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B

C

A

D

x

x

x

x

x

Revision – reading coordinate points on an axes

- Reading a set of coordinates :
- - read off values on the x-axis
- read off values on the y-axis
- put the two values in the order
- (x, y) and you get a coordinate pair

3

2

1

1

2

3

-3

-2

-1

-1

-2

-3

A

F

E

D

B

C

x

x

x

x

x

x

x

Revision – reading coordinate points on an axes

3

2

1

3 3 (3, 3)

2 -2 (2, -2)

1

2

3

-3

-2

-1

-1

-2 -1 (-2, -1)

-2

-1.5 1 (-1.5, 1)

-3

2 1 (2, 1)

Use the spacebar to see the next set

B

x

Join the two points

Produce a right-angled triangle

Mark a point C

(1, 2)

x

C

y

Length of Line Segment

Here, we want to find the

length of the line segment AB

12

We read off the coordinates of the

two points

10

8

6

4

x

2

A

x

2

4

6

8

10

12

Next we will

determine the length of AC and BC

B

x

(1, 2)

x

C(7, 2)

y

Length of Line Segment

Here, we want to find the

length of the line segment AC

12

Looking at the x-values of both

points A and C,

x-value of point A = 1

x-value of point C = 7

Therefore length AC = (7 – 1) = 6

10

8

6

4

Looking at the y-values of both

points B and C,

y-value of point B = 10

y-value of point C = 2

Therefore length BC = (10 – 2) = 8

x

2

A

x

2

4

6

8

10

12

By pythagorus’ theorem, AB2 = AC2 + BC2

B

x

(1, 2)

x

C(7, 2)

AC = 6 units, BC = 8 units

y

Length of Line Segment

By pythagorus’ theorem,

AB2 = AC2 + BC2

12

10

8

6

4

units

x

2

A

x

2

4

6

8

10

12

B

x

(x1, y1)

x

C(x2, y1)

Length of Line Segment – IN GENERAL

y

Replacing A with coordinates (x1, y1) and

B with coordinates (x2, y2)

Therefore point C will have (x2, y1)

12

y2

10

Looking at the x-values of both

points A and C,

x-value of point A = x1

x-value of point C = x2

Therefore length AC = (x2 – x1)

8

6

4

y1

x

2

Looking at the y-values of both

points B and C,

y-value of point B = y2

y-value of point C = y1

Therefore length BC = (y2 – y1)

A

x

2

4

6

8

10

12

x1

x2

B

x

(x1, y1)

x

C(x2, y1)

Length of Line Segment – IN GENERAL

y

Using Pythagorus’ theorem,

12

AB2 = AC2 + BC2

y2

10

8

6

4

y1

x

2

A

x

2

4

6

8

10

12

x1

x2

Applying Pythagorus’ theorem :

Distance =

(to 3 sig. fig.)

Example 1 : Find the distance between the points (2, 8) and (3, 16)

Let A be (2, 8) (x1, y1)

B be (3, 16) (x2, y2)

Applying Pythagorus’ theorem :

Distance =

units (to 3 sig. fig.)

Example 2 : Find the distance between the points (-3, 8) and (-8, 14)

Let A be (–3 , 5) (x1, y1)

B be (–8, 14) (x2, y2)

Complete the following exercise in foolscap papers

and hand up when you return to school.

- Find the distance between the following pairs of points:
- (a) (2, 3) and (9, 7)
- (b) (–1 , 4) and (8, 5)
- (c) (2, –5) and (6, 12)
- (d) (–10, 2) and (–4, –7)
- If the distance between the points A(k, 0) and B(0, k) is 10, find the possible values of k.
- The coordinates of two points are A(–2, 6) and B(9, 3). Find the coordinates of the point C such
- that AC = BC.
- 4. The distance between the points (1, 2t) and (1 – t , 1) is ,
- find the possible values of t.

For all Sec 4 E and 5 N(A), please also complete the following:

Ten Years’ Series Page 67

Q1 to 7.

End of Lesson

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