

1 CO-ORDINATE GEOMETRY

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3 MATHS SERIES CO-ORDINATE GEOMETRY By N.P. BALI FIREWALL MEDIA (An Imprint of Laxmi Publications Pvt. Ltd.) BANGALORE CHENNAI COCHIN GUWAHATI HYDERABAD JALANDHAR KOLKATA LUCKNOW MUMBAI RANCHI NEW DELHI BOSTON, USA VED 4-D:\L-gcorg\cogtit-cog IInd

4 Copyright 2013 by Laxmi Publications Pvt. Ltd. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher. Published by : FIREWALL MEDIA (An Imprint of Laxmi Publications Pvt. Ltd.) 113, Golden House, Daryaganj, New Delhi Phone : Fax : info@laxmipublications.com Price : ` Only. First Edition : 2008; Reprint : 2009, 2010, 2011; Second Edition : 2013 OFFICES Bangalore Jalandhar Chennai Kolkata Cochin , Lucknow Guwahati , Mumbai , Hyderabad Ranchi FCO G. CO-ORDINATE GEO-BAL Typeset at : Goswami Associates, Delhi. C 2023/10/08 Printed at : Ajit Printers, Delhi. VED 4-D:\L-gcorg\cogtit-cog IInd

5 CONTENTS Chapters Pages 1. The Point The Straight Line Two or More Straight Lines Change of Axes The Circle Two or More Circles The Parabola The Ellipse The Hyperbola The General Equation of the Second Degree and Tracing of Conics System of Conics Confocal Conics Polar Equation of a Conic (v) VED 4-D:\L-gcorg\cogtit-cog IInd

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7 PREFACE TO THE SECOND EDITION This book has been my aim to lay before the students a strictly rigorous and yet very simple exposition of Co-ordinate Geometry and its applications. Like the companion volumes Differential Calculus, Real Analysis, Differential Equations, Integral Calculus, Statics, Dynamics etc. The book has the following special features: 1. The development of the subject is systematics. 2. Almost every article is followed by properly graded illustrative examples. 3. It contains solutions of questions set in various Indian Universities and competitive examinations. Most of the examples are selected from the university papers. Errors or misprints, if any, are unintentional and regretted. Suggestions for further improvement will be warmly received. Author (vii) VED 4-D:\L-gcorg\cogtit-cog IInd

8 DO NOT MISS TO READ: MATHEMATICS SERIES ALGEBRA HIGHER TRIGONOMETRY DEFINITIONS AND FORMULAE IN MATHS DIFFERENTIAL EQUATIONS DIFFERENTIAL CALCULUS DYNAMICS INTEGRAL CALCULUS MATRIX ALGEBRA MODERN ALGEBRA NUMERICAL ANALYSIS REAL ANALYSIS SEQUENCES AND SERIES SOLID GEOMETRY STATICS THEORY OF NUMBERS VECTOR ALGEBRA VECTOR CALCULUS ETC. FOR ALL COLLEGE CLASSES PLEASE WRITE FOR FREE CATALOGUE (viii) VED 4-D:\L-gcorg\cogtit-cog IInd

9 LIST OF SYMBOLS AND ABBREVIATIONS Symbol Meaning belongs to or is an element of does not belong to Is a sub-set of Is not a sub-set of Is a super-set of Union of sets Intersection of sets A B Cartesian (or cross) product of sets A and B U (or X) Universal set A c Complement of A A B Difference of two sets A and B (or complement of B w.r.t. A) φ Null (empty or void) set : or or s.t. such that iff if and only if V for all there exists implies implies and is implied by (or iff) A Δ B Symmetric difference of A and B N the set of all natural numbers (or positive integers) Z the set of all integers Q the set of all rational numbers R the set of all real numbers is less than or equal to is greater than or equal to Sup. S (or l.u.b. S) supermum (or least upper bound) of S Inf. S (or g.l.b. s) infimum (or greatest lower bound) of S x absolute value of x (a, b) open interval a < x < b [a, b] closed interval a x b nbd neighbourhood A interior of A A derived set of A f : X Y closure of A f is a function from X to Y Σ = (or Σu n ) infinite series u 1 + u 2 + u Σ > alternating series u = 1 u 2 + u 3 u Π Π infinite product u = 1 u 2 u 3... (ix) VED 4-D:\L-gcorg\cogtit-cog IInd

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11 1 The Point 1.1. CO-ORDINATE GEOMETRY It is that branch of geometry which defines the position of a point in a plane by a pair of algebraic numbers. It is also called Algebraic Geometry or Analytical Geometry RECTANGULAR AXES AND ORIGIN Let X OX and Y OY be two perpendicular straight lines intersecting at the point O. Then (i) X OX is called the axis of x or the x-axis. (ii) Y OY is called the axis of y or the y-axis. (iii) Both X OX and Y OY taken together, in this very order, are called the rectangular axes or the axes of co-ordinates or the co-ordinate axes or simply the axes. They are called rectangular axes because the angle between them is a right angle. (iv) Their point of intersection O is called the origin. X Y y-axis O Y 90 (Origin) x-axis X 1.3. CARTESIAN CO-ORDINATES OF A POINT Let X OX and Y OY be two perpendicular straight lines intersecting at the point O. Let P be any point in the plane of the axes. From P, draw PM X OX. Then (i) OM is called the x-coordinate or abscissa of P and is Y denoted by x. P(x, y) (ii) MP is called the y-coordinate or ordinate of P and is denoted by y. (iii) The numbers x and y are called the Cartesian y Rectangular Co-ordinates or simply the co-ordinates of P. (Abscissa) (iv) The symbol P(x, y) is used to denote the point P. In this X O x M X symbolic representation, the abscissa is always written first and Y separated from the ordinate by a comma. Remember: 1. Abscissa is the distance (i.e. perpendicular distance) of a point from y-axis. 2. Ordinate is the distance of a point from x-axis. 3. Abscissa is +ve to the right of y-axis and ve to the left of y-axis. 4. Ordinate is +ve above x-axis and negative below x-axis. 5. Abscissa of any point on y-axis is zero. 6. Ordinate of any point on x-axis is zero. 7. Co-ordinates of the origin are (0, 0). 1 (Ordinate)

12 2 GOLDEN CO-ORDINATE GEOMETRY DISTANCE FORMULA Article 1.1. To Find the Distance Between Two Points Whose Co-ordinates are Given Let P(x 1, y 1 ) and Q(x 2, y 2 ) be the given points. Draw PL and QM OX and PR QM. Then PR = LM = OM OL Y = x 2 x 1 RQ = MQ MR = MQ LP = y 2 y 1 90 In right d. ΔPQR PQ 2 = PR 2 + RQ 2 [Pythagoras Theorem] P( x1, y1) Q( x2, y2) R = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 X O L M X PQ = + Remember. Distance between two points (x 1, y 1 ) and (x 2, y 2 ) = + b g b g = + Corollary. Distance of a point (x, y) from the origin (0, 0) = + = +. Example 1. Find the distance between the following pairs of points: (i) (am 2 1, 2 am 1 ); (am 2 2, 2 am 2 ) (ii) (a cos α, a sin α); (a cos β, a sin β). Sol. (i) Reqd. distance = + = + = + = + + = + + (Assuming m 1 > m 2 ) (ii) Reqd. distance = α β + α β = α β + α β Y = a α + α + β + β α β + α β = a + α β = a α β = a α β α β = 2 a.

13 THE POINT 3 Note 1. When three points are given and it is required to prove that: (i) they form an isosceles triangle, show that two of its sides are equal. (ii) they form an equilateral triangle, show that its three sides are equal. (iii) they form a right angled triangle, show that the sum of the squares of two sides is equal to the square of the third side. (iv) they are collinear, show that the sum of the distances between two point pairs is equal to the distance between the third point pair. Note 2. When four points are given and it is required to prove that: (i) they form a square, show that all sides are equal and diagonals are equal. (ii) they form a rhombus, show that all sides are equal and diagonals are unequal. (iii) they form a rectangle, show that the opposite sides are equal and diagonals are also equal. (iv) they form a parallelogram, show that the opposite sides are equal. Example 2. Show that the points ( 2, 3), (1, 2) and (7, 0) are collinear. [Note. Points are collinear means points lie on the same straight line.] Sol. Let A( 2, 3), B(1, 2), C(7, 0) be the given points. AB = + = + = BC = + = + = = = AC = + = + = = = so that AB + BC = + = = AC The points A, B, C are collinear i.e., lie on the same straight line. Example 3. Show that the points (0, 1), (2, 1), (0, 3) and ( 2, 1) are the corners of a square. Sol. Let A(0, 1), B(2, 1), C(0, 3), D( 2, 1) be the given points. AB = + = + = BC = + = + = CD = + + = + = DA = + + = + =. Thus the four sides AB, BC, CD, DA are equal. Hence ABCD is either a square of a rhombus. In a square the two diagonals are equal and in a rhombus they are unequal. Let us, therefore, calculate the diagonals. AC = + = + = 4 BD = + + = + = 4 Thus, AC = BD Hence ABCD is a square. Example 4. Show that the four points (7, 3), (3, 0), (0, 4) and (4, 1) are the vertices of a rhombus.

14 4 GOLDEN CO-ORDINATE GEOMETRY Sol. Let A(7, 3), B(3, 0), C(0, 4), D(4, 1) be the given points. AB = + = + = 5 BC = + + = + = 5 CD = + + = + = 5 DA = + = + = 5 AB = BC = CD = DA. ABCD is either a square or a rhombus. In a square the two diagonals are equal and in a rhombus they are unequal. Let us, therefore, calculate the diagonals. AC = + + = + = BD = + + = + = Thus, AC BD Hence ABCD is a rhombus. Example 5. Show that the points (3, 2), (11, 8), (8, 12) and (0, 6) are the vertices of a rectangle. Sol. Let A(3, 2), B(11, 8), C(8, 12) and D(0, 6) be the given points. AB = + = + = = 10 BC = + = + = = 5 CD = + = + = = 10 DA = + = + = = 5 Thus, AB = CD and BC = DA The opposite sides of quadrilateral ABCD are equal. So, ABCD is either a rectangle or a parallelogram. In a rectangle two diagonals are equal and in a parallelogram they are unequal. Let us, therefore calculate the diagonals i.e., AC = + = + = BD = + = + = Diagonals AC and BD are also equal. Hence ABCD is a rectangle. Example 6. Show that the points ( 1, 0), (0, 3), (1, 3) and (0, 0) are the vertices of a parallelogram. Sol. Let A( 1, 0), B(0, 3), C(1, 3) and D(0, 0) be the given points. AB = + = + = BC = + = + = 1

15 THE POINT 5 CD = + = + = DA = + + = + = 1 AB = CD and BC = DA The opposite sides of quadrilateral ABCD are equal. Hence ABCD is a parallelogram. [Note. For another method see examples with Section Formula.] Example 7. Show that the points (2a, 4a), (2a, 6a) and (2a + an equilateral triangle., 5a) are the vertices of Sol. Let A(2a, 4a), B(2a, 6a); C(2a +, 5a) be the given points. AB = + = + = 2a BC = + = + = 2a CA = + + = + = 2a AB = BC = CA. Hence ABC is an equilateral triangle. Example 8. Two points (0, 0), (3, ) form with another point an equilateral triangle. Find that point. Sol. Let A(0, 0); B(3, ) be the given points and C(x, y) the third point so that ABC is an equilateral triangle. AB = BC = CA or AB 2 = BC 2 = CA 2 (0 3) 2 + (0 ) 2 = (3 x) 2 + ( y) 2 = (x 0) 2 + (y 0) 2 or 12 = x 2 + y 2 6x 2 y + 12 = x 2 + y 2...(1) From the first and third members of (1), we get x 2 + y 2 = 12...(2) From second and third members of (1), we get x 2 + y 2 6x 2 y + 12 = x 2 + y 2 or 6x 2 y + 12 = 0 or 2 y = 6x + 12 = 6(2 x) or y = (2 x) = (2 x) Putting y = (2 x) in (2), we get x 2 + 3(2 x) 2 = 12 or 4x 2 12x + 12 = 12 or 4x 2 12x = 0 or x 2 3x = 0 or x(x 3) = 0 x = 0, 3 When x = 0, y = (2 x) = (2 0) = 2 When x = 3, y = (2 x) = (2 3) =. Hence the third point is (0, 2 ) or (3, ).

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