BMM10234 Chapter 7 - Coordinate Geometry.pdf

7/28/2019 BMM10234 Chapter 7 - Coordinate Geometry.pdf

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Chapter 7 | Coordinate Geometry | BMM10234 | 181 of 204

INTRODUCTION

npintinpnntinpsitinwithnttwutuppniuinsints

intstingthth.

Onhizntin-is

Onvtiin -is

O intintstinv

s OiginO sitivpt-isO

nngtivpt-is

On

ngtivpt-isO psitivpt-is

Oiginiskszwhthistnssung-isv-is-isv-isinnvnint

unitngth.

Cartesian Coordinates of a Point

LtpintinthEuinpn.nKppniuswnpintn-isn-issptiv.

K=istnng-is=O=

=istnng-is=OK=

Thnthpipsntsthtsinintspintnhninitsthtinpint

inthgivnpn.

-intsisspint.

-intintpint

y

yn

xn xo

y

x

K P

M

Quadrants

Thtwsivisthpnintuqutpnsthqunts.Thquntsnusshwninth

ig.

y

yn

xn

xO

Preface

Coordinate Geometry


2/24


Distance between two given points

Lt

nQ

twpintsuhtht

Q=istntwnnQ

2 2

2 1 2 1

| d | (x x ) (y y )= +

Area of a triangle

Lttingwhsvtis:


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1 2 3 2 3 1 3 1 2

1Area of ABC A [x (y y ) x (y y ) x (y y )]

2 = = + +

Condition for Collinearity of Three Points

Thpints

n

ini..inthsstightini =Z.

+

+

=0

Division of a line by a point

1. Internal division

Ltpintwhihivisintnthinjining

n

inthti:

AP l

PB m = thn

2 1

2 1

lx mxx

l m

ly myy

l m

+=

++

=+

intspintpintintnivisin

2. External division

Hpintivistnthinjinting

n

inthti:.

AP l

PB m = thn

2 1

2 1

lx mxx

l m

ly myy

l m

=

=


4/24


intspintpinttnivisin

Note:utninttiivisininp+tgtthintstnivisinit.

Mid-point of a line

Ltthipintinjining

n

ut=sinAM l 1

MB m 1= = inthvu

intnivisin

intsi-pint

1 2 1 2x x y y,2 2

+ +

Centroid of a Triangle

Thntiisthpintintstinthinsting.

Lttingwithvtis


5/24


DEnFthinsintsting

tpintGntithtingthn

intsntiG

1 2 3 1 2 3x x x y y y,3 3

+ + + +

Circumcentre of a Triangle

Thiuntisthpintwithinthtingsuhthtitisquiistntthvtis.

Lttingwithvtis

=

=

=

IOisthiuntthn

O=O=O=iu-ius

O=O=O

+

=

+

=

+

Thvtinisustinthunknwn

Incentre of a Triangle

Thinntisthpintwithinthtingwhthintniststhngsintst.Lttingwithvtis.

=

=

=


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nngthsis==n=

Thnintsinntis 1 2 3 1 2 3ax bx cx ay by cy

(x, y) ,a b c a b c

+ + + + + + + +

Slope or gradient of a line

Thspgintinptthinisquttngntthnginintinthintthpsitiv

itinth-isthngingsuinthunt-kwisitin.

Gintin=tn =Thinpsssthughpints

nQ

HnSpin==000000tn = 2 1

2 1

y y

x x

Intercepts of a line on the axes

Ltinintsting-isn-istnQsptiv.

ThnOQ=+vintptn-is

O=+vintptn-is


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Intptisviitintstsnthngtivsi

-is-is

Equation of a straight line

Thqutinstightinisthquitintwnnwhihisstisinthintshn

vpintingnthinnntthsnthpintutsiit.

Equation of a straight line in general formThqutinstightiningnis

++=0

Whnntsiutnusznnstnts.

Spthin=a Coefficient of x

b Coefficient of y

=

Reduction of general form to standard form

Gninthpnstightinthqutinstightinwipsntingn.i..++=

0intgs/tins.Stinutthspnintptgivninitisnssthtgn

qutinistutstns.

Gntsp-intpt=+

++=0 =

{ {

Slope Intercepton y-axis

a cy x

b b

= +

Gntis-intptsx y

1a b

+ =

++=0 +=

a bx y 1

c c

+ =

{ {

InterceptIntercepton y-axison x-axis

x y

c c

a b

+

=

Angle between two intersecting lines

Case 1:

Whenequations of two straight lines in slope-intercept from

EqutinLin

=

+

Wh=tn

EqutininD

=

+

Wh=tn


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Thngtwnthstwinsisgivn

1 1 2 1 2

1 2 1 2

m m m mtan tan

l m m l m m

= = + +

Corollary

Twinspi=

=00 tn=0

Twinsppniuththi

=

Case 2:

Whenequations of two straight lines in general form

Thin+

+

=0

+

+

=0

Twinsintstingi1 1 1

2 2 2

a b c

a b c

Twinspi1 1

2 2

a b

a b=

Twinsppniui

+

=0

Twinsintii1 1 1

2 2 2

a b ca b c

= =

Thngtwnthstwinsisgivn 1 2 1 1 2

1 2 1 2

a b a btan

a a b b

= +

Intersection of two lines

Twins +

+

=0

n +

+

=0

Ithvtwinintsttpintsthnthintsthpintnunutsvingthtw

qutinsn.

Thvusnthpintintstins.

intspintintstin

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

b c b c c a c a,

a b a b a b a b

Equation of a straight line under stated conditions

Equtinstightinwhihpsssthughgivnpint

nstisisnnthwingnitins.


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tgivnin

pniutgivnin

Case 1:

Equation of line passing through point (x1,y

1) and parallel to given line.

Ltthgivnin++=0.Tinutthqutinthquiinthwingstpshvnnut.

Step 1: Thqutinthquiinwhihispt++=0wi++k=0 .....i

Step 2: ut==

thintsthgivnpinttinthvuk

+

+k=0 k=

+

Step 3:utthvuktininstpinitgtthquiin.

Equtinquiinis++

=0

Case 2: Equation of a line passing through (x1, y

1) and is perpendicular to a given line.

Ltthgivnin++=0.Tinutthqutinthquiinstpshvnnutsw.

Step 1: Thqutinthquiinwhihisppniut ax + by + c = 0

b

a

wi+k=0.....i

Step 2: InEq.iput=n=

thintsthgivnpinttinvuk

+k=0

k=

Step 3: utthvuktininstpinqu.itgtthqutinquiin.Equtinquiinis

=0

Case 3: Equation of a line passing through intersections of the two given lines and parallel to a third given line ax + by + c =0

Ltthtwintstingingivn

+

+

=0

+

+

=0

Equtinthquiinntinssttw.

Step 1: Finthpintintstins

svingthvtwqutins

Step 2:FwthstpsttinCase 1.

Case 4: Equation of a line passing through intersections of the two given lines and perpendicular to a third given line ax + by

+ c = 0

Ltthtwintstinginsgivn

+

+

=0

+

+

=0

Equtinthquiinntinsws:

Step 1: Finthpintintstins

svingthvtwqutins.

Step 2: FwthstpsusinCase 2.


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SOLVED EXAMPLES

Example 1:

Finthistntwnthpintsn56

Solution:

Ltpintn56.

Thistntwnpintsis 2 2 2 2| AB | [1 ( 5)] [ 2 6] 6 8 10 units= + = + =

Example 2:

Thpints000086n86jintquit.Finthtpthquit.

Solution:

Ltthpints00=00=86nD=86

H 2 2AB 0 10 10= + =

2 2 2 2BC 8 (16 10) 8 6 10= + = + =

2 2 2CD 0 (16 6) 10 10= + = =

2 2DA 8 6 100 10= + = =

ginign 2 2AC 8 16 320= + = nignD

2 2 2 28 (10 6) 8 4 80= + = + =

S==D=Dsisqu

AC BD ignsntquSthquitisrhombus.

Example 3:

Finpintquiistntthpints6n.

Solution:

Ltthquipintquiistntthgivnpint6

=

6+=++

+6+4+4=+++6+9 7=5 .....in=

6+=+++

+6+4+4=+6+9+++ .....ii +=5Svinginiiwgt==

Hnthquipintis

Example 4:I4000thvtisthtingthninthshpthting.

Solution:

I400n0thn

2 2AB ( 4 0) (0 3) 25 5 units.= + = =

2 2BC (0 0) (3 3) 36 6 units.= + + = =


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2 2AC ( 4 0) (0 3) 25 5 units.= + + = =

Sin AB AC BC,= thtingisisosceles.

Example 5:

Ithpints797nthvtistingthninthsu.

Solution:797nthgivnvtisting.

=7+9+7=4+6=7

=++7=6+0=6

=7+9=0+6=6

S=+

isight-ngtingwith=900

Hnsus900

Example 6:

Finthtingwhsvtis846n5.

Solution:

Usingthuting

1 2 3 2 3 1 3 1 2

1[x (y y ) x (y y ) x (y y )]

2= + +

Wh =8

=

=4

=6

=

=5

1

[ 8( 6 5) ( 4)(5 2) ( 1)( 2 6)]2

= + + + +

1[88 28 4]

2= =8squunits.

Example 7:

Thintsn5n57sptivDEnFthipintsn

sptiv.utththtingDEF.

A (-1, 5)

B (3, 1) C (5, 7)

F (x ,y )3 3 E (x ,y )2 2

D (x ,y )1 1


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Solution:

Usingthu

i-pintD

=

3 5 1 7,

2 2

+ +=44

i-pintE

1 5 5 7, (2, 6)

2 2

+ +

i-pintF

1 3 5 1, (1, 3)

2 2

+ +

Nwusingthu

ting= 1 2 3 2 3 1 3 1 21

[x (y y ) x (y y ) x (y y )]2

+ +

DEF=1

[4(6 3) 2(3 4) 1(4 6)]2

+ + =4squnits

Example 8:

Fwhtvuthpints++n+in.

Solution:

Thgivnpintsnin.

Sththtingthspintswiz.

(+++++=0 =0 +=0

=1

2 =

Example 9:

Finthngthsthinstingwhsvtis4n.

A (3, 4)

B ( 1, 2) C (2, 3)

F (x , y )3 3E (x , y )2 2

D (x ,y )1 1

Solution:

LtDEnFthi-pintnsptiv.Thintsthi-pint

D

1 2 2 3 1 5, ,

2 2 2 2

+ +

E

3 2 4 3 5 7, ,

2 2 2 2

+ +


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F

3 1 4 2, (1,3)

2 2

+ = =

inD

2 21 5 17

3 4 units2 2 2

= + =

inE

2 25 7 29

1 2 units2 2 2

= + =

inF 2 2(2 1) (3 3) 1 unit= + =

Example 10:

Thvtispg+++.Finthuthvt.

A (a + b, a b) B (2a + b, 2a b)

C (a b, a + b)D (x, y)

O

Solution:

Lt+++nDthvtispg.

Thi-pintthignsnDhvthsintstpintO.

2a b x 2a b y a b a b a b a b, ,

2 2 2 2

+ + + + + + + =

2a b x a b a b

2 2

+ + + + =

1 3 5 1, (1, 3)

2 2

+ +

2a b x 2a x b + + = =

n2a b y a b a b

2 2

+ + +=

2a b y 2a y b + = =

HnthuthvtisD


14/24


Example 11:

Inwhihpintthstightinjiningthpints7isivitninthti5:

P (x, y)

A5

3B

Solution:

Ltthquipintiviingthintninthti5:

AP 5 l(say)

BP 3 m = =

Usingthu2 1lx mxxl m

=

Wh=

= 5 ( 3) 3 2 21x

5 3 2 = =

n2 1ly myyl m

=

Wh=

=7

5 7 3 ( 3)y 22

5 3

= =

Hnthquipintis21

, 222

=

Example 12:

Finthintsthpintwhihivisthinsgntjiningthpint5n96inthti:

Solution:

Hthquipintivisintnthinjining5n96inthti:

A (5, 2)

3 1

P B (9, 6)

AP 3 l(Say)

PB 1 m = =

Usingthu

2 1lx mx (3 9) (1 5)x= 8l m 3 1

+ + = =

+ +

2 1ly my (3 6) (1 2)y 4l m 3 1

+ + = = =+ +

Hnthintsthquipint84


15/24


Example 13:

Finthspthinthughthpintsn5

A (2, 3)

B ( 5, 1)

Solution:

Spin==2 1

2 1

y y

x x

Wh

=

=

=5

=

1 ( 3) 4m5 2 7

= =

Example 14:

Thinjining4n5isptthinjining17

, 12

n5 5

,2 2

.Finthvu.

Solution:

Sinthtwinspthispsqu.

Spstiny 4 y 4

5 ( 1) 6

= =

Spnin

51

125 17 4

2 2

+= =

Nwy 4 1 11

y6 4 2

= =

Example 15:

Finthqutinstightin

iWhsspisnwhihpsssthugh7.

iiWhsinintinis60 0npsssthugh4.

iiiWhihpsssthugh5npt-isivWhihpsssthughn68.

Solution:

iHsp==npsssthugh7s=

=7

usingsp-pint

=

7=


16/24


7=+6 +=0Hnthquiqutinthinis+=0

iiH=600

tn tn600= 3 =npsssthugh4S

=

=4nusingsp-pintin:

=

4= 3 +

3 + 3 +4=0

Hnthquiqutinthinis 3 + 3 +4=0

iiiSinthquiinispt-is

H=00 tn=tn00=0=npsssthugh5s=

=5

Usingsp-pint

=

+5=0Hnthquiqutinthinis+5=0

ivHthquiinpsssthughtwpintsn68

Susingtw-pint

2 11 1

2 1

y yy y (x x )

x x

=

8 3(y 3) (x 2)

6 2

= 4=5 54+=0Hnthquiqutinthinis54+=0

Example 16:

Finthqutininpssingthughpint5nptthin7+5=0.

B

P

A

C

D

7x 2y + 5

(5, 1)

Solution :

Ltthgivninwhsqutinis7+5=0

IDthquiinwhihisptnpsssthughpint5thnusingthqutinthquiin

is:++

=0

Wh=7==5

=

775+=0 7=0Hnthquiqutinthinis7=0


17/24


Example 17:

Fwhtvusthins=n2y

x 3 0m

+ + = wippniu?

Solution:

Ltthin=

=+ Spin=

IDthin2y

x 3 0m

+ + =

=

=m 3

x m2 2

inth=+

SspinD=m

2

SininninDppniuththspinspinD=

m2m 12 =

m 1 = Hnthvuis 1

Example 18:

Fwhtvukthstightin++4+k6+=0wi

it-is

iipniut7+5=4

Solution:

Thgivninis

+6k+k+4+k=0

Spthin Cofficient of xCofficient of y

= (2 6k)3 k+=

iNwthgivninispt-issitisppniut-is =900 tn =tn900=

S(2 6k)

3 k

+ =

k=0 k=iiThgivninisppniut7+54=0

Spthin7+54=0is7

5

Nw(2 6k) 7

1

3 k 5

+ =

1 2[ m m 1] = Q

29k

37 =


18/24


Example 19:

Whihthins+=0n+49=0isththigin?

Solution:

Thins

i+=0 Ingn

ii+49=0 Ingn

Tutthppniuistnthiginpthgivnqutinistnvttppnius+sin= .

2 2 2 3 2 2

2x 3y 13[perpendicular form]

2 3 2 3 2 3

=

+ + +

1

13p 13 3.6 units

13

= = =

Siiin

2 2 2 2 2 2

x 4y 9

1 4 1 4 1 4

+ =+ + +

29 9 17p 2.18 unit

1717 = = =

Sp>p

Hnthin+=0isththigin.

Example 20:

Finthqutinthstightinwhihpsssthughthintstinthstightins++=0n+4+

7=0nisptthin5

y x8

=

Solution:

Thpintintstinsins

x 2y 3 0

3x 4y 7 0

x y 1

14 12 9 7 4 6

+ + =+ + =

= =

Sthquiinpsssthughnpt=5

x8

Equtinthq.inwi8+5+k=0 .....i

sinitpsssthughs8+5+k=0 k=Hnthquiinsis5+8+=0.


19/24


. Thistntwnthpints o o(a cos35 , 0) and (a,cos65 ) is. ss-00700

4 45 5

. insgntisngth0units.Ithintsitsnn(2, 3) nthsissththnis0thn

itsintis.ss-00700

96 3, 9 3,9 49, 6 5Nnths

. I 4 4n5 8thvtistingthnthngththinthughvtis.ss-00700

65 117 93 4 113 5 85

4. Ithpints00(3, 3) n(3, ) nquittingthn . = ss-007004

3 4 45 5Nnths

5. If ( 1, 2), (2, 1) and (3,1) nthvtispgthn.ss-007005

a 2, b 0= = a 2, b 0= = a 2, b 6= = 4a 6, b 2= = 5Nnths

6. Ipintstt ( 2, 6) ninthnt=.ss-007006

3

4

4

3

5

34

3

55

2

5

7. Ipints00ninthn1 1

.a b

+ = ss-007007

0 4 1 5Nnths

8. Thinsgntjiningpints ( 3, 4), and (1, 2) isivi-isinthti.ss-007008

: : : 4: 5:

9. Th-ints6 54 sptivnispintthninthvuPBC

:ABC

ss-007009

x y 2

7

+ x y

4

x y 3

5

+ 4

x y

2

+5Nnths

0. Whtisthspthinppniutthinpssingthughthpints5n 4?ss-00700

3

5

2

3

5

3 4

2

5

5

7

3

Practice Exercise - 1


20/24


. 04nDthtupintsquit.Thquitis:ss-0070Squ Rhus g 4Nn 5Nnths

. Thintsthvtistingn .Finthintsthntithting:ss-0070

45 456 55

. Finthspnthintptnth-isthin 3x 3y 6:+ = ss-0070

1

2

1

5

1

34

1

3 5Nnths

4. Thiitingpsitinthpintintstinthins+4=n++=stnstis. ss

-00704

( 5,4) (5, 4) (4, 5) 445 5Nnths

5. Finthistntwnthtwpstightins=+n=+?ssu>.ss-00705

12 2

(c d)

(1 m )

+

12 2

(d c)

(1 m )

+

12 2

d

(1 m )

+

4 12

d

(1 m)

+

5Nnths

SORESHEET

Use HB pencil only. Abide by the time-limit

4 5

4 5

4 5

4

5

6

4 5

4 5

4 5

7

8

9

4 5

4 5

4 5

0

4 5

4 5

4 5

4

5

4 5

4 5

4 5


21/24


. ipus50unitsgRs.0n80unitsRs.80.Suppsingthtthstuvisstightin

stitthstpuing0units:ss-00706

Rs.0 Rs.665 Rs.440 4Rs.65 5Rs.56

. Whtwiththhus ax by c 0? = ss-00707

23c

ab

24c

ab

25c

ab4

2c

ab5

22c

ab

. IthinsTnRs.tingwithvtis0Q00nR0ppniuththwhihth

wingstisisthtinshiptwnn?ss-00708

2 24b a= 2 22b a= a 2b= 4 2 2a b 0+ = 5Nnths

4. Fuvtispgtknin ( 3,1), (a, b), (3,3) and (4,3). Whtwithtit?ss

-00709

4: : : 4: 5:

5. Itwvtisnquittinghvintgintsthnththivtwihv.ss-00700

intgints intswhihtin

tstnintitin 4intswhihitin 5Nnths

6. Ithinsgntjiningn isiviintninthti:4thin+=kthnkis.ss-0070

41

7

5

7

36

74

31

75Nnths

7. Thistntwnthins+4=9n6+8+5=0is.ss-0070

3

10

33

10

33

54

32

55Nnths

8. Ivtnquittingisthiginnthsippsittithsthqutin+=thnthntth

tingis.ss-0070

1 1

,3 3

2 2

,3 3

2 2

,3 3

42 2

,33

5Nnths

9. Ithntithtingthpintsnistthiginthn 3 3 3a b c+ + = ss-00704

0 ++ 4 5 a b c

0. inispsntthqutin4X 5Y 6+ = inthintsstwiththigin00.Yuquitinthqutinthstightinppniutthisinthtpsssthughthpint whihisinthintsstwhiginist .ss-00705

5x 4y 11 = 5x 4y 13 = 5x 4y 3 = 45x 4y 7 = 5Nnths



22/24


SORESHEET

4 5

4 5

4

4 5

4 5

5

6

4 5

4 5

7

8

4 5

4 5

9

0

4 5

4 5

Use HB pencil only. Abide by the time-limit


23/24


. . 5. 7. 9.4.5 4. 6. 8. 0.

Answers

.5 4.5 7. 0.5 .

. 5. 8.5 . 4.

.5 6. 9. . 5.




24/24


Space for rough work

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BMM10234 Chapter 7 - Coordinate Geometry.pdf