12. Cartesian Coordinate Geometry and Straight Lines-1

8/13/2019 12. Cartesian Coordinate Geometry and Straight Lines-1

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Mathematics


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Cartesian Coordinate Geometry

And

Straight Lines

Session


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1. Cartesian Coordinate system and

Quadrants2. Distance formula3. Area of a triangle4. Collinearity of three points5. Section formula

6. Special points in a triangle7. Locus and equation to a locus8. Translation of axes - shift of

origin9. Translation of axes - rotation of

axes

Session Objectives


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Ren Descartes


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Coordinates

XX

Y

Y

O

Origin

1 2 3 4

+ve direction

-1-2-3-4

-ve direction

-1

-2

-3

-ve

direction

1

2

3

+ve

directio

n

X-axis : XOX

Y-axis : YOY


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Coordinates

XX

Y

Y

O

1 2 3 4-1-2-3-4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(?,?)


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Coordinates

XX

Y

Y

O

1 2 3 4-1-2-3-4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(4,?)


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Coordinates

XX

Y

Y

O

1 2 3 4-1-2-3-4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(4,-2.5)


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Quadrants

XX O

Y

Y

III

III IV

(+,+)(-,+)

(-,-) (+,-)


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Quadrants

XX O

Y

Y

III

III IV

(+,+)(-,+)

(-,-) (+,-)Q : (1,0) lies in which Quadrant?

Ist? IInd?

A : None. Points which lie on the axes do not lie in anyquadrant.


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Distance Formula

x1

XX

Y

O

Y

x2

y1

y2

N PQN is a right angled .

PQ2= PN2+ QN2

2 22 1 2 1PQ x x y y

y2-y1

(x2-x1)

PQ2= (x2-x1)2+(y2-y1)

2


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Distance From Origin

Distance of P(x, y) from theorigin is

2 2x 0 y 0

2 2

x y


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Applications of Distance Formula

Parallelogram


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Rhombus


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Rectangle


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Square


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Area of a Triangle

XX

Y

O

Y A(x1, y1)

C(x3, y3)

B(x2,y2

)

M L N

Area of ABC =

Area of trapezium ABML + Area of trapezium ALNC- Area of trapezium BMNC


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Area of a Triangle

XX

Y

O

YA(x

1

, y1

)

C(x3, y3)

B(x2,y2

)

M L N

Area of trapezium ABML + Area of trapezium ALNC- Area of trapezium BMNC

1 1 1BM AL ML AL CN LN BM CN MN2 2 2

2 1 1 2 1 3 3 1 2 3 3 21 1 1

y y x x y y x x y y x x2 2 2

1 1

2 2

3 3

x y 11

x y 12

x y 1

Sign of Area : Points anticlockwise +ve

Points clockwise -ve


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Area of Polygons

Area of polygon with points Ai(xi, yi)

where i = 1 to n

2 21 1 n 1 n 1 n n

3 32 2 n n 1 1

x yx y x y x y1. . .

x yx y x y x y2

Can be used to calculatearea of Quadrilateral,

Pentagon, Hexagon etc.


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Collinearity of Three Points

Method II :

Use Area of Triangle

A (x1, y1)

B (x2, y2)

C (x3, y3)

Show that1 1

2 2

3 3

x y 1

x y 1 0

x y 1


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Section Formula Internal Division

XX

Y

O

Y

L N M

H

K

Clearly AHP ~ PKBAP AH PH

BP PK BK

1 1

2 2

x x y ym

n x x y y

2 1 2 1mx nx my nyP ,m n m n


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Midpoint

Midpoint of A(x1, y1) and B(x2,y2)

m:n 1:1

1 2 1 2x x y yP ,2 2


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Section Formula External Division

XX

Y

O

Y

L N M

H

K

Clearly PAH ~ PBKAP AH PH

BP BK PK

1 1

2 2

x x y ym

n x x y y

2 1 2 1mx nx my nyP ,m n m n

P divides AB externally in ratio m:n


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Centroid

Intersection of medians of atriangle is called the centroid.

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2

Centroid isalways denotedby G.


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Centroid

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2

Consider points L, M, N dividing AD, BEand CF respectively in the ratio 2:1

2 3 2 31 1

x x y yx 2 y 2

2 2L ,1 2 1 2


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Centroid

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2


1 3 1 32 2

x x y yx 2 y 2

2 2M ,1 2 1 2


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Centroid

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2


1 2 1 23 3

x x y yx 2 y 2

2 2N ,1 2 1 2


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Centroid

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2

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

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

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

We see that L M N G

Medians areconcurrent at thecentroid, centroiddivides medians in

ratio 2:1


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Centroid

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF G

2 3 2 3x x y yD ,2 2

1 3 1 3x x y yE ,2 2

1 2 1 2x x y yF ,2 2

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

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

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

We see that L M N G

Centroid

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

G ,3 3


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Incentre

Intersection of angle bisectors of atriangle is called the incentre

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF I

Incentre isthe centre ofthe incircle

Let BC = a, AC = b, AB = c

AD, BE and CF are the angle

bisectors of A, B and Crespectively.

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,b c b c


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Incentre

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF I

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,

b c b c

AI AB AC AB AC c bNow,

ID BD DC BD DC a

2 3 2 31 1

bx cx by cyax b c ay b c

b c b cI ,

a b c a b c

1 2 3ax bx cxIa b c

Similarly I can be derivedusing E and F also


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Incentre

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF I

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,

b c b c

AI AB AC AB AC c bNow,

ID BD DC BD DC a

2 3 2 31 1

bx cx by cyax b c ay b c

b c b cI ,

a b c a b c

1 2 3ax bx cxIa b c

Angle bisectors areconcurrent at the incentre


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Excentre

Intersection of external anglebisectors of a triangle is calledthe excentre

A(x1, y1)

B(x2, y2) C(x3, y3)D

EF

E

Excentre isthe centre ofthe excircleEA= Excentre opposite A

1 2 3 1 2 3A

ax bx cx ay by cyE ,

a b c a b c


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Excentre


A(x1, y1)

B(x2, y2) C(x3, y3)D

EF

E

Excentre isthe centre ofthe excircleEB= Excentre opposite B

1 2 3 1 2 3B


a b c a b c


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Excentre


A(x1, y1)

B(x2, y2) C(x3, y3)D

EF

E

Excentre isthe centre ofthe excircleEC= Excentre opposite C

1 2 3 1 2 3C


a b c a b c


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Cirumcentre

Intersection of perpendicularbisectors of the sides of a triangleis called the circumcentre.

OA = OB = OC

= circumradius

A

B

C

O

The above relation gives twosimultaneous linear equations. Theirsolution gives the coordinates of O.


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Orthocentre

Intersection of altitudes of a triangleis called the orthocentre.

A

B C

H

Orthocentreis always

denoted by H

We will learn to findcoordinates of Orthocentre

after we learn straight linesand their equations


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Cirumcentre, Centroid andOrthocentre

The circumcentre O, Centroid G and

Orthocentre H of a triangle arecollinear.

O

H

G

G divides OH in the

ratio 1:2


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Locus a Definition

The curve described by a point

which moves under a given conditionor conditions is called its locus

e.g. locus of a point having a

constant distance from a fixed point:

Circle!!


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Locus a Definition

The curve described by a point

which moves under a given conditionor conditions is called its locus

e.g. locus of a point equidistant from

two fixed points :

Perpendicular bisector!!


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Equation to a Locus

The equation to the locus of a point

is that relation which is satisfied bythe coordinates of every point on thelocus of that point

Important :A Locus is NOTanequation. But it isassociated with an

equation


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Equation to a Locus

Algorithm to find the equation to a

locus :

Step I : Assume the coordinatesof the point whose locus is to befound to be (h,k)

Step II : Write the given conditions in mathematicalform using h, k

Step III : Eliminate the variables, if any

Step IV : Replace h by x and k by y in Step III. Theequation thus obtained is the required equation to locus


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Illustrative Example

Find the equation to the locus of

the point equidistant fromA(1, 3) and B(-2, 1)

Let the point be P(h,k)

PA = PB (given)

PA2= PB2

(h-1)2+(k-3)2= (h+2)2+(k-1)2

6h+4k = 5equation of locus of (h,k) is 6x+4y = 5

Solution :


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Illustrative Example

A rod of length l slides with its

ends on perpendicular lines. Findthe locus of its midpoint.

Let the point be P(h,k)

Let the

lines be the axesLet the rod meet the axes at

A(a,0) and B(0,b)

h = a/2, k = b/2

Also, a2+b2= l2

4h2+4k2= l2

equation of locus of (h,k) is 4x2+4y2= l2

B(0,b)

A(a,0)O

P(h,k)

Solution :


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Shift of Origin

XX

Y

O

Y

O(h,k)

P(x,y)

x

yX Y

Consider a point P(x, y)

Let the origin be shifted toO with coordinates (h, k)relative to old axes

Let new P (X, Y)x = X + h, y = Y + k

X = x - h, Y = y - kO (-h, -k) with reference to new axes


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Illustrative Problem

Show that the distance between two

points is invariant undertranslation of the axes

Let the points have vertices

A(x1, y1), B(x2, y2)

Let the origin be shifted to (h, k)

new coordinates : A(x1-h, y1-k), B(x2-h, y2-k)

2 21 2 1 2Old dist. (x x ) (y y )

2 21 2 1 2& New dist. (x h x h) (y h y h)

= Old dist.

Solution :


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Rotation of Axes

XX

Y

O

YP(x,y)

x

y

Consider a point P(x, y)

Let the axes be rotatedthrough an angle .

Let new P (X, Y) make

an angle with the newx-axis

xcos ,R

ysin ,R

Y

sin ,R

X

cosR


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Rotation of Axes

xcos ,R

ysin ,R

Y

sin ,R

X

cosR

xcos cos sin sin

R

ysin cos cos sin

R

X Y x

cos sinR R R

X Y ysin cos

R R R

x X cos Y sin

y X sin Y cos

X x cos y sin

Y y cos x sin


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Class Exercise


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Class Exercise - 1

If the segments joining the points

A(a,b) and B(c,d) subtend an angle

at the origin, prove that

2 2 2 2ac bd

cosa b c d


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Solution

On simplifying,

2 2 2 2ac bd

cos a b c d

Let O be the origin.

OA2= a2+b2, OB2= c2+d2, AB2= (c-a)2+(d-b)2

Using Cosine formula in OAB, we have

AB2= OA2+OB2-2OA.OBcos

2 2 2 2 2 2 2 2 2 2c a d b a b c d 2 a b c d cos


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Class Exercise - 2

Four points A(6,3), B(-3,5), C(4,-2)and D(x,3x) are given such that

Find x.DBC 1

ABC 2

Given that ABC = 2DBC

6 3 1 x 3x 1

3 5 1 2 3 5 1

4 2 1 4 2 1

6 5 2 3 4 3 1 6 20 2 x 5 2 3x 4 3 1 6 20 2 28x 14 49

4928x 14

2

11 3x or x

8 8

Solution :


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Class Exercise - 3

If a b c, prove that (a,a2), (b,b2)and (c,c2) can never be collinear.

Let, if possible, the three points becollinear.

2

2

2

a a 11b b 1 0

2c c 1

R2R2-R1, R3R3- R22

2 2

2 2

a a 1

b a b a 0 0

c b c b 0

2a a 1

b a c b 1 b a 0 0

1 c b 0

Solution :


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Solution Cont.R2R2-R3

2a a 1

b a c b 0 a c 0 0

1 c b 0

b a c b c a 0

This is possible only if a = b or b = c or c = a.

But a b c. Thus the points can never be collinear.

Q.E.D.


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Class Exercise - 4

Three vertices of a parallelogramtaken in order are (a+b,a-b),(2a+b,2a-b) and (a-b,a+b). Find thefourth vertex.

Let the fourth vertex be (x,y).

Diagonals bisect each other.

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

and2 2 2 2

the required vertex is (-b,b)

Solution :


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Class Exercise - 5

If G be the centroid of ABC and Pbe any point in the plane, prove that

PA2+PB2+PC2=GA2+GB2+GC2+3GP2.

Let A (x1,y1), B (x2,y2), C (x3,y3), P (p,0)

LHS = (x1-p)2+y1

2+(x2-p)2+y2

2+(x3-p)2+y3

2

= (x12+y12)+(x22+y22)+(x32+y32)+3p2-2p(x1+x2+x3)

=GA2+GB2+GC2+3GP2

=RHS

Choose a coordinate system such that G isthe origin and P lies along the X-axis.

Q.E.D.

Solution :


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Class Exercise - 6

The locus of the midpoint of the portionintercepted between the axes by the

line xcos+ysin= p, where p is aconstant, is

2 2 2

2 2 2

2 2

2 2 2 2

1 1 4(a) x y 4p (b)

x y p

4 1 1 2(c) x y (d)p x y p


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Solution

Let the line intercept at theaxes at A and B. Let R(h,k) bethe midpoint of AB.

p pR h,k ,2 cos 2 sin

p psin , cos

2k 2h

2 2

2 2

p p1

4k 4h 2 2 2

1 1 4Locus

x y p

Ans : (b)


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Class Exercise - 7

A point moves so that the ratio of itsdistance from (-a,0) to (a,0) is 2:3.Find the equation of its locus.

Let the point be P(h,k). Given that

2 2

2 2

h a k 2

3h a k

2 2

2 2

h a k 4

9h a k

2 2 2

2 2 2

h 2ah a k 4

9h 2ah a k

2 2 25h 26ah 5k 5a 0

2 2 2

the required locus is

5x 26ax 5y 5a 0

Solution :


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Class Exercise - 8

Find the locus of the point such thatthe line segments having end points

(2,0) and (-2,0) subtend a right angleat that point.

Let A (2,0), B (-2,0)Let the point be P(h,k). Given that

2 2 2PA PB AB

2 2 22 2h 2 k h 2 k 2 2 2 22h 2k 8 16

2 2

the required locus is

x y 4

Solution :


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Class Exercise - 9

Find the coordinates of a point where theorigin should be shifted so that the equation

x2+y2-6x+8y-9 = 0 will not contain terms inx and y. Find the transformed equation.

Let the origin be shifted to (h,k). The given equation becomes(X+h)2+(Y+k)2-6(X+h)+8(Y+k)-9 = 0

Or, X2+Y2+(2h-6)X+(2k+8)Y+(h2+k2-6h+8k-9) = 0

2h-6 = 0; 2k+8 = 0 h = 3, k = -4.

Thus the origin is shifted to (3,-4).

Transformed equation is X2+Y2+(9+16-18-32-9) = 0

Or, X2+Y2= 34

Solution :


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Class Exercise - 10

Through what angle should the axesbe rotated so that the equation11x2+4xy+14y2= 5 will not haveterms in xy?

Let the axes be rotated through anangle . Thus equation becomes

2

2

11 X cos Y sin 4 X cos Y sin X sin Y cos

14 X sin Y cos 5

Solution :


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Solution Cont.

Therefore, the required angle is

cos 2 sin 2 cos sin 0 1

tan or tan 22

1 11tan or tan 22

2 2

2 2

2 2

Or, 11cos 4 sin cos 14 sin X

4 cos 6 sin cos 4 sin XY

11sin 4 sin cos 14 cos 5

2 22 cos 3 sin cos 2 sin 0


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Thank you

Documents

12. Cartesian Coordinate Geometry and Straight Lines-1