Mathematics. Cartesian Coordinate Geometry And Straight Lines Session

Mathematics

Cartesian Coordinate Geometry

And

Straight Lines

Session

1. Cartesian Coordinate system and Quadrants

2. 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

René Descartes

Coordinates

XX’

Y

Y’

O

Origin

1 2 3 4

+ve direction

-1-2-3-4

-ve direction

-1

-2

-3 -ve d

irect

ion

1

2

3

+ve d

irect

ion

X-axis : X’OX

Y-axis : Y’OY

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

(?,?)

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,?)

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)

Quadrants

XX’ O

Y

Y’

III

III IV

(+,+)(-,+)

(-,-) (+,-)

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 any quadrant.

Distance Formula

P(x 1, y 1

)

Q(x 2, y 2

)

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-y

1(x2-x1)

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

Distance From Origin

Distance of P(x, y) from the origin is

2 2x 0 y 0

2 2x y

Applications of Distance Formula

Parallelogram


Rhombus


Rectangle


Square

Area of a Triangle

XX’

Y’

O

Y A(x1, y1)

C(x3, y3)

B(x

2,

y2)

M L N

Area of ABC =

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

Area of a Triangle

XX’Y’O

YA(x1, y1)

C(x3, y3)

B(x

2,

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 MN

2 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

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 calculate area of Quadrilateral,

Pentagon, Hexagon etc.

Collinearity of Three Points

Method I :

Use Distance Formula

a b

c

Show that a+b = c

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

Section Formula – Internal Division

A(x 1, y 1

)

B(x 2, y 2

)

XX’

Y’

O

Y

P(x, y)

m

n:

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

Midpoint

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

m:n 1:1

1 2 1 2x x y yP ,

2 2

Section Formula – External Division

XX’

Y’

O

Y

A(x 1, y 1

)

P(x, y)

B(x 2, y 2

)

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

Centroid

Intersection of medians of a triangle 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 is always denoted by G.

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, BE and 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

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

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

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 are concurrent at the centroid, centroid divides medians in

ratio 2:1

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 yG ,

3 3

Incentre

Intersection of angle bisectors of a triangle is called the incentre

A(x1, y1)

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

EF I

Incentre is the centre of the incircle

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

AD, BE and CF are the angle bisectors of A, B and C respectively.

BD AB bDC AC c

2 3 2 3bx cx by cy

D ,b c b c

Incentre

A(x1, y1)

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

EF I

BD AB bDC AC c

2 3 2 3bx cx by cy

D ,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 cxI

a b c

Similarly I can be derived using E and F also

Incentre

A(x1, y1)

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

EF I

BD AB bDC AC c

2 3 2 3bx cx by cy

D ,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 cxI

a b c

Angle bisectors are concurrent at the incentre

Excentre

Intersection of external angle bisectors of a triangle is called the excentre

A(x1, y1)

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

EF

E

Excentre is the centre of the excircleEA = Excentre opposite A

1 2 3 1 2 3A

ax bx cx ay by cyE ,

a b c a b c

Excentre


A(x1, y1)

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

EF

E

Excentre is the centre of the excircleEB = Excentre opposite B

1 2 3 1 2 3B


a b c a b c

Excentre


A(x1, y1)

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

EF

E

Excentre is the centre of the excircleEC = Excentre opposite C

1 2 3 1 2 3C


a b c a b c

Cirumcentre

Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre.

OA = OB = OC

= circumradius

A

B

CO

The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.

Orthocentre

Intersection of altitudes of a triangle is called the orthocentre.

A

B C

H

Orthocentre is always denoted by H

We will learn to find coordinates of Orthocentre after we learn straight lines

and their equations

Cirumcentre, Centroid and Orthocentre

The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear.

O

H

G

1 : 2

G divides OH in the ratio 1:2

Locus – a Definition

The curve described by a point which moves under a given condition or conditions is called its locus

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

Circle!!

Locus – a Definition

The curve described by a point which moves under a given condition or conditions is called its locus

e.g. locus of a point equidistant from two fixed points :

Perpendicular bisector!!

Equation to a Locus

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

Important :A Locus is NOT an equation. But it is associated with an

equation

Equation to a Locus

Algorithm to find the equation to a locus :

Step I : Assume the coordinates of the point whose locus is to be found to be (h,k)

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

Step III : Eliminate the variables, if any

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

Illustrative Example

Find the equation to the locus of the point equidistant from A(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 = 5

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

Solution :

Illustrative Example

A rod of length l slides with its ends on perpendicular lines. Find the locus of its midpoint.

Let the point be P(h,k)

Let the lines be the axes

Let 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 :

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 to O’ 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

Illustrative Problem

Show that the distance between two points is invariant under translation 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 :

Rotation of Axes

XX’

Y’

O

Y

P(x,y)

x

yConsider a point P(x, y)

Let the axes be rotated through an angle .Let new P (X, Y) make an angle with the new x-axis

X

X’

Y’

O

Y

X

X’Y’

O

Y

X

X’ Y’

O

Y

X

X’Y’

O

Y

Y

X

xcos ,

R

R

ysin ,

R

Ysin ,

R

Xcos

R

Rotation of Axes

xcos ,

R y

sin ,R

Y

sin ,R

X

cosR

xcos cos sin sin

R

ysin cos cos sin

R

X Y xcos sin

R 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

Class Exercise

Class Exercise - 1

If the segments joining the pointsA(a,b) and B(c,d) subtend an angle at the origin, prove that

2 2 2 2

ac bdcos

a b c d

Solution

On simplifying, 2 2 2 2

ac bdcos

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

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 1ABC 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 49

28x 142

11 3x or x

8 8

Solution :

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 be collinear.

2

2

2

a a 11

b b 1 02

c c 1

R2 R2-R1, R3 R3- R2

2

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 :

Solution Cont.R2 R2-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.

Class Exercise - 4

Three vertices of a parallelogram taken in order are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b). Find the fourth 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 yand

2 2 2 2

the required vertex is (-b,b)

Solution :

Class Exercise - 5

If G be the centroid of ABC and P be 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+y12+(x2-p)2+y2

2+(x3-p)2+y32

= (x12+y1

2)+(x22+y2

2)+(x32+y3

2)+3p2-2p(x1+x2+x3)

=GA2+GB2+GC2+3GP2

=RHS

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

Q.E.D.

Solution :

Class Exercise - 6

The locus of the midpoint of the portion intercepted between the axes by the line xcos+ysin = p, where p is a constant, 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

Solution

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

p pR h,k ,

2cos 2sin

p psin , cos

2k 2h

2 2

2 2

p p1

4k 4h 2 2 2

1 1 4Locus

x y p

Ans : (b)

Class Exercise - 7

A point moves so that the ratio of its distance 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 :

Class Exercise - 8

Find the locus of the point such that the line segments having end points (2,0) and (-2,0) subtend a right angle at that point.

Let A (2,0), B (-2,0)

Let the point be P(h,k). Given that2 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 :

Class Exercise - 9

Find the coordinates of a point where the origin should be shifted so that the equation x2+y2-6x+8y-9 = 0 will not contain terms in x 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 :

Class Exercise - 10

Through what angle should the axes be rotated so that the equation 11x2+4xy+14y2 = 5 will not have terms in xy?

Let the axes be rotated through an angle . 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 :

Solution Cont.

Therefore, the required angle is

cos 2sin 2cos sin 0

1tan or tan 2

2

1 11tan or tan 2

2

2 2

2 2

2 2

Or, 11cos 4sin cos 14sin X

4cos 6sin cos 4sin XY

11sin 4sin cos 14cos 5

2 22cos 3sin cos 2sin 0

Thank you

Documents

Mathematics. Cartesian Coordinate Geometry And Straight Lines Session