123328034 6 coordinate geometry doc
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6. COORDINATE GEOMETRY
Unit 6.1 : To Find the distance between two oints [BACK TO BASICS]
A( 11 ,yx ) and B( 22 ,yx ) : AB =2 2
2 1 2 1( ) ( )x x y y + .
Eg. 1 Given two points A(2,3) and B(4,7)
Distance of AB = 2 2(4 2) (7 3) +
= 4 1+
= 2! "nit.
E1. #(4,$) and %(3,2)
#% =
& 1! '
E2. ($,!) and ($,2)
&2'
E3. *(7,1) and +(2,$)
& 41 '
E4. (1!,) and -(4,2)
& $2 '
E$. (/4,/1) and 0(/2,1)
& 1 '
oe ca55enging %"estions6.E1. *e distance etween two points A(1, 3) and
B(4, 8) is $. 9ind te possi5e va5es of 8.
7, /1
E2. *e distance etween two points #(/1, 3) and
%(8, ) is 1!. 9ind te possi5e va5"es of 8.
7, /
E3. *e distance etween two points (/2, $) and
(1, 8) is 1! . 9ind te possi5e va5es of 8.
, 4
E4. *e distance etween two points ;(/1, p) and
(!, !) and #(8, 28) is te
sa?e as te distance etween te points A(/4, 3) and
B(1, /7), find te possi5e va5"es of 8.
k = 5, -5
7
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Unit 6.2 : Division of a Line Segment
6.!.1 To "ind the #id$oint o" Two Gi%en &oints.
9o?"5a : idpoint =
++
2,
2
2121 yyxx
Eg. #(3, 2) and %($, 7)
idpoint, =
++
2
72,
2
$3
= (4 ,2
:)
E1 #(/4, ) and %(, !)
(2, 3)
E2 #(, 3) and %(2, /1)
(4, 1)
E3 #(!,/1), and %(/1, /$)
(/ @ , /3)
6.!.! Di%ision o" a 'ine (e)#ent
% divides te 5ine seg?ent # in te atio PQ :QR= ? : n. #(x,y), (x,y)
% (x,y)= ++++ nmmyny
nm
mxnx2121 ,
(NOT : St!"ents a#e st#ong$% a"vise" to s&et' a $ine segment )efo#e a**$%ing te fo#m!$a+
Eg1. *e point # intena55 divides te 5ine seg?ent
oining te point (3,7) and C(,2) in te atio 2 : 1.
9ind te coodinates of point #.
# =
++
++
12
)2(2)7(1,
12
)(2)3(1
=
3
11,
3
1$=
11$,
3
E1. *e point # intena55 divides te 5ine seg?ent
oining te point (4,$) and C(/,/$) in te atio
1 : 3. 9ind te coodinates of point #.
$1,
2
More Exercise : The Ratio Theorem
n?
#(1,
1) (
2,
2),(- %+
n
?
(2,
2)
#(1,
1)
,(- %+
1
2
C(, 2)
(3, 7)
/(- %+
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(NOT : St!"ents a#e st#ong$% a"vise" to s&et' a $ine segment )efo#e a**$%ing te fo#m!$a+
E1. divides #% in te atio 2 : 1. 9ind te
coodinates of if
(a) #(1, 2) and %( /$, 11)
() #(/4, 7) and %(, /$)
(a) (/3, ) () (4 , /1)
E2. # divides AB in te atio 3 : 2. 9ind te
coodinates of # if
(c) A(2, /3) and B( /, 7)
(d) A(/7, $) and B(, /$)
(a) (/4, /3) () (2 , /1)
E3. is a point tat 5ies on te staigt 5ine s"c
tat 3 = . f te coodinates of te points
and ae (4,$) and (/,/$) espective5, find te
coodinates of point .
3 =
MS
RM=3
1, : = 1 : 3
Ans :
2
$,1
E4. # is a point tat 5ies on te staigt 5ine *+ s"c
tat 3*# = 2#+. f te coodinates of te points * and
+ ae (/,7) and (1,/3) espective5, find te
coodinates of point #.
(/$, 3)
E$. *e points #(3, p), B(/1, 2) and F(,7) 5ie on a
staigt 5ine. f # divides BF intena55 in te atio? : n , find (a) ? : n , () te va5"e of p.
(a) 2 : 3 () p = 4
E. (x, y) , divides te points #(28, 8) and
%(2x, 4y) in te atio 3 : $. Epessxin te?s of y.
(Ans : = 4)
Unit 6.* To Find A+eas o" &o,-)ons
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Aea of a po5gon =2
1
1 2 3 1
1 2 3 1
...
...
x x x x
y y y y
Note : Te a#ea fo!n" 0i$$ )e *ositive if te 'oo#"inates of te *oints a#e 0#itten in te anti1'$o'&0ise
o#"e# an" negative if te% a#e 0#itten in te '$o'&10ise o#"e#.
Ea?p5e 1 : Fa5c"5ate te aea of a tiang5e given :
E1. #(!, 1), %(1, 3) and (2,7)
Aea of H #% =2
1 ! 1 2 !
1 3 7 1
=
= 1 "nit 2
1. #(2,3), %($,) and (/4,4)
Aea of H #% =
17
2"nit2
2. *e coodinates of te tiang5e ABF ae ($, 1!), (2,1)and (, 8) espective5. 9ind te possi5e va5"es of 8,
given tat te aea of tiang5e ABF is 24 "nits2.
8 = 3 , 3$
3. *e coodinates of te tiang5e * ae (4, 3), (/1, 1) and(t, /3) espective5. 9ind te possi5e va5"es of t , given tat
te aea of tiang5e * is 11 "nits2.
t = ! , /22
ii) Aea of a I"adi5atea5 =2
1
14321
14321
yyyyy
xxxxx
1. #(1,$), %(4,7), (,) and (3,1).
Aea of #% =
= "nit 2
2. #(2, /1), %(3,3), (/1, $) and (/4, /1).
&27'
Note : If te a#ea is e#o ten te *oints a#e 'o$$inea#.
1. Given tat te points #($, 7), %(4, 3) and (/$, 8) ae
co55inea, find te va5"e of 8.
2. ow tat te points ;(4, ),
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8= 33
Unit 6. : E/0ations o" (t+ai)ht 'ines
*e EI"ation of a taigt 5ine ?a e epessed in te fo55owing fo?s:
i) *e genea5 fo? : a J J c = !
ii) *e gadient fo? : y = mx c ! m = "ra#ient , c = y-interce$t
iii) *e intecept fo? :a
xJ%
y= 1 , a = x-interce$t , % = y-interce$t
a+ If given te g#a"ient an" one *oint:
1yy = )( 1xxm
Eg. 9ind te eI"ation of a staigt 5ine tat passes
to"g te point (2,/3) and as a gadient of4
1.
1yy = )( 1xxm
)2(4
1)3( = xy
144 =xy
E1. 9ind te eI"ation of a staigt 5ine tat passes
to"g te point ($,2) and as a gadient of /2.
= /2 J 12
E2. 9ind te eI"ation of a staigt 5ine tat passes
to"g te point (/,3) and as a gadient of4
3.
4 = 3 J 3
b I" two oints a+e )i%en :
Cote : 0o" ?a find te gadient fist, ten "se
eite (a) y = mx c
> () y & y'= m( x & x')
>
(c)1
1
xx
yy
=12
12
xx
yy
Eg. 9ind te eI"ation of a staigt 5ine tat passes
to"g te points (/3, /4) and (/$,)
)3(
)4(
xy = )3($ )4(
1:$ = xyE1. 9ind te eI"ation of a staigt 5ine tat passes
to"g te points (2, /1) and (3,!)
E2. 9ind te eI"ation of a staigt 5ine tat passes
to"g te points (/4,3) and (2,/$)
1
Gadient = m
/(-3 %3+
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= / 3 4 J 3 J7 = !
c+ Te -1inte#'e*t an" te %1inte#'e*t a#e given:
m= /
errce$tx
erce$ty
int
int
EI"ation of taigt
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x 2y & = 0 1x =
Unit 6.7 /a#a$$e$ Lines an" /e#*en"i'!$a# $ines
6.2.1 &a+a,,e, ,ines3 1m 4 2m6.2.! &e+endic0,a+ ,ines3 21mm 4 $1
Unit 6.7.3 Dete#mine 0ete# ea' of te fo$$o0ing *ai#s of $ines a#e *a#a$$e$.
Eg. = 3 2 and 3 = 4
= 3 2 , ?1 = 3
3 = 4
= 3 4 , ?2 = 3
ince ?1= ?2 , te two 5ine ae paa55e5 .
1. = 2 J$ and 4 J 2 = $
C
2. 3 3 = 7 and J = $
C
3. 2 3 = $ and = 4 J
0
4. 3 = 12 and = 3 J 2
0
$. 43 2
x y = and = / 3
C
Unit 6.7.2 Dete#mine 0ete# ea' of te fo$$o0ing *ai#s of $ines a#e *e#*en"i'!$a#.
Eg. = 3 2 and J 3 = 4
= 3 2 , ?1 = 3
J 3 = 43 = J 4
1 4
3 3y x= + , ?2 =
1
3
ince ?1. ?2 = 13 13 =
,
*e two given 5ines ae pependic"5a .
1. = 2 J$ and 4 J 2 =
C
2. 3 = 2 2 and 2 J 3 = 1
C
3. 3 = 2 and J 2 = $
0
4. = 2 / 3 and 43
x y
=
$. 13 4
x y
= and J 3 = !
3
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0 0
6.7.2 A**$i'ations (m3.m25 8 3+
E5.1 (&M !77. Diaga? 1 sows a staigt
5ine #% wit te eI"ation 12 4
x y+ = . 9ind te
eI"ation of te staigt 5ine pependic"5a to #%
and passing to"g te point %.
Answe:
y = 3 x 4
E.2. Diaga? 2 sows a staigt 5ine #% wit te
eI"ation 1 2
x y+ = . 9ind te eI"ation of te
staigt 5ine pependic"5a to #% and passing
to"g te point #.
Answe:
y = x & '/
E.3 Diaga? 3 sows a staigt 5ine wit te
eI"ation J 2 = . 9ind te eI"ation of te
staigt 5ine pependic"5a to and passing
to"g te point .
Answe:
E.4. Diaga? 4 sows a staigt 5ine AB wit te
eI"ation 2 3 = . 9ind te eI"ation of te
staigt 5ine pependic"5a to AB and passing
to"g te point B.
Answe:
4
x
y
.
Q
P
Diaga? 1
x
y
.
Q
P
Diaga? 2
x
y
.
R
S
Diaga? 3x
y
.
Diaga? 4
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y = 2x & '2 2y = x & 6
6.7.2 A**$i'ations (m3.m25 8 3+ 8 mo#e e-e#'ises
E.$ Diaga? $ sows a staigt 5ine #% wit teeI"ation 4 J 3 = 12. 9ind te eI"ation of te
staigt 5ine pependic"5a to and passing
to"g te ?idpoint of .
Answe:
4xy = /
E.. Diaga? sows a staigt 5ine AB wit teeI"ation 1
4
x y = . 9ind te eI"ation of te
pependic"5a isecto of te 5ine AB.
Answe:
2x y =
E.7. 9ind te eI"ation of te staigt 5ine tatpasses to"g te point ( 1, 2) and is pependic"5a
to te staigt 5ine J 3 J = !.
y = x & '
E. 9ind te eI"ation of te staigt 5ine tatpasses to"g te point (3, !) and is pependic"5a
to te staigt 5ine 3 2 = 12.
2xy =
E. 9ind te eI"ation of te staigt 5ine tat
passes to"g te oigin > and is pependic"5a to
te staigt 5ine tat passes to"g te points
#(1, 1 ) and %(/3,7).
y = 3 x
E. 1! 9ind te eI"ation of te staigt 5ine tat
passes to"g te point (/2,4) and is pependic"5a
to te staigt 5ine wic passes to"g te oigin
> and te point (, 2).
y = -x
$
x
y
.
R
S
Diaga? $ x
y
.
Diaga?
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Unit 6.6 9!ation of a Lo'!s
Note : St!"ents UST )e a)$e to fin" "istan'e )et0een t0o *oints [ !sing /%tago#as Teo#em]
*A; : *o 9ind te eI"ation of te 5oc"s of te ?oving point # s"c tat its distances of # fo? te points
% and ae eI"a5.
Eg 1. %(, /$) and (1,)
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y =2x - ' x y = 4
TA(8 : To "ind the e/0ation o" the ,oc0s o" the #o%in) oint & s0ch that its distances "+o#
the oints A and 9 a+e in the +atio # : n(Note : S&et' a "iag#am to e$* %o! !sing te "istan'e fo#m!$a 'o##e't$%+
Eg 1. A(/2,3), B(4,) and ? : n = 1: 2
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x2y2 '0x & '4y 2 = 0 x2y2 'x 2y 5 = 0
S/
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1
2$
k=
1
2 1$
k=
7. (2!!) Diaga? $ sows te staigt 5ine AB
wic is pependic"5a to te staigt 5ine FB at te
point B.
*e eI"ation of FB is = 2 1 .
9ind te coodinates of B.
(2, )
.Diaga? sows te staigt 5ine #% wic is
pependic"5a to te staigt 5ine % at te point %.
*e eI"ation of % is = 4 .
9ind te coodinates of %
%($, 1)
.(2!!4) *e point A is (/1, 2) and B is (4, ). *e
point # ?oves s"c tat #A : #B = 2 : 3. 9ind te
eI"ation of 5oc"s of #. &3 ?a8s'
15x25y250x'2y & '=0
1!. *e point is (3, /$) and is (!, 1). *e point
# ?oves s"c tat # : # = 2 : 1. 9ind te
eI"ation of 5oc"s of #. &3 ?a8s'
1x2y22x & y & '0 = 0
11.*e point A is (, /2) and B is (4, ). 9ind te
eI"ation of te pependic"5a isecto of AB.
&3 ?a8s'
12. *e point is (2, /3) and is (4, $). *e point
# ?oves s"c tat it is a5was te sa?e distance
fo? and fo? . 9ind te eI"ation of 5oc"s of
#. &3 ?a8s'
x
y
O
A( >+
C
Diaga? $B
?
?
?
x
y
O
/( 6+
=
Diaga? ,
?
?
?
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2y = x & 2
x4y = ?
SPM Questions (Coordinate Geometry) Paper 2
Cote : So$!tions to tis 9!estion )% s'a$e "#a0ing 0i$$ not )e a''e*te".
3. (S/ 26 /2 ,@+
Diaga? 3 sows te tiang5e.wee .is te oigin. #oint @5ies on te staigt 5ine.
(a) Fa5c"5ate te aea, in "nits2, of tiang5e.. &2 marks'
() Given tat@ : @ = 3 : 2, find te coodinates of @. &2 marks'
(c) A pointP?oves s"c tat its distance fo? pointis a5was twice its distance fo? point.
(i) 9ind te eI"ation of 5oc"s ofP,
(ii) Lence, dete?ine wete o not tis 5oc"s intecepts tey/ais. & marks'
2. (S/ 27 /2 ,@+n Diaga? $, @= !! and te eI"ation of te staigt 5ine@is 2yJxJ = !.
(a) 9ind
(i) te eI"ation of te staigt 5ine
(ii) te coodinates of &$ marks'
() *e staigt 5ineis etended to a pointAs"c tat : A = 2 : 3.
9ind te coodinates ofA. &2 marks'
(c) A pointP?oves s"c tat its distance fo? pointis a5was $ "nits.9ind te eI"ation of te 5oc"s ofP. &3 marks'
1!!
x
y
O
A(1 >+
Diaga? 3
C
?
?
?
B(6 12+
x
y
O
(-4, 6)
Diaga? $B
?
?
?
C
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. (S/ 2> /2 ,2+
Diaga? 1 sows a staigt 5ine @Awic ?eets a staigt 5ine at te point D.
*e point @5ies on tey/ ais.
(a) -ite down te eI"ations ofin te fo? of intecepts. & 1 mark '
() Given tat 2A = A, find te coodinates ofA> & 2 marks'
(c) Given tat @Ais pependic"5a to,find tey/intecept of @A. & 3 marks'
>. (S/ 2 /2 ,33+
A pointP?oves a5ong te ac of a cic5e wit cente(2, 3). *e ac passes to"g Q(/2, !) and
($, 8).
(a) 9ind
(i) te eI"ation of te 5oc"s ofP,
(ii) te va5"es of k> & marks'
() *e tangent to te cic5e at point Q intesects tey/ is at point T.
9ind te aea of tiang5e .QT> &4 marks'
7. (S/ 22 /2 ,33+
Given tat( 1 , 2 ) and(2, 1) ae two fied points. #ointP?oves s"c tat te atio ofPtoPis
a5was 1 : 2.
(a) ow tat te 5oc"s ofPis 2J 2J4 J J$ = !. &2 marks'
() ow tat te point @(!, /$) 5ies on tis 5oc"s. &2 marks'
(c) 9ind te eI"ation of te 5ine@> &3 marks'
(d) Given tat te 5ine@c"ts te 5oc"s ofPagain at pointA> 9ind te coodinates ofA> &3 marks'
1!1
x
y
O
A( 6+
Diaga? 1
C
?
?
?
B(@ +
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SPM Questions (Coordinate Geometry) Paper 1
1. !77**e eI"ations of two staigt 5ines ae 1
$ 3
y x+ = and $ = 3 J 24. Dete?ine wete te
5ines ae pependic"5a to eac ote. &'
! !77 Diaga? 4 sows a staigt 5ine #% wit te eI"ation 12 3
x y+ = . 9ind te eI"ation of te
staigt 5ine pependic"5a to #% and passing to"g te point %. &3 marks'
12
33
y x= +
*.!77 *e point A is (/1, 2) and B is (4, ). *e point # ?oves s"c tat #A : #B = 2 : 3. 9ind te
eI"ation of 5oc"s of #. & marks'15x25y250x'2y'=0
Epess$in te?s of k.
1!2
. !772*e fo55owing info?ation efes to te eI"ations of two staigt 5ines, K; and *, wic ae
pependic"5a to eac ote.
1
2$
k=
x
y
O
,
/
Diaga? 4
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1!3
2. !776Diaga? $ sows te staigt 5ineABwic is pependic"5a to te staigt 5ine CBat te point.
*e eI"ation of CBis % 5 2- 8 3 .
9ind te coodinates ofB. (2, )
x
y
O
A( >+
C
Diaga? $B
?
?
?
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