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Session

Cartesian Coordinate Geometry And Straight Lines

Session Objectives

- Cartesian Coordinate system and Quadrants
- Distance formula
- Area of a triangle
- Collinearity of three points
- Section formula
- Special points in a triangle
- Locus and equation to a locus
- Translation of axes - shift of origin
- Translation of axes - rotation of axes

René Descartes

Coordinates

Y-axis YOY

X-axis XOX

Coordinates

(2,1)

Abcissa

Ordinate

(-3,-2)

Coordinates

(2,1)

Abcissa

Ordinate

(-3,-2)

(4,?)

Coordinates

(2,1)

Abcissa

Ordinate

(-3,-2)

(4,-2.5)

Quadrants

(,)

(-,)

I

II

III

IV

(,-)

(-,-)

Quadrants

Ist? IInd?

Q (1,0) lies in which Quadrant?

A None. Points which lie on the axes do not lie

in any quadrant.

Distance Formula

PQN is a right angled ?. ? PQ2 PN2 QN2

? PQ2 (x2-x1)2(y2-y1)2

Distance From Origin

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

Applications of Distance Formula

Parallelogram

Applications of Distance Formula

Rhombus

Applications of Distance Formula

Rectangle

Applications of Distance Formula

Square

Area of a Triangle

Area of ? ABC Area of trapezium ABML

Area of trapezium ALNC

- Area of trapezium BMNC

Area of a Triangle

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

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 ab c

Collinearity of Three Points

Method II Use Area of Triangle

A ? (x1, y1) B ? (x2, y2) C ? (x3, y3) Show that

Section Formula Internal Division

Clearly ?AHP ?PKB

Midpoint

Midpoint of A(x1, y1) and B(x2,y2) mn ? 11

Section Formula External Division

P divides AB externally in ratio mn

Clearly ?PAH ?PBK

Centroid

Intersection of medians of a triangle is called

the centroid.

Centroid is always denoted by G.

Centroid

Consider points L, M, N dividing AD, BE and CF

respectively in the ratio 21

Centroid

Consider points L, M, N dividing AD, BE and CF

respectively in the ratio 21

Centroid

Consider points L, M, N dividing AD, BE and CF

respectively in the ratio 21

Centroid

Medians are concurrent at the centroid, centroid

divides medians in ratio 21

We see that L ? M ? N ? G

Centroid

We see that L ? M ? N ? G

Incentre

Intersection of angle bisectors of a triangle is

called the incentre

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.

Incentre

Similarly I can be derived using E and F also

Incentre

Angle bisectors are concurrent at the incentre

Excentre

Intersection of external angle bisectors of a

triangle is called the excentre

Excentre is the centre of the excircle

Excentre

Intersection of external angle bisectors of a

triangle is called the excentre

Excentre is the centre of the excircle

Excentre

Intersection of external angle bisectors of a

triangle is called the excentre

Excentre is the centre of the excircle

Cirumcentre

Intersection of perpendicular bisectors of the

sides of a triangle is called the circumcentre.

OA OB OC circumradius

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.

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.

G divides OH in the ratio 12

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)

Solution

Let the point be P(h,k) PA PB (given) ?

PA2 PB2 ? (h-1)2(k-3)2 (h2)2(k-1)2 ?

6h4k 5 ? equation of locus of (h,k) is 6x4y

5

Illustrative Example

A rod of length l slides with its ends on

perpendicular lines. Find the locus of its

midpoint.

Solution

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, a2b2

l2 ? 4h24k2 l2 ? equation of locus of (h,k)

is 4x24y2 l2

Shift of Origin

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 - k

O ? (-h, -k) with reference to new axes

Illustrative Problem

Show that the distance between two points is

invariant under translation of the axes

Solution

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)

Old dist.

Rotation of Axes

R

Consider 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

Rotation of Axes

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Class Exercise - 1

Solution

Let O be the origin. ? OA2 a2b2, OB2 c2d2,

AB2 (c-a)2(d-b)2 Using Cosine formula in ?OAB,

we have AB2 OA2OB2-2OA.OBcos?

Class Exercise - 2

Solution

Given that ?ABC 2?DBC

Class Exercise - 3

If a ? b ? c, prove that (a,a2), (b,b2) and

(c,c2) can never be collinear.

Solution

Let, if possible, the three points be collinear.

R2 ? R2-R1, R3 ? R3- R2

Solution Cont.

R2 ? R2-R3

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 (ab,a-b), (2ab,2a-b) and (a-b,ab).

Find the fourth vertex.

Solution

Let the fourth vertex be (x,y). Diagonals

bisect each other.

? the required vertex is (-b,b)

Class Exercise - 5

If G be the centroid of ?ABC and P be any

point in the plane, prove that PA2PB2PC2GA2GB2

GC23GP2.

Solution

Choose a coordinate system such that G is the

origin and P lies along the X-axis.

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

(p,0) ? LHS (x1-p)2y12(x2-p)2y22(x3-p)2y32

(x12y12)(x22y22)(x32y32)3p2-2p(x1x2x3

) GA2GB2GC23GP2 RHS

Q.E.D.

Class Exercise - 6

Solution

Let the line intercept at the axes at A and B.

Let R(h,k) be the midpoint of AB.

? Ans (b)

Class Exercise - 7

A point moves so that the ratio of its

distance from (-a,0) to (a,0) is 23. Find the

equation of its locus.

Solution

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

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.

Solution

Let A ? (2,0), B ? (-2,0) Let the point be

P(h,k). Given that

Class Exercise - 9

Find the coordinates of a point where the

origin should be shifted so that the equation

x2y2-6x8y-9 0 will not contain terms in x and

y. Find the transformed equation.

Solution

Let the origin be shifted to (h,k). The given

equation becomes (Xh)2(Yk)2-6(Xh)8(Yk)-9

0 Or, X2Y2(2h-6)X(2k8)Y(h2k2-6h8k-9)

0 ? 2h-6 0 2k8 0 ? h 3, k -4. Thus

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

equation is X2Y2(916-18-32-9) 0 Or, X2Y2

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Class Exercise - 10

Through what angle should the axes be rotated

so that the equation 11x24xy14y2 5 will not

have terms in xy?

Solution

Let the axes be rotated through an angle ?. Thus

equation becomes

Solution Cont.

Therefore, the required angle is

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