1. Distance between Any Two Points on a Plane

2. Well done. Now, try to find the distance between A and B.
Do you remember how to calculate the distance between P and Q? x y
P( , 5) 1
Q( , 5)
B( , 5) 5
A( , 2) 1
AB is neither a horizontal line nor a vertical line. I don’t know how to calculate the distance.
The distance between P and Q is (5 
1) units
= 4 units.

3. Distance between Any Two Points on a Plane
BC is a vertical line.
Consider two points A(1, 2) and B(5, 5) on a rectangular coordinate plane. 2 3 4 1 x y 5
B(5, ) 5
Draw a horizontal line from A and
Coordinates of C =
( , ) 5 2
a vertical line from B.
AC = (5 – 1) units =
4 units
4 units
The two lines intersect at C. C
( , ) 1 5 2
BC = (5 – 2) units =
3 units
3 units
A( , 2)
By Pythagoras’ theorem, 2 + = BC AC AB
AC is a horizontal line.
units 3 4 2 + =
units 5 =
5 units

4. It is known as the distance formula between two points.
In general, for any two points A(x1, y1) and B(x2, y2) on a rectangular coordinate plane, x y
A(x1, y1)
B(x2, y2)
y2 – y1
C( , ) x2 y1
x2 – x1 AB ( ) 2 1 y x - + =
It is known as the distance formula between two points.

5. Find the length of PQ in the figure.y
Find the length of PQ in the figure.
Q( , ) 9 6
P( , ) 3 1 x = PQ [9 -
3)] ( - 2 + (6 - 1) 2
units
Remember to write the ‘units’. = 12 2 + 5 2
units =
144 + 25
units
units
169 =
units 13 =

6. Follow-up question 1
In each of the following, find the distance between the two given points.
(a) A(2, 1) and B(5, 5) (b) C(1, 2) and D(7, 6)
(Leave your answers in surd form if necessary.)
Solution
(a) = AB ) 2 5 ( - 2 + ) 1 5 ( - 2
units
units 16 9 + =
units 25 =
units 5 =

7. Follow-up question 2
In each of the following, find the distance between the two given points.
(a) A(2, 1) and B(5, 5) (b) C(1, 2) and D(7, 6)
(Leave your answers in surd form if necessary.)
Solution
(b) = CD 1) 7 ( - 2 +
2)] ( [6 -
units 2
units 8 8) ( 2 + - =
units 64 + =
units) 2 8
(or
units
128 =

8. Example 1

9. Solution

10. Example 2

11. Solution

13. Example 3

14. Solution

16. Slope and Inclination of a Straight Line

17. How about the steepness of these two lines?
You are right! In fact, in coordinate geometry, we use or to describe the steepness of a straight line.
How about the steepness of these two lines?
Let’s consider the two paths below. Which path is steeper?
slope
inclination x y
Straight line B
Straight line A
Path A
Path B
It seems that straight line B is steeper.
Of course, path B is steeper.

18. Slope of a Straight Line
The slope of a straight line is the ratio of the vertical change to the horizontal change between any two points on the straight line,
horizontal change
vertical change x y A B
i.e.
line
straight a of
slope =
change
vertical
horizontal change

19. Consider a straight line L passing through A(x1, y1) and B(x2, y2), where x1  x2.
Coordinates of C =
(x2, y1) y L
B( , ) x2 y2
change
horizontal
vertical
line
straight a of
Slope =
vertical change x
A( , ) x1 y1 C
( , ) x2 y1 1 2 y - =
horizontal change 1 2 x -

20. If we use the letter m to represent the slope of the straight line L, thenx y
A(x1, y1)
B(x2, y2) L 1 2 x y m - = 2 1 x y + - 2 1 x y - 2 1 x
(x y -
(y ) or m = 2 1 x y m -  1 2 x y m - 
Note:
and

21. Let’s find the slope of AB.y x
A(–1, –1)
B(4, 3) 1 2 - x y of
Slope = AB 1) ( 3 - =
(x1, y1) = (1, 1)
(x2, y2)= (4, 3) 1) ( 4 - 5 4 =

22. Let’s find the slope of AB.x y
A(–1, –1)
B(4, 3)
Alternatively, 2 1 - x y of
slope = AB 3 1 - =
(x1, y1) = (1, 1)
(x2, y2)= (4, 3) 4 1 - 5 4 =

23. Follow-up question 3
In each of the following, find the slope of the straight line passing through the two given points.
(a) A(2, 4) and B(3, –2) (b) C(1, 1) and D(3, 5)
Solution 3 2 ) ( 4 of
Slope - = AB 1 3 5 of
Slope - = CD
(a)
(b) 6 - = 2 =
The slopes of AB and CD are in opposite sign.
What does this mean?

24. In fact,
for straight lines sloping upwards from left to right, their slopes are positive.
for straight lines sloping downwards from left to right, their slopes are negative. x y
Slope = 2
Slope = 1
Slope = 0.5 x y
Slope = –0.5
Slope = –1
Slope = –2
Slope
Slope
> 0
< 0

25. The greater the value of the slope, the steeper is the straight line.
In fact, x y
Slope = 2
Slope = 1
Slope = 0.5
for straight lines sloping upwards from left to right, their slopes are positive. x y
Slope = –0.5
Slope = –1
Slope = –2
for straight lines sloping downwards from left to right, their slopes are negative.
The steepest line
The steepest line
Slope > 0
Slope < 0
The greater the numerical value of the slope, the steeper is the straight line.
The greater the value of the slope, the steeper is the straight line. 2 2 1
0.5
> 1
>
0.5
>
>

26. What are the slopes of a horizontal line and a vertical line?

27. The slope of a horizontal line is .x y
A(x1, y1)
B(x2, y1)
For a line that is parallel to the x-axis, of
Slope 1 2 - = x y AB =
2. The slope of a vertical line is
undefined x y
D(x1, y1)
C(x1, y2)
For a line that is parallel to the y-axis, of
Slope 1 2 - = x y CD 1 2 - = y
 It is meaningless to divide a number by 0.

28. Follow-up question 4
On the rectangular coordinate plane as shown, L1, L2, L3 and L4 are four straight lines. Given that their slopes are 0, 0.5, 1 and 2 (not in the corresponding order), determine the slopes of each line according to their steepness. x y L1
Straight Line L1 L2 L3 L4
Slope L2 L3 2 1
0.5 L4
L1 and L2 are sloping upwards from left to right and L1 is steeper.
L4 is sloping downwards from left to right.
L3 is a horizontal line.

29. Inclination
We can also describe the steepness of a straight line by its inclination. y
 is the angle that the straight line L makes with the positive x-axis (measured anti-clockwise from the x-axis to L)
Straight line L  x
positive x-axis
 is called the inclination of L.
Note: For 0 <  < 90, when  increases,
the steepness of L also increases.

30. Is there any relationship between the inclination of a straight line and its slope?

31. Draw a horizontal line from A and
Consider a straight line L passing through A and B with inclination  . y x A B L a
Draw a horizontal line from A and C 
a vertical line from B.
They intersect
at C.
Let BAC = a.

32. Consider a straight line L passing through A and B with inclination  .y L B A a C  AC BC AC BC x
L = of
Slope
L = of
Slope
 = a ∵
  and a are corresponding angles. ∴ a =
tan
tan q
tan q
Note that ACB = 90. AC BC =
 By the definition of tangent ratio ∴
tan  =

33. If the inclination of a straight line L is 50, slope of L = tan 50
The relationship between the inclination  and the slope of a straight line L is
slope of L = tan 
For example:
If the inclination of a straight line L is 50,
slope of L = tan 50
= 1.19 (cor. to 3 sig. fig.) y L
50 x

34. Let’s find the inclination  of L.y L
Slope of L = tan  ∵
slope = ∴
= tan 
60 x  60 = q

35. Follow-up question 5
(a) Given that the inclination of a straight line L is 35, find the slope of L correct to 3 significant figures.
Solution
(a) Slope of L = tan 35
fig.)
sig. 3 to
(cor.
700 . =

36. Follow-up question 5
(b) Given that the slope of a straight line L is 2, find the inclination  of L correct to the nearest degree.
Solution
(b) Slope of L = tan 
2 = tan  ∴  63 = q
(cor. to the nearest degree)

37. Example 4
Solution

38. Example 5
Solution

39. Example 6
Solution

40. Example 7
Solution

41. Example 8
Solution

42. Example 9
Solution

43. Example 10
Solution

44. Parallel and Perpendicular Lines

45. The figure shows two parallel horizontal lines. What are their slopes?
Good. Actually, the slopes of parallel lines are always equal. Let me show you the proof.
Yes. If now, we rotate the lines to the same extent, what do you think about their steepness and their slopes?
The figure shows two parallel horizontal lines. What are their slopes? y
slope = 0
slope = 0 x
By observation, it seems that their steepness are always the same, so they have the same slopes.
Both of the slopes are 0.

46. Parallel Lines
The figure shows two straight lines L1 and L2, whose inclinations are 1 and 2 respectively. y L1 L2 1 2
If L1 // L2, then ∴
i.e. slope of L1= slope of L2 x
1 = 2 (corr. s, L1 // L2)
tan 1 = tan 2

47. From the above result, we have
If L1 // L2, then m1 = m2.
The converse of the above result is also true:
If m1 = m2, then L1 // L2.

48. Determine whether two lines AB and CD are parallel.y
A(–1, 3)
B(1, 6)
D(8, 5)
C(6, 2) ) 1 ( 3 6 of
Slope - = AB 2 3 = 6 8 5 of
Slope 2 CD - = 2 3 = x
∵ Slope of AB = slope of CD
∴ AB // CD

49. Follow-up question 6
The figure shows four points P(6, 3), Q(2, 8), R(2, 5) and S(6, 6). Prove that PQ is parallel to RS. y
R(2, 5)
P(6, 3) x
Solution
S(6, 6) of
Slope PQ = 6) ( 2 3 8 - 4 11 - =
Q(2, 8) of
Slope = RS 2 6 5 - 4 11 - =
∵ Slope of PQ = slope of RS
∴ PQ // RS

50. Example 11
Solution

51. Example 12

52. Solution

54. Perpendicular Lines Consider a point A(2, 1) in the figure.
Rotate 90
Consider a point A(2, 1) in the figure.
In fact, the slopes of two perpendicular lines are also related.
We have learnt the relationship between the slopes of parallel lines. 1 O 1 3 2 x y 2 A
Rotate A anti-clockwise about O through 90 to A. A
Let me show you.
Then, the coordinates of A
are (1, 2).
Slope of OA = 2 1 - 2 1 =
Slope of OA = 1 2 - 2 - =
Slope of OA  slope of OA = 2) ( 2 1 -  ∴ 1 - =

55. The converse of the above result is also true:
In general, we have:
If L1  L2, then m1  m2 = –1.
Proof
The converse of the above result is also true:
If m1  m2 = –1, then L1  L2.
Note: The results are not applicable when one of the straight lines is vertical.

56. In the figure, AB ⊥ CD. Find x.2 1 1) ( 3 of
Slope AB - = 3 4 - = y
B(2, 1)
A(1, 3)
C(1, 0)
D(3, x) 1) ( 3 of
Slope x CD - = 4 x = ∵
AB ⊥ CD x ∴ 1 of
slope
Slope - =  CD AB 1 4 3 - =  x 3 = x

57. Follow-up question 7 In the figure, AB ⊥ CD. Find x. Solution ) 1 ( 4 x y
A(–1, –1)
B(4, 2)
C(3, –1)
D(x, 2)
Solution ) 1 ( 4 2 of
Slope - = AB 5 3 = 3 ) 1 ( 2 CD of
Slope - = x 3 - = x
∵ AB  CD
∴ Slope of AB  slope of CD = –1 1 3 5 - =  x 15 5 9 + - = x 5 6 = x

58. Example 13

59. Solution

61. Example 14

62. Solution

68. Point of Division

69. Mid-point
If M is a point on the line segment AB such that M bisects AB
(i.e. AM = MB),
then M is called the mid-point of AB. x y
(x1, y1)
(x2, y2) A M B
Let’s try to find the coordinates of the mid-point M of AB.

70. Draw a horizontal line segment AE and a vertical line segment ME.y
Draw a horizontal line segment AE and a vertical line segment ME.
B(x2, y2)
x2 – x
y2 – y
M(x, y)
Obviously, AEM = 90.
y – y1 F
(x2, y)
Similarly, draw a horizontal line segment MF and a vertical line segment BF.
A(x1, y1) E
(x, y1)
x – x1 x
Obviously, MFB = 90.
Coordinates of E =
Now, we need to find the coordinates of E and F. Then, calculate the lengths of AE, ME, MF and BF.
(x, y1)
Coordinates of F =
(x2, y)
AE =
x  x1
ME =
y  y1
MF =
x2  x
BF =
y2  y

71. ∵ △AEM  △MFB (AAS) ∴ (corr. sides,  △s) MF AE = x - = 2 x + =y
B(x2, y2)
∵ △AEM  △MFB (AAS)
x2 – x
y2 – y ∴
(corr. sides,  △s) MF AE =
M(x, y)
y – y1 F x 2 1 - = 2 x 1 + =
A(x1, y1) E
x – x1 x
(corr. sides,  △s) BF ME =
and y 2 1 - = 2 y 1 + = ∴ ç è æ = , of s
Coordinate M + 2 x 1 + 2 1 y

72. This is known as the mid-point formula.
If M(x, y) is the mid-point of the line segment joining A(x1, y1) and B(x2, y2), then
A(x1, y1)
M(x, y)
B(x2, y2) x y
and
This is known as the mid-point formula.

73. Find the coordinates of the mid-point M of AB in the figure.x y
B(6, 1)
A(–4, 3) M
Let (x, y) be the coordinates of M.
By the mid-point formula, we have 2 6 4 + - = x 2 1 3 + = y
and 1 = 2 =
(1, 2) of
Coordinates = ∴ M

74. \ Follow-up question 8 2 8) ( 12 - + = x 2 10) ( - + = y 2 = 6 - =
In the figure, if M is the mid-point of AB, find the coordinates of M. x y
A (12, –2)
B (–8, –10) M
Solution
Let (x, y) be the coordinates of M.
By the mid-point formula, we have 2 8) ( 12 - + = x 2
10) ( - + = y
and 2 = 6 - = 6)
(2, of
Coordinates - = \ M

75. Example 15
Solution

76. Example 16
Solution

77. Example 17
Solution

78. Internal Point of Division
If a point P divides AB into AP and PB such that AP : PB = : , r s
then P is called the internal point of division of AB. r s A P B x y :
Also, we can say that ‘P divides AB internally’.

79. Using the property of similar triangles, we can find the coordinates of P.
Then, we find the coordinates of E and F,
vertical line segments PE and BF.
Similar to the case for the mid-point, we can also derive the coordinates of internal point of division.
First, we construct horizontal line segments AE and PF, and
and the lengths of AE, PE,
PF and BF.

80. ∵ △AEP ~ △PFB (AAA)
∴ Consider their corresponding sides.

81. \ + = s r ry sy rx sx P , ç è æ of Coordinate ∵ △AEP ~ △PFB (AAA)
∴ Consider their corresponding sides. + = \ s r ry sy rx sx P 2 1 , ç è æ of
Coordinate

82. This is known as the section formula for internal division.
If P(x, y) is the internal point of division of the line segment joining A(x1, y1) and B(x2, y2) such that AP : PB = r : s, then r s
A(x1, y1)
P(x, y)
B(x2, y2) :
and .
This is known as the section formula for internal division.

83. In the figure, if P(x, y) is a point on the line segment AB such that AP : PB = 2 : 3, find the coordinates of P.
By the section formula for internal division, we have x y
B(2, 4)
A(–9, 1)
P(x, y) 3 : 2 9) 3( + -
2(2)
3(1) +
2(4) = x
and = y 3 2 + 3 2 +
4.6 - =
2.2 =
2.2)
4.6, ( of s
Coordinate - = \ P

84. Follow-up question 9
The figure shows two points A(13, 5) and B(3, 2). If P(x, y) lies on AB such that AP : PB = 1 : 4, find the coordinates of P.
Solution x y
B(3, –2)
A(–13, –5)
P(x, y)
By the section formula for internal division, we have
Coordinates of P = (–9.8, –4.4) \

85. Example 18
Solution

86. Example 19
Solution

87. Example 20
Solution

88. Using Analytic Approach to Prove Results Relating to Rectilinear Figures

89. Analytic Approach We call this the analytic approach.
Do you remember the deductive approach for proofs you learnt in S2? Here is an example.
Actually, we can also perform the proofs by introducing a rectangular coordinate system to the figures. y x
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. A O B P
Solution
AO = BO given
∠AOP = ∠BOP = 90 given
PO = PO common side
∴ △AOP  △BOP SAS
∴ AP = BP corr. sides,  △s

90. Introduce a rectangular coordinate system to the figure.
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
Solution
Introduce a rectangular coordinate system to the figure.
With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.
Which coordinate system would you introduce? The orange one or the red one?
Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question. A O B P A O B P

91. Introduce a rectangular coordinate system to the figure.
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
(0, b)
Solution
Introduce a rectangular coordinate system to the figure.
(a, 0)
(a, 0)
Let a be the length of AO and b be the length of PO, then the coordinates of A, B and P are
With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.
Which coordinate system would you introduce? The orange one or the red one?
Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question.
(a, 0),
(a, 0) and
(0, b) respectively.

92. In the figure, AOB is a straight line, AO = BO and PO ⊥ AB
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
(0, b)
Solution 2 0) ( )] [0 b a AP - + =
(a, 0)
(a, 0) 2 b a + = 2 0) ( ) (0 b a BP - + = 2 ) ( b a + - = 2 b a + =
∴ AB = BP

93. Follow-up question 10
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution
Let AE = BE = a and CE = DE = b. E x y
Introduce a rectangular coordinate system as shown in the figure.
C(0, b)
A(a, 0)
B(a, 0)
The coordinates of A, D, B and C are
A(a, 0),
B(a, 0),
C(0, b) and
D(0, b)
D(0, b)
respectively.

94. Follow-up question 10 (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution 2 b a AD - + = )] ( [ )
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y 2 b a + - = ) ( 2 b a + = 2 b a BD - + = )] ( [ ) 2 b a + =

95. Follow-up question 10 (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution 2 b a AC - + = ) ( )] [
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y 2 b a + = 2 b a BC - + = ) ( 2 b a + - = ) ( 2 b a + =

96. Follow-up question 10 (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution
∵ AD = BD = BC = AC
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y
∴ ADBC is a rhombus.

97. Example 21
Solution

99. Example 22

100. Solution

104. Extra Teaching Example

105. Teaching Example 8.3 (Extra)

106. Solution

111. Teaching Example 8.7 (Extra)
Solution

112. Teaching Example 8.8 (Extra)
Solution

114. Teaching Example 8.10 (Extra)
Solution

116. Teaching Example 8.12 (Extra)

117. Solution

122. Teaching Example 8.14 (Extra)

123. Solution

128. Teaching Example 8.17 (Extra)

129. Solution

130. Teaching Example 8.20 (Extra)

131. Solution

134. Teaching Example 8.22 (Extra)

135. Solution