Coordinate Geometry session 3 SECTION FORMULA · We sayFrom Q divides A and BAB is inm the: n ratio m : n externally B n m A Vikasana – Bridge Course 2012. Section Formula To find

W lWelcomeB id Bridge course

Coordinate Geometry session 3Coordinate Geometry session ‐ 3SECTION FORMULA

Vikasana – Bridge Course 2012

AP B1 4

AWe say P divides AB in the ti i t ll

D2ratio 1 : 4 internally

P2

Here P divides 3CD in the ratio

3 : 2 internally

C

3 : 2 internally

Vikasana – Bridge Course 2012C

AP Bm n

A

We say P divides AB in the D

We say P divides AB in the ratio m: n internally

P1

Here P divides

λ

Here P divides CD in the ratio λ : 1 internally λλ : 1 internally λ

Vikasana – Bridge Course 2012C

Here we say M divides AB in the ratio 3 : 3

3

M divides AB in the ratio 1 : 1 or M is mid point of AB

B

3

3M

3

AVikasana – Bridge Course 2012

We say Q divides AB in the ratioWe say Q divides AB in the ratio6 : 3 or 2 : 1 externally

The ratio of distances of QFrom A and B is 6 : 3

Q3

From A and B is 6 : 3

B3

3

3

AVikasana – Bridge Course 2012

The ratio of distances of Q

Q

The ratio of distances of QFrom A and B is m : nWe say Q divides AB in the ratio

m : n externally

B

Qn

m : n externally

B

mA

m


Section Formula

To find the co-ordinates of a point internal Division

To find the co ordinates of a point which divides the line joining A(x1 , y1) and B(x2 y2) in the ratio m : n and B(x2 , y2) in the ratio m : n internally B

Pm

n

A

P

Vikasana – Bridge Course 2012A

Y Clearly

KK ∆AHP ~ ∆PKB

X

HH

XX’

Y’

O L N ML N M


External division


Find the point which divides the line joiningthe points and B(‐4, 2) in the ratio 3 : 2 internally and externallyratio 3 : 2 internally and externally

==

m = 3 n = 2Vikasana – Bridge Course 2012

Case (i) When the division is internal


Case (i) When the division is external


Find the coordinates of middle point of li j i i h i A( ) d B(6 ) line joining the points A(2, ‐3) and B(6, 5)

(x1 y1 ) = (2, ‐3) (x y ) = (6 5)(x1, y1 ) (2, 3) (x2, y2 ) = (6, 5)

(4 1)= (4, 1)=


If one end of diameter of a circle If one end of diameter of a circle with centre at (2, 3) is (‐1, 5) find the other endthe other end

(x y ) = (‐1 5) (x 1 , y 1 ) = ( 1 , 5)

(x , y ) = (2 , 3) (x2 , y2 ) = ? ‐1 + x2 = 4 ; 5 + y2 = 6 , x2 = 5 y2 =1


CentroidA(x1, y1)

Centroid is always denoted by G.

EF G

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

Point of Intersection of the medians of a triangle is called centroid of the triangle

Vikasana – Bridge Course 2012centroid of the triangle

A(x y )

CentroidA(x1, y1)

EF G

B(x2, y2) C(x3, y3)DL divides AD internally in the ratio 2 : 1

Suppose that L , M and N divide the

in the ratio 2 : 1

pp ,medians AD, BE and CF internally in the ratio 2 : 1

Vikasana – Bridge Course 2012the ratio 2 : 1

CentroidA(x1, y1)

EF G

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


Find the centroid of a triangle whose i (1 2) ( 2 3) d ( 3 8)vertices are (1 , 2), ( 2, 3) and ( 3, ‐8)


To find the ratio where the line joining i di id h di two points divide the coordinate axes

λy2+y1 = 0A(x y )λ

y2 y1λ =

A(x1, y1)

11B(x2, y2)


Find the ratio in which the line joining (3, ‐1) and (2, 5) is divided by x‐axis.

λ =

Therefore requeired ratio is 1:5 internallyinternally


IncenterA(x1, y1)

The point of intersection of the angle bisectors( 1, y1)

EF I

of the angle bisectorsIncentre is the

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

is the centre of the

Let BC = a, AC = b, AB = cthe incircle

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

Vikasana – Bridge Course 2012,

respectively.

IncenterA(x1, y1)

EF I

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

Incenter divides the bisector AD internally in the ratio (c + b) : ainternally in the ratio (c + b) : a


Documents

Coordinate Geometry session 3 SECTION FORMULA · We sayFrom Q divides A and BAB is inm the: n ratio m : n externally B n m A Vikasana – Bridge Course 2012. Section Formula To find