Distance between Any Two Points on a Plane. Do you remember how to calculate the distance between P and Q? Well done. Now, try to find the distance between

Distance between Any TwoPoints on a Plane

Do you remember how to calculate the distance between P and Q?Well done. Now, try to find the distance between A and B.

B( , 5)

x

y

0

P( , 5) Q( , 5)1 5

(5 1) units = 4 units.The distance between P and Q is

A( , 2)1

AB is neither a horizontal line nor a vertical line. I don’t know how to calculate the distance.

a vertical line from B.

The two lines intersect at C.

Draw a horizontal line from A and ( , )

Distance between Any Two Points on a Plane

2 3 40

1

3

4

x

y

1

5

5

2 C

Consider two points A(1, 2) and B(5, 5) on a rectangular coordinate plane.

Coordinates of C =

By Pythagoras’ theorem,

units 34 22

22 BCACAB

AC = (5 – 1) units =

3 unitsBC = (5 – 2) units =

4 units

units 5

( , )

AC is a horizontal line.

BC is a vertical line.

5 2

5 2

B(5, )

A( , 2)

5

1

4 units

3 units

5 units

In general, for any two points A(x1, y1) and B(x2, y2) on a rectangular coordinate plane,

0x

y

A(x1, y1)

B(x2, y2)

C( , )

y2 – y1

x2 – x1

It is known as the distance formula between two points.

x2 y1

AB 212

212 yyxx

Find the length of PQ in the figure.

x

y

0

Q( , )69

units 25144

units 5

units

2122

22

units 13

units 169

1)(6 [9 3)]( PQRemember to write the ‘units’.

P( , )

3 1

)25( )15(

Follow-up question 1In 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)

units 25=

units 169 =

units 22 =AB

units 5=

Follow-up question 2In 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)


Solution

(b)

units) 28 (or units 128

units 6464

units 88)( 22

units22)]([6 2 1)7( CD

Example 1

In each of the following, find the distance between the two given

points.

(a) A(–8, –2) and B(–4, 0)

(b) P (2a, –5a) and Q (7a, 7a)


Solution

(a)

units) 52(or units 20

units 24

units )]2(0[)]8(4[

22

22

AB

(b)

units 31

units 169

units )12()5(

units )]5(7[)27(

2

22

22

a

a

aa

aaaaPQ

Example 2

In the figure, A(1, –2), B(5, –2) and

)322 ,3( C are the vertices of a triangle. (a) Find the lengths of AB, BC and CA.

(b) What kind of triangle is ABC?

(c) Find ABC.

Solution

units 4

units 16

units )32()2(

units )]322(2[)31(

22

22

CA

(b) Since the lengths of AB, BC and CA are all equal, △ ABC is an equilateral triangle.

(c) ∵ △ ABC is an equilateral triangle. (proved in (b))

∴ 60ABC

Example 3

The figure shows a parallelogram EFGH on a

rectangular coordinate plane.

(a) Find, in surd form, the lengths of EF,

EH and IH.

(b) Find the perimeter of parallelogram

EFGH, correct to 3 significant figures.

(c) Find the area of parallelogram EFGH.

Solution(a)


units 55

units )]1(4[)16(

22

22

EF

units 26

units 51

units )]1(4[)12(

22

22

EH


units 2)2(

units )24()42(

22

22

IH

(b) Perimeter of parallelogram

fig.) sig. 3 to(cor. units 24.3

units )2650(2

)(2

EHEFEFGH

(c) Area of parallelogram

units sq. 20

units sq. 850

IHEFEFGH

Slope and Inclination of a Straight Line

It seems that straight line B is steeper.Of course, path B is steeper.

Let’s consider the two paths below. Which path is steeper?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.

Path A Path Bx

y

0

Straight line A

Straight line B

slope inclination

horizontal change

vertical change

0x

y

A

B

horizontal change

i.e.

change vertical

line straight a of slope

=

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,

Consider a straight line L passing through A(x1, y1) and B(x2, y2), where x1x2.

0x

y

A( , )

B( , )

( , )

vertical change

horizontal change

LCoordinates of C =

Cchange horizontal

change vertical

line straight a of Slope

=

12 xx 12 yy

=

x1 x2

y2

y1

(x2, y1)

y1

x2

21

21

xx

yy

21

21

x(xy

(y )

)21

21

xxyy

m =

If we use the letter m to represent the slope of the straight line L, then

12

12

xxyy

m

=

0x

y

A(x1, y1)

B(x2, y2)L

or

21

12

xxyy

m

Note:12

21

xxyy

m

and

y

0x

A(–1, –1)

B(4, 3)

Let’s find the slope of AB.

5

4

1)(4 1)(3

12

12

xx

yy of Slope AB

(x1, y1) = (1, 1)

(x2, y2)= (4, 3)

0x

y

A(–1, –1)

B(4, 3)

Let’s find the slope of AB.

5

4

21

21

xx

yy of slope AB

(x1, y1) = (1, 1)

(x2, y2)= (4, 3)

Alternatively,

41 31

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)

6

(a)32

)2(4 of Slope

AB

21315

of SlopeCD(b)

Solution

The slopes of AB and CD are in opposite sign. What does this mean?

In fact,

0x

y Slope = 2Slope = 1

Slope = 0.5

0x

y

Slope = –0.5

Slope = –1 Slope = –2

Slope

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.

Slope > 0 < 0

0x

y

Slope = –0.5

Slope = –1 Slope = –2

for straight lines sloping downwards from left to right, their slopes are negative.

0x

y Slope = 2Slope = 1

Slope = 0.5

for straight lines sloping upwards from left to right, their slopes are positive.

In fact,

2 > 1 > 0.5 2 > 1 > 0.5

The greater the numericalvalue of the slope, the

steeper is the straight line.

The steepest line

The greater the value of the slope, the steeper

is the straight line.

Slope > 0

The steepest line

Slope < 0

What are the slopes of a horizontal line and

a vertical line?

1. The slope of a horizontal line is .

2. The slope of a vertical line is .

0

of Slope12

11

xx

yyAB

of Slope11

12

xxyy

CD

For a line that is parallel to the x-axis,

For a line that is parallel to the y-axis,

0

undefined

12 0

yy It is meaningless to divide

a number by 0.

0x

y

A(x1, y1) B(x2, y1)

0x

yD(x1, y1)

C(x1, y2)

L3 is a horizontal line.L4 is sloping downwards from left to right.

L1 and L2 are sloping upwards from left to right and L1 is steeper.


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.

0x

y L1

L2

L3

L4

StraightLine

L1 L2 L3 L4

Slope 2 1 0 0.5

is called the inclination of L.

Inclination

We can also describe the steepness of a straight line by its inclination.

y

0

Straight line L

x

is the angle that the straight line L makes with the positive x-axis (measured anti-clockwise from the x-axis to L)

positive x-axis

Note: For 0 < < 90, when increases, the steepness of L also increases.

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

Consider a straight line L passing through

A and B with inclination .

Draw a horizontal line from A and

a vertical line from B. They intersect

Let BAC = a.

0

y

x

A

B

C

at C.

a

L

L of Slope L of SlopeAC

BC

Consider a straight line L passing through

A and B with inclination .

x

0

y

A

B

C

a

∵ and a are corresponding angles. a

AC

BC

AC

BC

∴ a tan tan

Note that ACB = 90.

By the definition of tangent ratio

∴ tan

tan

L

The relationship between the inclination and the slope of a straight line L is

slope of L = tan

y

0

L

50

For example:

If the inclination of a straight line L is 50,

slope of L = tan 50

= 1.19 (cor. to 3 sig. fig.) x

Let’s find the inclination of L.

0x

yL

slope =

60=

Slope of L = tan ∵3

∴ = tan 360

(a) Given that the inclination of a straight line L is 35, find the

slope of L correct to 3 significant figures.


(a) Slope of L = tan 35

fig.) sig. 3 to (cor. 700.0=

Solution

(b) Given that the slope of a straight line L is 2, find the

inclination of L correct to the nearest degree.


(b) Slope of L = tan

63=∴ (cor. to the nearest degree)

Solution

2 = tan

Example 4In each of the following, find the slope of the straight line passing

through the two given points on a rectangular coordinate plane.

(a) P (−2, 5) and Q (6, 8)

(b) R (5, 0) and S (0, −6)

(c) T (1, −2) and U (5, −4)

(a)

8

3

)2(6

58 of Slope

PQ

Solution

Example 5In the figure, A(–2, 7), B(3, 2) and

C (9, 5) and D (0, 8) are four points

on the same rectangular coordinate

plane.

(a) Find the slopes of AB and CD.

(b) Which line, AB or CD, is steeper?

3

19

390

58 of Slope

CD

Solution

Example 6

∵ Slope of RS –1

∴

4

73

17

3

161

)3(

b

b

b

b

Solution

Given that the slope of the straight line passing through R(6, –3) and

S(–1, b) is –1, find the value of b.

Example 7

Prove that the three points D(–5, –1), E(–3, 2) and F(1, 8) on a

rectangular coordinate plane are collinear.

Solution

Example 8Given that A(6, –4), B(–2, 6) and C(1, d) are three points on a straight

line, find the value of d.

Solution

4

58

1062

)4(6 of Slope

AB

Example 9

A straight line L passes through A(5, –4) and B(–1, –8).

(a) Find the slope of L.

(b) Find the inclination of L, correct to 3 significant figures.

Solution

Example 10

If the inclination of the line segment joining )3 ,1(C and ) ,3( mD is 60°, find the value of m.

Solution∵ 60 tan of Slope CD

∴

3

333

313

3

m

m

m

Parallel and Perpendicular Lines

The figure shows two parallel horizontal lines. What are their slopes?

Yes. If now, we rotate the lines to the same extent, what do you think about their steepness and their slopes?

Good. Actually, the slopes of parallel lines are always equal. Let me show you the proof.

Both of the slopes are 0.By observation, it seems that their steepness are always the same, so they have the same slopes.

x

y

0

slope = 0

slope = 0

Parallel Lines

L1 L2

21x

y

0

The figure shows two straight lines L1 and L2, whose inclinations are 1 and 2 respectively.

If L1 // L2, then

∴

i.e. slope of L1= slope of L2

tan 1 = tan 2

1 = 2 (corr. s, L1 // L2)

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.

Determine whether two lines AB and CD are parallel.

x

y

0

A(–1, 3)

B(1, 6)

D(8, 5)

C(6, 2)

23

=)1(1

36 of Slope

=AB

685

of Slope2

CD

=23

=

∵ Slope of AB = slope of CD

∴ AB // CD


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.

Solutionx

y

0

P(6, 3)

Q(2, 8)

R(2, 5)

S(6, 6)

∵ Slope of PQ = slope of RS

∴ PQ // RS

4

11= of Slope =RS26

56

4

11= of Slope PQ =6)(2

38

The figure shows two points A(7, 2) and B(5, 5). C is a point on the y-axis

whose y-coordinate is

7

25. Prove that

CB // OA.

Example 11

Solution

7

2

57

10

057

255

of Slope

CB

7

2

07

02 of Slope

OA

∵ Slope of CB slope of OA ∴ CB // OA

Example 12

In the figure, W(–6, 3), X(–3, –6),

Y(m, –7) and Z(1, 2) are four points on a

rectangular coordinate plane. It is given

that WX // ZY.

(a) Find the value of m.

(b) Is WXYZ a parallelogram?

Give the reason.

Solution

1

91

27 of Slope

m

mZY

∵ WX // ZY ∴

4

123

9331

93

of slope of Slope

m

m

mm

ZYWX

(b)

units 90

units )9(3

units )36()]6(3[

22

22

WX

units 90

units )9(3

units )27()14(

22

22

ZY

∵ WX // ZY and WX ZY ∴ WXYZ is a parallelogram. (opp. sides equal and //)

We have learnt the relationship between the slopes of parallel lines.In fact, the slopes of two perpendicular lines are also related. Let me show you.

Consider a point A(2, 1) in the figure.

Perpendicular Lines

Rotate A anti-clockwise about O through

90 to A.

2

1=Slope of OA =

02

01

2=Slope of OA = 01

02

1 O 13

2

1

x

y

2

2

21

3

A

A

1=

Slope of OA slope of OA = 2)(2

1 ∴

Rotate 90

Then, the coordinates of A are (1, 2).

If L1 L2, then m1 m2 = –1.

In general, we have:

The converse of the above result is also true:

If m1 m2 = –1, then L1 L2.

Proof

Note: The results are not applicable when one of the straight lines is vertical.

In the figure, AB ⊥ CD. Find x.

x

y

0∵ AB ⊥ CD

4

x=

1)(3

0 of Slope

xCD

=

3

4=21

1)(3 of Slope AB

=

3=x

143

4 = x

∴ 1 of slope of Slope = CDAB

C(1, 0)

D(3, x)

B(2, 1)

A(1, 3)

Follow-up question 7In the figure, AB ⊥ CD. Find x.

x

y

0

A(–1, –1)

B(4, 2)

C(3, –1)

D(x, 2)Solution

∵ AB CD

∴ Slope of AB slope of CD = –1

53

=)1(4)1(2

of Slope

=AB

33

=x3

)1(2CD of Slope

=x

1559 = x

13

353 =

x

56

=x

Example 13

In the figure, the vertices of △ ABC

are A(5, 5), B(–3, 3) and C(–2, –1).

(a) Find the slopes of AB, BC and

CA.

(b) Hence, prove that △ ABC is a

right-angled triangle.

Solution(a)

4

18

253

53 of Slope

AB

41

4

)3(2

31 of Slope

BC

7

6

)2(5

)1(5 of Slope

CA

(b) ∵

1

)4(4

1 of slope of Slope

BCAB

∴ AB BC ∴ △ ABC is a right-angled triangle.

Example 14In the figure, the vertices of a quadrilateral ABCD are A(0, a),

2

21 ,6B , C(2c + 1, 2 – c) and

D(5, 1). It is given that ADC 90°

and the slope of AD is

2

3.

(a) (i) Find the slope of CD.

(ii) Hence, find the value of c.

(b) (i) Find the value of a.

(ii) Prove that AB DA.

(c) Determine whether ABCD is a rectangle.

Solution(a) (i) ∵ ADC 90° ∴ AD CD ∴

3

2 of Slope

1 of slope2

3

1 of slope of Slope

CD

CD

CDAD

(ii) ∵ 3

2 of Slope CD

∴

5

48333

2

24

1

3

2

)12(5

)2(1

c

ccc

c

c

c

(b) (i) ∵ 2

3 of Slope AD

∴

2

13

132

15222

3

5

12

3

05

1

a

a

a

a

a

(ii)

3

26

406

2

13

2

21

of Slope

AB

∵

12

3

3

2 of slope of Slope

DAAB

∴ AB DA

(c) By substituting c = 5 into C )2 ,12( cc ,we have coordinates of C = )3 ,11(

2

352

15611

2

213

of Slope

BC

∵

12

3

3

2 of slope of Slope

BCCD

∴ CD BC

∵

12

3

3

2 of slope of Slope

BCAB

∴ AB BC

Since AD CD, AB DA, AB BC and CD BC,

i.e. ADC DAB ABC BCD 90°,

ABCD is a rectangle.

Point of Division

A

M

B

Mid-point

If M is a point on the line segment AB such that M bisects AB then M is called the mid-point of AB.(i.e. AM = MB),

Let’s try to find the coordinates of the mid-point M of AB.

x

y

0

(x1, y1)

(x2, y2)

Now, we need to find the coordinates of E and F. Then, calculate the lengths of AE, ME, MF and BF.

A(x1, y1)

M(x, y)

B(x2, y2)

x

y

0

Draw a horizontal line segment AE and a vertical line segment ME.

E

F

Similarly, draw a horizontal line segment MF and a vertical line segment BF.

Coordinates of E = Coordinates of F =

AE = ME = MF = BF =

(x, y1) (x2, y)

x x1 y y1 x2 x y2 y

(x, y1)

(x2, y)

x – x1

y – y1

y2 – y x2 – x

Obviously, AEM = 90.

Obviously, MFB = 90.

A(x1, y1)

M(x, y)

B(x2, y2)

x

y

0

F

x – x1

x2 – x ∵ △AEM △MFB (AAS)

2

xxx 21

=

xxxx 21 =

∴ (corr. sides, s)△MFAE =

2

yyy 21

=

yyyy 21 =

(corr. sides, s)△BFME =and

221 yy

2

x21x∴

, of sCoordinate M

y – y1

y2 – y

E

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

0

221 xx

x

.2

21 yyy

and

This is known as the mid-point formula.

Find the coordinates of the mid-point M of AB in the figure.

x

y

0

B(6, 1)

A(–4, 3)

M

Let (x, y) be the coordinates of M.

By the mid-point formula, we have

and

and12

64x

22

13y

(1, 2) of Coordinates ∴ M


Solution

x

y

0

A (12, –2)

B (–8, –10)

M

In the figure, if M is the mid-point of AB, find the coordinates of M.

Let (x, y) be the coordinates of M.


and

and22

8)(12 x

62

10)(2 y

6)(2, of Coordinates M

Example 15In the figure, a straight line cuts the x-axis

and the y-axis at A(–6, 0) and B(0, 8)

respectively. If M is the mid-point of AB,

find the coordinates of M.

Solution

Example 16In the figure, ABC is a straight line and AB BC.

Find the coordinates of A.

Let (x, y) be the coordinates of A. By the mid-point formula, we have

∴ 8 and 82

)10(1 and

2

42

yx

yx

∴ Coordinates of )8 ,8(A

Solution

Example 17In the figure, ABC is a straight line and

AB BC. Find the values of a and b.


∴ 5.5and 32

38 and

2

93

ba

ba

Solution

r

s

A

P

B

x

y

0

:

Internal Point of Division

If a point P divides AB into AP and PB such that AP : PB = : , then P is called the internal point of division of AB.

r s

Also, we can say that ‘P divides AB internally’.

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

vertical line segments PE and BF.Then, we find the coordinates of E and F, and the lengths of AE, PE, PF and BF.

Using the property of similar triangles, we can find the coordinates of P.

∵ △AEP ~ △PFB (AAA)

∴ Consider their corresponding sides.

∵ △AEP ~ △PFB (AAA)

∴ Consider their corresponding sides.

srrysy

srrxsx

P 2121 , of sCoordinate

This is known as the section formula for internal division.

and .

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):

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.

x

y

0

B(2, 4)

A(–9, 1)

P(x, y)

By the section formula for internal division, we have

and

and4.6

2(2)

329)3(

x

2.2

2(4)

323(1)y

2.2) 4.6,( of sCoordinate P

2

3:


Solution

x

y

0

B(3, –2)

A(–13, –5)P(x, y)

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.


9.841

1(3)13)4(

x

4.441

2)1(5)4(

y

Coordinates of P = (–9.8, –4.4)

Example 18P is a point lying on the line segment joining A(–3, –4) and B(7, 8),

such that AP : PB 3 : 2.

Find the coordinates of P.

SolutionLet (x, y) be the coordinates of P. By the section formula for internal division, we have

35

21623

)7(3)3(2

x

5

165

24823

)8(3)4(2

y

∴ Coordinates of

5

16 ,3P

Example 19The figure shows two points P(–12, –9)

and Q(–8, –7). If R is a point on PQ

produced such that PQ : QR 1 : 4,

find the coordinates of R.

Solution

Solution

In the figure, the line segment joining

M(–1, –4) and N(–5, 8) cuts the x-axis at P.

Find (a) MP : PN, (b) the coordinates of P.

Example 20

Using Analytic Approach to Prove Results Relating to Rectilinear Figures

Actually, we can also perform the proofs by introducing a rectangular coordinate system to the figures.

We call this the analytic approach.Do you remember the deductive approach for proofs you learnt in S2? Here is an example.

Analytic Approach

AO

B

P

In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP.

Solution

AO = BO given

∠AOP = ∠BOP = 90 given

PO = PO common side

∴ △AOP △BOP SAS

∴ AP = BP corr. sides, s△

y

x

Which coordinate system would you introduce? The

orange one or the red one?

With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.

Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question.

In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove thatAP = BP.

AO

B

y

x

Solution

Introduce a rectangular coordinate system to the figure.

P

AO

B

P

AO

B

P

Which coordinate system would you introduce? The

orange one or the red one?

With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.

Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question.

In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove thatAP = BP.

AO

B

y

x

Solution

Introduce a rectangular coordinate system to the figure.

Let a be the length of AO and b be the length of PO, then the coordinates of A, B and P are(a, 0), (a, 0) and (0, b) respectively.

(a, 0) (a, 0)

P(0, b)

In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP.

Solution

AO

B

y

x

(a, 0) (a, 0)

P(0, b)

∴ AB = BP

22 ba

22 0)()]([0 baAP

22 ba

22)( ba

22 0)()(0 baBP


A

D

B

C

E

Solution

Let AE = BE = a and CE = DE = b.

Introduce a rectangular coordinate system as shown in the figure.

The coordinates of A, D, B and C are A(a, 0), B(a, 0), C(0, b) and D(0, b)

respectively.

A(a, 0)

D(0, b)

B(a, 0)

C(0, b)

Ex

y

0

In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus.

A(a, 0)

D(0, b)

B(a, 0)

C(0, b)

Ex

y

0

Follow-up question 10 (cont’d)

A

D

B

C

E

Solution


22 ba

22 ba )(

22 00 baAD )]([)(

22 ba

22 00 baBD )]([)(

A(a, 0)

D(0, b)

B(a, 0)

C(0, b)

Ex

y

0


A

D

B

C

E

Solution


22 ba

22 00 baAC )()]([

22 ba

22 ba )(

22 00 baBC )()(



Solution

A

D

B

C

E

A(a, 0)

D(0, b)

B(a, 0)

C(0, b)

Ex

y

0

∵ AD = BD = BC = AC

∴ ADBC is a rhombus.

Example 21

In the figure, OABC is a parallelogram.

Prove by the analytic approach that

OB2 + AC2 2(OA2 + OC2).

Solution

∵

)(2

22

)()(

222

222222

222222

tsr

trrsstsrsr

trstsrACOB

and )(2)(2 22222 tsrOCOA

∴ )(2 2222 OCOAACOB

Example 22In the figure, OABC is a square. D is the

mid-point of AB. E is a point on CB such that

CE : EB 3 : 1.

(a) Let OA a units. By introducing a

rectangular coordinate system to the

square with O as the origin, express the

coordinates of A, B, C, D and E in terms of a.

(b) (i) Hence, prove by the analytic approach that

OE2 OD2 + DE2.

(ii) What kind of triangle is ODE?

Solution(a) By introducing a rectangular coordinate system to the square with

O as the origin, we have )0 ,(aA , ) ,( aaB and ) ,0( aC .

Let (x1, y1) be the coordinates of D and (x2, y2) be the coordinates of E.


2 and

2

0 and

2 11

aa

ay

aax

∴ Coordinates of

2 ,a

aD


aa

aay

ax

and 4

313

)(1)(3 and

13

)0(1)(322

∴ Coordinates of

a

aE ,

4

3

(b) (i)

16

25

4

3

)0(04

3

2

22

22

2

a

aa

aa

OE

∵

4

5

2

02

)0(

2

22

222

a

aa

aaOD

and

16

5

24

24

3

2

22

222

a

aa

aaa

aDE

∴

16

25

16

5

4

5

2

2222

a

aaDEOD

∴ OE2 OD2 + DE2

(ii) ∵ OE2 OD2 + DE2 ∴ ODE 90° (converse of Pyth. theorem) ∴ △ ODE is a right-angled triangle.

Extra Teaching Example

Teaching Example 8.3 (Extra)The figure shows a quadrilateral

ABCD on a rectangular coordinate

plane.

(a) Find the lengths of AB, BC,

CD, DA and CA.

(b) Show that △ ABC and

△ ADC are right-angled

triangles.

(c) Find the area of the quadrilateral ABCD.


Solution(a)


units )2()2(

units )53()]3(5[

22

22

AB


units )4(4

units )31()]5(1[

22

22

BC


units 42

units )]1(3[)]1(1[

22

22

CD


units 2)4(

units )35()13(

22

22

DA


units 6)2(

units )]1(5[)]1(3[

22

22

CA

(b) Consider △ ABC.

∵

units sq. 40

units sq. ])32()8[( 2222

BCAB

and

units sq. 40

units sq. )40( 22

CA

∴ AB2 + BC2 CA2 ∴ △ ABC is a right-angled triangle. (converse of Pyth. theorem)

Consider △ ADC.

∵

units sq. 40

units sq. ])20()20[( 2222

DCAD

and

units sq. 40

units sq. )40( 22

CA

∴ AD2 + DC2 CA2 ∴ △ ADC is a right-angled triangle. (converse of Pyth. theorem)

(c) Area of the quadrilateral ABCD

units sq. 18

units sq. 10)(8

units sq. 20202

1328

2

1

of area of area

ADCABC △△

Teaching Example 8.7 (Extra)Prove that the three points X(a, b), Y(3a, 4b) and Z(–3a, –5b) on a

rectangular coordinate plane are collinear.

Let mXY and mYZ be the slopes of XY and YZ respectively.

a

ba

baa

bbmYZ

2

36

933

45

a

baa

bbmXY

2

33

4

∵ mXY mYZ and Y is the common end-point of line segments XY and YZ.

∴ X, Y and Z are collinear.

Solution

Teaching Example 8.8 (Extra)In the figure, B(8, 0) is a vertex of △ OAB. It is given that slope of

2

1OA and slope of

2

3AB .

Find the coordinates of A.

Let (x, y) be the coordinates of A.

∵

(1) 22

12

1

0

02

1 of Slope

yxx

yx

y

OA

Solution

and

32422

3

8

2

3

8

02

3 of slope

xyx

yx

y

AB

xx 324 (∵ yx 2 )

(2) 6

244

x

x

∴ By substituting (2) into (1), we have

3

26

y

y

∴ Coordinates of A = 3) ,6(

Teaching Example 8.10 (Extra)In the figure, the inclination of the straight

line DE is 45°.

(a) Express m in terms of n.

(b) If the area of △ DEO is 18 sq. units,

find the values of m and n.

(a) ∵ 45 tan of Slope DE

∴

nmm

nm

n

2

12

10

20

Solution

(b) ∵ units sq. 18 of Area DEO△

∴

9

(a)) (from 18)2()2(2

1

18 )2(2

1

units sq. 182

1

2

n

nn

nm

ODOE

(rejected) 3or 3n (∵ 0n )

By substituting n 3 into the result in (a), we have

6

)3)(2(

m

Teaching Example 8.12 (Extra)The figure shows four points A(–6, –5),

B(4, b), D(5, 5) and E(3, 7) on a

rectangular coordinate plane. The straight

line AE cuts the y-axis at C and

CD // AB.

(a) Find the coordinates of C.

(b) Find the value of b.

(c) (i) Prove that BC // DE.

(ii) Determine whether CBDE is a parallelogram.

Give the reason.

Solution(a) Let (0, y) be the coordinates of C.

6

5

)6(0

)5( of Slope

y

yAC

3

703

7 of Slope

y

yCE

∵ A, C and E are collinear. ∴

3

279

6421533

7

6

5

of slope of Slope

y

y

yy

yy

CEAC

∴ Coordinates of C = )3 ,0(

(b)

5

205

35 of Slope

CD

10

5

)6(4

)5( of Slope

b

bAB

∵ CD // AB

∴

1

5410

5

5

2

of slope of Slope

b

b

b

ABCD

(c) (i)

14

440

)1(3 of Slope

BC

12

253

57 of Slope

DE

∵ Slope of BC slope of DE ∴ BC // DE

(ii)


units )4(4

units )31()04(

22

22

CB


units )2(2

units )75()35(

22

22

ED

Since the lengths of CB and ED are not equal, CBDE is not a parallelogram.

Teaching Example 8.14 (Extra)In the figure, D(2m, m) is a point on the

line joining

0 ,

2

15A and B(0, 15).

C is a point on the x-axis such that AB BC. (a) Find the value of m.

(b) (i) Find the coordinates of C.

(ii) Hence, prove that BC // DO.

(c) Find the area of ODBC.

Solution(a)

2

02

15150

of Slope

AB

m

mm

mBD

2

1502

15 of Slope

∵ A, D and B are collinear. ∴

3

155

1542

152

of slope of Slope

m

m

mmm

m

BDAB

(b) (i) Let (x, 0) be the coordinates of C.

x

xBC

150

015 of Slope

∵ AB BC ∴

30

130

115

)2(

1 of slope of Slope

xx

x

BCAB

∴ Coordinates of C = )0 ,30(

(ii) From (a), coordinates of D

)3 ,6(

)3 ,32(

2

106

03 of Slope

OD

2

130

15 of Slope

BC

∵ Slope of OD slope of BC ∴ BC // DO

(c) Since BC // DO, ODBC is a trapezium.


units 36

units )03()06(

22

22

OD


units )15()30(

units )150()030(

22

22

BC


units )12(6

units )153()06(

22

22

BD

units sq. 270

units sq. 565182

1

units sq. 56)5155(32

1

units sq. (2

1 of Area

BDBC)ODODBC

Teaching Example 8.17 (Extra)

In the figure, A(2, 8), B(−3, 5),

C(4, −2) and D are the vertices of a

quadrilateral. The diagonals AC and

BD intersect at K(a, b), where

AK KC and BK KD.

(a) Find the coordinates of K.

(b) Hence, find the coordinates

of D.

Solution


In the figure, the line segment joining A(2, 11)

and B(9, 3) intersects the line segment

joining C(1, 4) and D at K(k, 5).

If AK : KB CK : KD m : n, find

(a) m : n,

(b) the value of k,

(c) the coordinates of D.

Solution(a) According to the information, K is the internal point of division of AB. Consider the y-coordinate of K. By the section formula for internal division, we have

∴ 1:3:

3

2

6

62

31155

)3()11(5

nmn

mn

m

nm

mnnmnm

mn

(b) Consider the x-coordinate of K. By the section formula for internal division, we have

4

2913

)9(3)2(1

k

(c) According to the information, K is the internal point of division of CD. Let (x, y) be the coordinates of D. By the section formula for internal division, we have

∴ 3

16 and

3

284

345 and

4

31

4

2913

)(3)4(15 and

13

)(3)1(1

4

29

yx

yx

yx

∴ Coordinates of

3

16 ,

3

28D


In the figure, O is the centre of the circle with

radius r. A, B and P are points on the circle.

By introducing a rectangular coordinate system

to the circle, let the coordinates of P be (a, b),

(a) prove that a2 + b2 r2,

(b) prove by the analytic approach that AP PB.

Solution(a) Introduce a rectangular coordinate system

to the circle with O as the origin. ∵ OA, OP and OB are radii. ∴ OA OP OB r

∴ 222

22 )0()0(

bar

baOP

i.e. 222 rba

(b) ∵ OA = OB = r ∴ Coordinates of A = 0) ,( r and coordinates of B = (r, 0)

ra

b

ra

bAP

)(

0 of Slope

ra

bar

bPB

0

of Slope

∵

1

(a)) (from )(

))((

of slope of Slope

2

2

222

2

22

2

2

b

b

baa

b

ra

b

rara

b

ra

b

ra

bPBAP

∴ AP PB

Documents

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