Cartesian Coordinate Geometry and Straight Lines-1

8/10/2019 Cartesian Coordinate Geometry and Straight Lines-1

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Mathematics


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Cartesian Coordinate GeometryAnd

Straight Lines

Session


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1. Cartesian Coordinate system and

Quadrants2. Distance formula3. Area of a triangle4. Collinearity of three points5. Section formula6. 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

Origin1 2 3 4

+ve direction

-1-2-3-4

-ve direction

-1

-2

-3

- v e

d i r

e c t i o n

1

2

3

+ v e

d i r

e c t i o 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

X X

Y

O

Y

x2

y 1

y 2

N PQN is a right angled .

PQ 2 = PN 2 + QN 2

2 22 1 2 1PQ x x y y

y 2

- y 1

(x 2-x 1) PQ 2 = (x 2-x 1) 2+(y 2-y 1) 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

X X

Y

O

Y A(x 1 , y 1)

C(x 3 , y 3)

B ( x 2 ,

y 2

)

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

X X Y O

Y A(x

1, y

1)

C(x 3 , y 3)

B (

x 2 ,

y 2

)

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 A i (x i, y i)

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 I :

Use Distance Formula

a b

c

Show that a+b = c


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Collinearity of Three Points

Method II :

Use Area of Triangle

A (x 1 , y 1)

B (x 2 , y 2)

C (x 3 , y 3)

Show that1 1

2 2

3 3

x y 1x y 1 0x y 1


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Section Formula Internal Division

X X

Y

O

Y

L N M

H

K

Clearly AHP ~ PKBAP AH PHBP PK BK

1 1

2 2

x x y ymn x x y y

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


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Midpoint

Midpoint of A(x 1 , y 1) and B(x 2 ,y 2)

m:n 1:1

1 2 1 2x x y yP ,2 2


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Section Formula External Division

X X

Y

O

Y

L N M

H

K

Clearly PAH ~ PBKAP AH PHBP BK PK

1 1

2 2

x x y ymn 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(x 1 , y 1)

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

B(x 2 , y 2) C(x 3 , y 3)D

EF G2 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(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF G2 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(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF G2 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(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF G2 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(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF G2 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 yG ,3 3


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Incentre

Intersection of angle bisectors of atriangle is called the incentre

A(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF I

Incentre isthe centre ofthe incircle

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

AD, BE and CF are the anglebisectors of A, B and Crespectively.

BD AB bDC AC c

2 3 2 3bx cx by cy

D ,b c b c


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Incentre

A(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF I

BD AB b

DC AC c

2 3 2 3bx cx by cy

D ,b c b c

AI AB AC AB AC c b

Now,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(x 1 , y 1)

B(x 2 , y 2) C(x 3 , y 3)D

EF I

BD AB b

DC AC c

2 3 2 3bx cx by cy

D ,b c b c

AI AB AC AB AC c b

Now,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(x 1 , y 1)

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

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

B(x 2 , y 2) C(x 3 , y 3)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

CO

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 Orthocentreafter we learn straight lines

and their equations


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Cirumcentre, Centroid andOrthocentre

The circumcentre O, Centroid G andOrthocentre 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 pointwhich moves under a given conditionor conditions is called its locus

e.g. locus of a point having aconstant distance from a fixed point:

Circle!!


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Locus a Definition

The curve described by a pointwhich moves under a given conditionor conditions is called its locus

e.g. locus of a point equidistant fromtwo fixed points :

Perpendicular bisector!!


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Equation to a Locus

The equation to the locus of a pointis that relation which is satisfied bythe coordinates of every point on thelocus of that point

Important :A Locus is NOT anequation. But it isassociated with an

equation


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Equation to a Locus

Algorithm to find the equation to alocus :

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 ExampleFind 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)

PA 2 = PB 2

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

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

Solution :


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Illustrative ExampleA 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, a 2+b 2 = l 2

4h 2+4k 2 = l 2

equation of locus of (h,k) is 4x 2+4y 2 = l 2

B(0,b)

A(a,0)O

P(h,k)

Solution :


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Shift of Origin

X X

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 - k

O (-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(x 1 , y 1), B(x 2 , y 2)

Let the origin be shifted to (h, k)

new coordinates : A(x 1-h, y 1-k), B(x 2-h, y 2-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

X X

Y

O

Y P(x,y)

x

yConsider 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

Ysin ,R

XcosR


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Rotation of Axes

xcos ,R

ysin ,R

Ysin ,R

XcosR

xcos cos sin sin

R

ysin cos cos sin

R

X Y xcos sinR R R

X Y ysin cos

R R R

x X cos Y sin

y X sin Y cos

X x cos y sinY 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 = a 2+b 2 , OB 2 = c 2+d 2 , AB 2 = (c-a) 2+(d-b) 2

Using Cosine formula in OAB, we have

AB2 = OA 2+OB 2-2OA.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 - 2Four points A(6,3), B(-3,5), C(4,-2)and D(x,3x) are given such that

Find x. DBC 1ABC 2

Given that ABC = 2 DBC

6 3 1 x 3x 13 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,a 2), (b,b 2)and (c,c 2) can never be collinear.

Let, if possible, the three points becollinear.

2

2

2

a a 11b b 1 0

2c c 1

R2 R 2-R 1 , R 3 R 3- R 22

2 2

2 2

a a 1b a b a 0 0c b c b 0

2a a 1

b a c b 1 b a 0 01 c b 0

Solution :


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Solution Cont.R2 R 2-R 3

2a a 1

b a c b 0 a c 0 01 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 - 4Three 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 - 5If G be the centroid of ABC and Pbe any point in the plane, prove thatPA2+PB 2+PC 2=GA 2+GB 2+GC 2+3GP 2 .

Let A (x 1 ,y 1), B (x 2 ,y 2), C (x 3 ,y 3), P (p,0)

LHS = (x 1-p) 2+y 12+(x 2-p) 2+y 22+(x 3-p) 2+y 32

= (x 1 2+y 12)+(x 2 2+y 22)+(x 32+y 3 2)+3p 2-2p(x 1+x 2+x 3)=GA 2+GB 2+GC 2+3GP 2

=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 - 6The locus of the midpoint of the portionintercepted between the axes by theline xcos +ysin = p, where p is aconstant, is

2 2 22 2 2

2 22 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 p 14k 4h

2 2 21 1 4

Locusx y p

Ans : (b)


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Class Exercise - 7A 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 23h a k

2 2

2 2

h a k 49h a k

2 2 2

2 2 2

h 2ah a k 49h 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 - 8Find 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 - 9Find the coordinates of a point where theorigin should be shifted so that the equationx2+y 2-6x+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, X 2+Y 2+(2h-6)X+(2k+8)Y+(h 2+k 2-6h+8k-9) = 0

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

Thus the origin is shifted to (3,-4).Transformed equation is X 2+Y 2+(9+16-18-32-9) = 0

Or, X 2+Y 2 = 34

Solution :


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Class Exercise - 10Through what angle should the axesbe rotated so that the equation11x 2+4xy+14y 2 = 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 2cos 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 3sin cos 2 sin 0


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Thank you

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Cartesian Coordinate Geometry and Straight Lines-1