2 Unit Coordinate Geometry

7/24/2019 2 Unit Coordinate Geometry

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First published by

John Kinny-Lewis in 2012

c John Kinny-Lewis 2012

National Library of Australia

Cataloguing-in-publication data

ISBN: 978-0-9872782-1-0

This book is copyright. Apart from any fair dealing for the purposes

of private study, research, criticism or review as permitted under theCopyright Act 1968, no part may be reproduced, stored in a retrievalsystem, or transmitted, in any form by any means, electronic, mechan-ical, photocopying, recording, or otherwise without prior written per-mission. Enquiries to be made to John Kinny-Lewis.

Copying for educational purposes.

Where copies of part or the whole of the book are made under Sec-tion 53B or Section 53D of the Copyright Act 1968, the law requiresthat records of such copying be kept. In such cases the copyright owner

is entitled to claim payment.

Typeset by Christopher Hines

Edited by Christopher Hines


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Preface

This revision of Mathematics Topics has been writtento follow the syllabus of the NSW Higher School Cer-tificate course of Mathematics.

It is assumed that the student is familiar with the con-tent of the corresponding Mathematics syllabus.


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2 COORDINATE GEOMETRY

Coordinate Geometry

Example 1

T

A

B

C

x

y

The line passes through A (1, 3) and has the equation x + 2y 5 = 0.The point B has the coordinates (1,2) and the line CB is perpendicularto .

(i) Find the length of the interval BA.

(ii) Write down the slope of BA and use your calculator to find the angleBA makes with the positive direction of the x-axis. Give your answerto the nearest degree.

(iii) Show that BChas the equation 2x y 4 = 0.

(iv) IfT is the intersection of and BC, find the coordinates ofT.(v) Find the exactperpendicular distance from Tto the line BA.

(vi) Find the area of the triangle AT B.

Solution:

(i)

BA =

(1 1)2 + (3 2)2

=

4 + 25

=

29 units


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COORDINATE GEOMETRY 3

(ii)

gradient ofBA =2 311=

5

2

tan = 52

= 112

(iii) Equation of is x + 2y 5 = 0,

y = 12

x +5

2 gradient is 1

2

Gradient ofBC= 2 (since BC )Equation ofBC is y 2 = 2(x 1)

y + 2 = 2x 2

2x y 4 = 0 is the equation ofBC.

(iv)Solving simultaneously:

2x y 4 = 0 (1)x + 2y 5 = 0 (2)

(1) 2 4x 2y 8 = 0 (3)

(3) + (2) 5x 13 = 0

x=13

5

In (2) 13

5 + 2y 5 = 0

2y =12

5

y = 11

5

T =

2

3

5, 1

1

5


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(v)The equation ofBA is y 3 = 5

2(x + 1).

2y 6 = 5x 55x + 2y 1 = 0

d =

5 135 + 2 6

5 1

52 + 22=

72

529

= 72

529=

72

29

145 units

(vi)

Area of the triangle AT B =1

2

29 725

29= 7.2 cm2

Example 2

O

A

x

y

B

C

OC is parallel to AB and OA is parallel to BC.A (0,

2), B = (1,

4) and O is the origin.

(i) Find the coordinates ofC.

(ii) Find the area ofOABC.


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(i)

gradient ofAB =42

1 0 = 2

Since OC is parallel to AB then from O run out 1 and fall by 2

co-ordinates ofC= (1,2)(ii) Area ofOABC= 2 1 = 2 units2.

Example 3

Plot the points A (1, 4), B (7, 2) and C(1,2) on the number plane.(i) Show that AB AC.(ii) Find the area of triangle ABC.

(iii) Write down the three inequalities that define the region enclosed by thesides of the triangle ABC.

(i)

A

B

C

x

y

gradient ofAB =2 47 1=

1

3=m1

gradient ofAC=2 41 1= 3 = m2

AB AC (m1m2= 1)


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1)

P O

x

y

Q R

The diagram shows the points P(2, 0), Q (3, 2) and R (2, 2). isthe acute angle between QR and P R.

(i) Show that P and Q lie on the line 2x + y + 4 = 0.

(ii) Show that the gradient ofP R is 1

2.

(iii) Show that the length ofP R is 2

5 units.

(iv) Show that P Q and P R are perpendicular.

(v) Find tan .

(vi) Find the midpoint (M) ofQR.

(vii) Find the equation of the circle with centre M

that passes through R.

(viii) Copy this diagram and shade the region which satisfies the in-equality

2x + y + 4 02) The lines L1 and L2 have equations L1 : x4y 13 = 0 and L2 :

2x + y + 1 = 0.

(i) Find the shortest distance from P(1, 2) to L1.

(ii) Show that the lines intersect at Q (1,

3).

(iii) Find the distance P Q.


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3) Let A and B be the points (0, 2) and (

2, 5) respectively.

(i) Find the co-ordinates of the midpoint ofAB.

(ii) Find the slope of the line AB.

(iii) Find the equation of the perpendicular bisector ofAB.

(iv) C lies on the line y = x+ 1 and is equidistant from A and B.Find the co-ordinates ofC.

4)

A

x

y

D

C

B

ABCD is a parallelogram where A = (1,3), B = (5, 2), C= (3, 5)and D= (1, 0).(i) Find the equation ofAD.

(ii) Find the perpendicular distance from C to AD.

(iii) Find the distance AD.

(iv) Find the area ofABCD.

(v) What fraction of the area ofABCDlies above the x-axis?


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5)

(i) Find the equation of the straight line AB, given that A has co-ordinates (2,1) and that AB is parallel to the line whose equa-tion is 3x y + 1 = 0.

(ii) The line ABin (i) intersects the y-axis at C. Find the co-ordinatesof the midpoint ofAC.

6)

(i) Show that the points (3, 4), (1, 2) and (5, 0) are collinear.

(ii) Show that the line 3x + 4y + 5 = 0 is a tangent to the circle withcentre at the origin and radius 1 unit.

7)

(i) Find the co-ordinates of the point of intersection of the lines

x + y 1 = 0 x y= 0

(ii) Draw these lines on the number plane and indicate, by shading,the region of intersection of

x + y 1 0, x y 0 and y 0

(iii) Find the area of the shaded region.

8) Plot the points A (3, 2), B (-1, -1) and C(0, 3).

(i) Find the equation of the line through Cand parallel to AB.

(ii) Find the co-ordinates ofD, the point where the line in (i) meetsthe x-axis.

(iii) Prove that ABCDis a parallelogram.

9) Find the co-ordinates ofA, on the line x = 2, such that the line joiningA to B (4, 7) is perpendicular to 3x y + 1 = 0.

10) Given that P, Q and S are the points (-1, -2), (2, 5) and (4, 1) re-spectively, and R lies in the first quadrant, find R so that PQRS is aparallelogram.

END OF CHAPTER


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10 COORDINATE GEOMETRY - SOLUTIONS

Coordinate Geometry - Solutions1)

P O

x

y

Q R

The diagram shows the points P(2, 0), Q (3, 2) and R (2, 2). is the acuteangle between QR and P R.

(i) Show that P and Q lie on the line 2x + y+ 4 = 0.

gradient ofP Q= 0 223 = 2

equation ofP Q is y 0 = 2(x2)

y= 2x 4

2x + y+ 4 = 0 is the equation ofP Q, hence P and Q lie on the line.

(ii) Show that the gradient ofP R is 12

.

gradient ofP R= 2 022=

1

2

(iii) Show that the length ofP R is 2

5 units.

P R=

(2 2)2 + (0 2)2

=

16 + 4

= 20

= 2

5 units


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COORDINATE GEOMETRY - SOLUTIONS 11

(iv) Show that P Q and P R are perpendicular.

mPQ= 2 and mPR= 12

mPQ mPR= 2 12

= 1

P Q P R

(v) Find tan .

QR is parallel to the x-axis.

tan = mPR=12

(alternate angles = in parallel lines QR, PO)

(vi) Find the midpoint (M) ofQR.

M =

3 + 22

, 2

=

1

2, 2

(vii) Find the equation of the circle with centre Mthat passes through R.

QR= 5 units

M R= 21

2 units

the circle has a radius of 21

2units and centre

1

2, 2

.

the equation is

x +

1

2

2

+ (y 2)2 =

5

2

2

x +1

2

2+ (y 2)2 = 6 1

4


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(viii) Copy this diagram and shade the region which satisfies the inequality

2x + y+ 4 0

P

x

y

Q R

2)The linesL1and L2have equationsL1 : x4y13 = 0 andL2 : 2x+y +1 = 0.

(i) Find the shortest distance from P(1, 2) to L1.

L1: x 4y 13 = 0

Nowp=|ax1+ by1+ c|

a2 + b2

a= 1, b= 4, c= 13, x1 = 1 andy1= 2

p =|1 1 4 2 + 13|

12 + (4)2 = 20

17

p = 20

17units.

(ii) Show that the lines intersect at Q (1,3).

x 4y= 13 (1)

2x + y= 1 (2)


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(2)

4 8x + 4y=

4 (3)

(1) + (3) 9x= 9

x= 1

In (2) 2 1 + y= 1

y = 3

Q (1,3)

(iii) Find the distance P Q.

P Q= 23

P Q= 5 units (P Q lies on the line x= 1)

3) Let A and B be the points (0, 2) and (2, 5) respectively.

(i) Find the co-ordinates of the midpoint ofAB.

M=

0 + 2

2 ,

2 + 5

2

=

1, 3 1

2

(ii) Find the slope of the line AB.

mAB= 2 502=

3

2

(iii) Find the equation of the perpendicular bisector ofAB.

The perpendicular passes through the midpoint M

1, 3 1

2

and has gradient 2

3.

the equation is y 72

=23

(x1)

3y 212

= 2x + 2

6y 21 = 4x + 4

4x 6y+ 25 = 0


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(iv) C lies on the line y = x+ 1 and is equidistant from A and B. Find theco-ordinates ofC.

If C is equidistant from A and B then it must lie on the perpendicular bisec-tor, as well as the line y = x + 1. Solving these two conditions simultaneously,

y= x + 1 (1)

4x 6y+ 25 = 0 (2)

Substituting for y in (2)

4x 6(x + 1) + 25 = 0

4x 6x 6 + 25 = 0

2x + 19 = 0

x= 91

2 and y= 10

1

2 (from i)

C=

9

1

2, 10

1

2

4)

A

x

y

D

C

B

E

ABCD is a parallelogram where A = (1,3), B = (5, 2), C = (3, 5) andD= (1, 0).


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(i) Find the equation ofAD.

gradient ofAD=3 011=

3

2

equation ofAD is y 0 = 32

(x1)

2y= 3x 3

3x + 2y+ 3 = 0

(ii) Find the perpendicular distance from C to AD.

p=|3 3 + 2 5 + 3|

32 + 22

= 22

13

=22

13

13 units

(iii) Find the distance AD.

AD=

(11)2 + (3 0)2

=

13 units

(iv) Find the area ofABCD.

area ofABCD= 22

13

13 = 22 units2

(v) What fraction of the area ofABCD lies above the x-axis?

gradient ofAB=

2

3

5 1 =5

4

equation ofAB is y 2 = 54

(x 5)


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Let y= 0 on the line AB to determine the co-ordinates ofE

E=

3

2

5, 0

E D= 32

5 1 = 4 2

5

the height ofAED = 3 (the y value ofA)

area ofAED = 12 4 2

5 3 = 6 3

5

area ofABCD= 22 from part (iv)

the area above the x-axis = 22 6 35

= 152

5

the fraction of area above the x-axis = 152

5 22

= 7

10

5)

(i)Find the equation of the straight lineAB , given thatAhas co-ordinates (2,1)and that AB is parallel to the line whose equation is 3x y+ 1 = 0.

IfAB is parallel to 3x y +1 = 0 then it will be of the form 3x y + k= 0. SinceA(2,1) lies on 3x y+ k= 0,

3 21 + k= 0 k= 7

the equation ofAB is 3x y 7 = 0

(ii) The line AB in (i) intersects the y-axis at C. Find the co-ordinates of themidpoint ofAC.

When x= 0, y= 7

C= (0,7)

midpoint ofAC=

2 + 02 ,1 72

= (1,4)


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6) (i) Show that the points (3, 4), (

1, 2) and (

5, 0) are collinear.

let A= (3, 4), B = (1, 2) and C= (5, 0)

gradient ofAB = 4 231=

1

2

gradient ofBC= 2 015 =

1

2

gradient ofAB = gradient ofBC

A, B and Care collinear.

(ii) Show that the line 3x+ 4y+ 5 = 0 is a tangent to the circle with centre atthe origin and radius 1 unit.

If the perpendicular distance from the centre O (0, 0) to the line 3x + 4y+ 5 = 0is 1 unit (the circle radius) then this line is a tangent to the circle.

d=|3 0 + 4 0 + 5|

32 + 42

=5

5= 1

the line 3x + 4y+ 5 is a tangent to this circle.

7) (i) Find the co-ordinates of the point of intersection of the lines

x + y 1 = 0 x y= 0

Solving these simultaneously

x + y= 1 (1)

x y= 0 (2)

(1) + (2) 2x= 1

x=1

2 and y=

1

2

the point of intersection is

1

2, 1

2


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(ii) Draw these lines on the number plane and indicate, by shading, the region ofintersection of

x + y 1 0, x y 0 and y 0

x

y

B

O A

(iii) Find the area of the shaded region.

x + y= 1 intersects the x-axis at A (1, 0)

base is 1 unit and the height is 1

2 units.

area of the region is = 1

2 11

2=

1

4 units2

8) Plot the points A (3, 2), B (-1, -1) and C (0, 3).

(i) Find the equation of the line through Cand parallel to AB.

A

x

y

D

C

B

gradient ofAB =

2

1

31=3

4

the equation of the line through C is y=3

4x + 3


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(ii) Find the co-ordinates ofD, the point where the line in (i) meets the x-axis.

when y= 0, 3

4x= 3 D (4, 0)

(iii) Prove that ABCD is a parallelogram.

gradient ofAC= 2 33 0=

1

3

gradient ofBD = 1 014=

1

3

AC

||BD and we are given that AB

||CD

ABCD is a parallelogram since it has two opposite sides parallel.

9)Find the co-ordinates ofA, on the line x = 2, such that the line joining Ato B(4, 7) is perpendicular to 3x y+ 1 = 0.

clearly the gradient of the line is 3

the gradient ofAB = 13

let Ahave the co-ordinates (2, b)

gradient ofAB = b 72 4=

1

3

b 7 = 23 b= 7 2

3

A =

2, 7

2

3


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10) Given that P, Q and Sare the points (-1, -2), (2, 5) and (4, 1) respectively,and R lies in the first quadrant, find R so that PQRSis a parallelogram.

x

y

P

Q

S

R

gradient ofP Q=522

1

=7

3

The run = 3 units and the rise = 7 units, from S(4, 1) the coordinates ofR are

R= (4 + 3, 1 + 7)

= (7, 8) which lies in the first quadrant.

END OF CHAPTER

Documents

2 Unit Coordinate Geometry