11 x1 t05 01 division of an interval (2013)

Coordinate Geometry
Distance Formula

Coordinate Geometry
Distance Formula

d  x2  x1    y2  y1 
2 2

Coordinate Geometry
Distance Formula

d  x2  x1    y2  y1 
2 2

e.g. Find the distance between
(–1,3) and (3,5)

Coordinate Geometry
Distance Formula

d  x2  x1    y2  y1 
2 2

(–1,3) and (3,5)
d  5  3   3  1
2 2

Coordinate Geometry
Distance Formula

d  x2  x1    y2  y1 
2 2

(–1,3) and (3,5)
d  5  3   3  1
2 2

 22  42
 20
 2 5 units

Coordinate Geometry
Distance Formula

d  x2  x1    y2  y1 
2 2

(–1,3) and (3,5)
d  5  3   3  1
2 2

 22  42
 20
 2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras

Coordinate Geometry
Distance Formula Midpoint Formula

d  x2  x1    y2  y1 
2 2

(–1,3) and (3,5)
d  5  3   3  1
2 2

 22  42
 20
 2 5 units
using Pythagoras

Coordinate Geometry
 x1  x2 y1  y2 
d  x2  x1    y2  y1 
2 2
M  , 
 2 2 
(–1,3) and (3,5)
d  5  3   3  1
2 2

 22  42
 20
 2 5 units
using Pythagoras

Coordinate Geometry
 x1  x2 y1  y2 
d  x2  x1    y2  y1 
2 2
M  , 
 2 2 
e.g. Find the distance between e.g. Find the midpoint of
(–1,3) and (3,5) (3,4) and (–2 ,1)
d  5  3   3  1
2 2

 22  42
 20
 2 5 units
using Pythagoras

Coordinate Geometry
 x1  x2 y1  y2 
d  x2  x1    y2  y1 
2 2
M  , 
 2 2 
(–1,3) and (3,5) (3,4) and (–2 ,1)
d  5  3   3  1
2 2
 3  2 , 4 1
M  
 2 2 
 22  42
 20
 2 5 units
using Pythagoras

Coordinate Geometry
 x1  x2 y1  y2 
d  x2  x1    y2  y1 
2 2
M  , 
 2 2 
(–1,3) and (3,5) (3,4) and (–2 ,1)
d  5  3   3  1
2 2
 3  2 , 4 1
M  
 2 2 
 22  42
 , 
1 5
 20  
2 2
 2 5 units
using Pythagoras

Coordinate Geometry
 x1  x2 y1  y2 
d  x2  x1    y2  y1 
2 2
M  , 
 2 2 
(–1,3) and (3,5) (3,4) and (–2 ,1)
d  5  3   3  1
2 2
 3  2 , 4 1
M  
 2 2 
 22  42
 , 
1 5
 20  
2 2
 2 5 units
The midpoint formula
averages the x and y values
using Pythagoras

Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1

In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.


X Y


X A Y


A divides XY internally
in the ratio m:n
X A Y


in the ratio m:n
X A Y
 
 
m n


in the ratio m:n
X A Y OR
 
 
m n


in the ratio m:n
X A Y OR
 
  A divides YX internally
m n
in the ratio n:m


in the ratio m:n
X A Y OR
 
m n
in the ratio n:m

X Y


in the ratio m:n
X A Y OR
 
m n
in the ratio n:m

X Y A


in the ratio m:n
X A Y OR
 
m n
in the ratio n:m

A divides XY externally
X Y A in the ratio m:n


in the ratio m:n
X A Y OR
 
m n
in the ratio n:m

 n 
 
A divides XY externally
X Y A in the ratio m:n

 
m

Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining  3,4  and
5,6 internally in the ratio 1 : 3

• Write down the endpoints of the interval in the same order as
they are mentioned.

they are mentioned.

 3,4 5,6

they are mentioned.
• Write down the ratio.

 3,4 5,6

they are mentioned.

 3,4 5,6

1: 3

they are mentioned.
• Draw a cross joining the ratio to the two points

 3,4 5,6

1: 3

they are mentioned.
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma

 3,4 5,6

1: 3

they are mentioned.

 3,4 5,6  
P , 
 

1: 3

they are mentioned.
• Add the numbers in the ratio together to get the denominator

 3,4 5,6  
P , 
 

1: 3

they are mentioned.

 3,4 5,6  
P , 
 4 4 

1: 3

they are mentioned.
• Multiply along the cross and add to get the numerator
 3,4 5,6 P    ,


 4 4 

1: 3

they are mentioned.
 3,4 5,6 P   3  3  1 5 ,



 4 4 

1: 3

they are mentioned.
 3,4 5,6 P   3  3  1 5 , 3  4  1 6 
 
 4 4 

1: 3

they are mentioned.
 3,4 5,6 P   3  3  1 5 , 3  4  1 6 
 
 4 4 
  4 , 18 

1: 3 
 4 4

they are mentioned.
 3,4 5,6 P   3  3  1 5 , 3  4  1 6 
 
 4 4 
  4 , 18 
   1, 9 
1: 3  
 4 4  2

Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.

• Done exactly the same as internal division, except make one of
the numbers in the ratio negative


3,1 9,2


3,1 9,2

5:2


3,1 9,2

5:2
 
P , 
 


3,1 9,2

5:2
 
P , 
 3 3 


3,1 9,2

5:2
 2 3  5 9 
P , 
 3 3 


3,1 9,2

5:2
 2  3  5  9 2  1  5  2 
P , 
 3 3 


3,1 9,2

5:2
 2  3  5  9 2  1  5  2 
P , 
 3 3 
  39 ,  12 
 
 3 3 


3,1 9,2

5:2
 2  3  5  9 2  1  5  2 
P , 
 3 3 
  39 ,  12 

 3 3 
 13,4 

Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4  divides the line segment joining A 1,8 and B x, y 
internally in the ratio 2 : 3.
Find the coordinates of the point B.

• Draw the endpoints, ratio and cross the same as previously


 1,8  x, y 


 1,8  x, y 

2:3

• Create the fraction for the x value and equate it with the known value

 1,8  x, y 

2:3


 1,8  x, y 

3  1  2  x 2:3
1
5


 1,8  x, y 

3  1  2  x 2:3
1
5
5  3  2 x


 1,8  x, y 

3  1  2  x 2:3
1
5
5  3  2 x
2x  8
x4

• Repeat for the y value
 1,8  x, y 

3  1  2  x 2:3
1
5
5  3  2 x
2x  8
x4

 1,8  x, y 

3  1  2  x 2:3 3 8  2  y
1 4
5 5
5  3  2 x
2x  8
x4

 1,8  x, y 

3  1  2  x 2:3 3 8  2  y
1 4
5 5
5  3  2 x 20  24  2 y
2x  8
x4

 1,8  x, y 

3  1  2  x 2:3 3 8  2  y
1 4
5 5
5  3  2 x 20  24  2 y
2x  8 2 y  4
x4 y  2

 1,8  x, y 

3  1  2  x 2:3 3 8  2  y
1 4
5 5
5  3  2 x 20  24  2 y
2x  8 2 y  4
x4  B  4,2  y  2

Alternative

Alternative

A B

Alternative

If P divides AB internally in the ratio 2 : 3
A P B
  
 
2 3

Alternative

A P B Then B divides AP externally in the ratio
  
 
2 3 5:3

Alternative

  
 
2 3 5:3

 1,8 1,4

Alternative

  
 
2 3 5:3

 1,8 1,4

5:3

Alternative

  
 
2 3 5:3

 
 1,8 1,4 B , 
 

5:3

Alternative

  
 
2 3 5:3

 
 1,8 1,4 B , 
 2 2 

5:3

Alternative

  
 
2 3 5:3

 3  1  5  1 
 1,8 1,4 B , 
 2 2 

5:3

Alternative

  
 
2 3 5:3

 3  1  5  1 3  8  5  4 
 1,8 1,4 B , 
 2 2 

5:3

Alternative

  
 
2 3 5:3

 3  1  5  1 3  8  5  4 
 1,8 1,4 B , 
 2 2 
 8, 4 

5:3 
  2  2

Alternative

  
 
2 3 5:3

 3  1  5  1 3  8  5  4 
 1,8 1,4 B , 
 2 2 
 8, 4 
5:3 
  2  2
 4,2 

Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4  find the value of k.
,

,
• Draw the endpoints, ratio and cross the same as usual

,

6,4 0,4

,

6,4 0,4

 k :1

,
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4

 k :1

,
6,4 0,4

 k :1
1 6   k  0
3 
 k 1

,
6,4 0,4

 k :1
1 6   k  0
3 
 k 1
3k  3  6

,
6,4 0,4

 k :1
1 6   k  0
3 
 k 1
3k  3  6
3k  9
k 3

,
6,4 0,4

 k :1
1 6   k  0
3 
 k 1
3k  3  6
3k  9
k 3  P divides AB externally in the ratio 3 : 1

,
6,4 0,4
Exercise 5A;
1ad, 2ad,
 k :1 3 i, iii in all, 4ace,
1 6   k  0
3  5 i, ii ace, 6bd,
 k 1 8, 9, 11, 13b, 16,
3k  3  6 17, 20, 21, 23, 24
3k  9
k 3  P divides AB externally in the ratio 3 : 1

11 x1 t05 01 division of an interval (2013)

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11 x1 t05 01 division of an interval (2013)