Distance between Any Two Points on a Plane

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Distance between Any Two Points on a Plane. y. A ( , 2). 1. x. 0. Well done. Now, try to find the distance between A and B. Do you remember how to calculate the distance between P and Q ?. P ( , 5). 1. B ( , 5). Q ( , 5). 5.

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## Distance between Any Two Points on a Plane

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### Distance between Any TwoPoints on a Plane

y

A( , 2)

1

x

0

Well done. Now, try to find the distance between A and B.

Do you remember how to calculate the distance between P and Q?

P( , 5)

1

B( , 5)

Q( , 5)

5

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.

y

5

4

3

2

1

x

0

1

2

3

4

5

2

2

=

+

AB

AC

BC

2

2

=

+

4

3

units

=

5

units

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.

B(5, )

5

( , )

Coordinates of C =

2

5

Draw a horizontal line from A and

a vertical line from B.

4 units

AC = (5 – 1) units =

4 units

The two lines intersect at C.

( , )

C

5

2

1

A( , 2)

BC = (5 – 2) units =

3 units

3 units

By Pythagoras’ theorem,

AC is a horizontal line.

5 units

y

B(x2, y2)

y2 – y1

x

0

A(x1, y1)

x2 – x1

(

)

(

)

-

2

2

=

-

y

y

+

AB

x

x

2

1

2

1

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

x2

y1

C(, )

It is known as the distance formulabetween two points.

=

PQ

-

(

3)]

=

2

12

=

=

169

units

=

13

units

y

Find the length of PQ in the figure.

Q( , )

6

9

3

1

P( , )

x

0

2

2

+

-

-

[9

(6

1)

units

Remember to write the ‘units’.

2

+

5

units

+

144

25

units

=

-

-

AB

(

5

2

)

(

5

1

)

=

+

9

16

units

=

25

units

=

5

units

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)

Solution

2

2

(a)

+

units

2

2

=

-

-

+

-

-

CD

(

7

1)

[6

(

2)]

units

2

2

=

-

+

(

8)

8

units

=

+

64

64

units

=

128

units

(or

8

2

units)

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)

Solution

(b)

### Slope and Inclination of a Straight Line

y

x

0

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

Straight line B

Straight line A

Path B

Path A

It seems that straight line B is steeper.

Of course, path B is steeper.

y

B

slope

of

a

straight

line

vertical

change

vertical change

=

x

0

A

horizontal change

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,

i.e.

horizontal change

Slope

of

a

straight

line

vertical

change

=

horizontal

change

-

y

y

=

2

1

-

x

x

2

1

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

Coordinates of C =

(x2, y1)

y

L

y2

x2

B( , )

vertical change

x

0

y1

x1

A(, )

y1

x2

C

(, )

horizontal change

y

L

B(x2, y2)

x

0

A(x1, y1)

-

+

y

y

-

(y

y

)

1

2

1

2

m

=

-

+

-

(x

x

x

x

)

1

2

1

2

-

-

-

y

y

y

y

y

y

1

2

2

1

1

2

-

m

m

m

=

y

y

1

2

-

-

-

x

x

x

x

x

x

1

2

2

1

2

1

-

x

x

1

2

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

or

and

Note:

B(4, 3)

y

-

y

=

2

1

Slope

of

AB

x

0

x

-

x

A(–1, –1)

2

1

-

-

3

(

1)

=

(x1, y1) = (1, 1)

(x2, y2)= (4, 3)

-

-

4

(

1)

4

=

5

Let’s find the slope of AB.

y

y

B(4, 3)

x

0

A(–1, –1)

=

slope

of

AB

y

-

y

1

2

x

-

x

-

-

1

3

1

2

=

(x1, y1) = (1, 1)

(x2, y2)= (4, 3)

-

-

1

4

4

=

5

Let’s find the slope of AB.

Alternatively,

The slopes of AB and CD are in opposite sign.

-

5

1

=

Slope

of

CD

-

3

1

-

-

4

(

2

)

=

2

=

Slope

of

AB

=

-

6

-

2

3

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

(b)

(a)

What does this mean?

Slope = 2

y

y

Slope = 1

Slope = 0.5

Slope = –2

Slope = –1

x

x

0

0

Slope = –0.5

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.

Slope

Slope

< 0

> 0

Slope = 2

y

y

Slope = 1

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

Slope = 0.5

Slope = –2

Slope = –1

x

x

0

0

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

Slope = –0.5

In fact,

The steepest line

The steepest line

Slope > 0

Slope < 0

The greater the numericalvalue of the slope, the steeper is the straight line.

The greater the value of the slope, the steeperis the straight line.

2

1

0.5

>

>

2

1

0.5

>

>

The slope of a horizontal line is .

y

y

B(x2, y1)

A(x1, y1)

-

y

y

1

1

=

Slope

of

AB

-

x

x

2

1

=

0

x

x

0

0

-

y

y

2

1

=

0

D(x1, y1)

-

y

y

2

1

=

Slope

of

CD

-

x

x

1

1

C(x1, y2)

0

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

2. The slope of a vertical line is .

undefined

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

 It is meaningless to divide a number by 0.

y

x

0

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.

L1

L2

L3

2

1

0.5

0

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.

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

0

 is called the inclination of L.

Note: For 0 <  < 90, when  increases,

the steepness of L also increases.

y

L

B

a

A

x

0

Consider a straight line L passing through A and B with inclination  .

Draw a horizontal line from A and

C

a vertical line from B.

They intersect

at C.

Let BAC = a.

y

B

a

A

x

BC

BC

0

Slope

Slope

of

of

L =

L =

AC

AC

q

q

=

tan

tan

tan

a

BC

AC

Consider a straight line L passing through A and B with inclination  .

L

C

=a

 and aare corresponding angles.

Note that ACB = 90.

=

By the definition of tangent ratio

=

tan 

y

L

50

0

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.)

x

Slope of L = tan 

= tan 

q

=

60

Let’s find the inclination  of L.

y

L

slope =

60

x

0

=

.

700

(cor.

to

3

sig.

fig.)

0

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

q

=

63

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 

(cor. to the nearest degree)

Example 4

Solution

Example 5

Solution

Example 6

Solution

Example 7

Solution

Example 8

Solution

Example 9

Solution

Example 10

Solution

### Parallel and Perpendicular Lines

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?

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

y

slope = 0

slope = 0

x

0

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

Both of the slopes are 0.

Parallel Lines

y

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

L1

L2

1

2

If L1 // L2, then

i.e. slope of L1= slope of L2

x

0

1 = 2 (corr. s, L1 // L2)

tan 1 = tan 2

From the above result, we have

If L1 // L2, then m1 = m2.

The converse of the above result is also true:

If m1 = m2, thenL1 // L2.

-

6

3

=

Slope

of

AB

-

-

1

(

1

)

3

3

=

=

2

2

-

5

2

B(1, 6)

=

Slope

of

CD

-

8

6

D(8, 5)

A(–1, 3)

C(6, 2)

Determine whether two lines AB and CD are parallel.

y

x

0

∵ Slope of AB = slope of CD

∴ AB // CD

-

-

6

5

11

-

=

=

Slope

of

RS

-

6

2

4

-

-

8

3

11

-

=

=

Slope

of

PQ

-

-

-

2

(

6)

4

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

0

Solution

S(6, 6)

Q(2, 8)

∵ Slope of PQ = slope of RS

∴ PQ // RS

Example 11

Solution

A

y

A

2

1

-

1

0

1

Rotate 90

Slope of OA =

=

x

-

2

0

2

O

3

2

1

1

2

1

-

2

0

2

Slope of OA =

-

=

2

3

-

-

1

0

1

-

-

Slope of OA slope of OA =

=

(

2)

1

2

Perpendicular Lines

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

We have learnt the relationship between the slopes of parallel lines.

In fact, the slopes of two perpendicular lines are also related.

Rotate A anti-clockwise about O through 90 to A.

Let me show you.

Then, the coordinates of A

are (1, 2).

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, thenL1L2.

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

-

-

3

(

1)

4

-

=

=

Slope

of

AB

D(3, x)

-

-

1

2

3

A(1, 3)

-

x

0

x

=

=

Slope

of

CD

-

-

3

(

1)

4

C(1, 0)

AB ⊥ CD

B(2, 1)

-

=

Slope

of

AB

slope

of

CD

1

4

x

-

-

=

1

3

4

=

x

3

In the figure, AB ⊥ CD. Find x.

y

x

0

3

3

-

=

1

-

5

x

3

-

+

=

9

5

x

15

y

-

-

2

(

1

)

3

=

=

Slope

of

AB

6

-

-

4

(

1

)

5

D(x, 2)

=

x

5

B(4, 2)

-

-

2

(

1

)

3

=

=

Slope

of

CD

-

-

x

x

3

x

3

0

C(3, –1)

A(–1, –1)

Follow-up question 7

In the figure, AB ⊥ CD. Find x.

Solution

∵ ABCD

∴ Slope of AB  slope of CD = –1

### Point of Division

y

B

(x2, y2)

M

(x1, y1)

A

x

0

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.

x2 – x

y2 – y

y – y1

x – x1

y

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

B(x2, y2)

M(x, y)

Obviously, AEM = 90.

(x2, y)

F

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

E

(x, y1)

A(x1, y1)

x

0

Obviously, MFB = 90.

(x, y1)

Coordinates of E =

(x2, y)

Coordinates of F =

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

x x1

y y1

MF =

AE =

ME =

BF =

x2x

y2y

x2 – x

y2 – y

(corr. sides, △s)

AE

MF

=

y – y1

-

-

x

x

x

x

=

1

2

+

x

x

1

2

x

=

x – x1

2

and

(corr. sides, △s)

ME

BF

=

-

-

y

y

y

y

=

1

2

+

y

y

1

2

y

=

æ

æ

+

+

x

x

y

y

2

ç

ç

=

1

2

1

2

Coordinate

s

of

M

,

è

è

2

2

y

B(x2, y2)

∵ △AEM △MFB (AAS)

M(x, y)

F

E

A(x1, y1)

x

0

y

B(x2, y2)

M(x, y)

A(x1, y1)

x

0

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

and

This is known as the mid-point formula.

y

A(–4, 3)

M

B(6, 1)

-

+

+

4

6

3

1

and

=

=

x

y

x

2

2

0

and

=

=

1

2

=

Coordinates

of

M

(1, 2)

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

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

By the mid-point formula, we have

y

x

0

A (12, –2)

M

B (–8, –10)

-

+

-

+

-

2

(

10)

12

(

8)

=

=

y

x

and

2

2

=

-

=

6

2

and

\

=

-

Coordinates

of

M

(2,

6)

Follow-up question 8

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

Solution

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

By the mid-point formula, we have

Example 15

Solution

Example 16

Solution

Example 17

Solution

y

B

s

:

r

P

A

x

0

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.

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

and the lengths of AE, PE,

PF and BF.

Similar to the case for the mid-point, we can also derive the coordinates of internal point of division.

vertical line segments PE and BF.

Then, we find the coordinates of E and F,

First, we construct horizontal line segments AE and PF, and

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

∵ △AEP~ △PFB (AAA)

∴ Consider their corresponding sides.

+

+

æ

æ

sx

rx

sy

ry

\

=

ç

ç

Coordinate

s

of

P

,

1

2

1

2

+

+

r

s

r

s

è

è

∵ △AEP~ △PFB (AAA)

∴ Consider their corresponding sides.

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

and

.

s

:

B(x2, y2)

r

P(x, y)

A(x1, y1)

This is known as the section formula for internal division.

y

B(2, 4)

and

+

-

+

3(1)

2(4)

3(

9)

2(2)

=

=

y

x

P(x, y)

+

+

2

3

2

3

and

A(–9, 1)

=

=

-

x

2.2

4.6

0

\

=

-

Coordinate

s

of

P

(

4.6,

2.2)

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

3

:

2

y

x

0

B(3, –2)

P(x, y)

A(–13, –5)

\

Coordinates of P = (–9.8, –4.4)

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

By the section formula for internal division, we have

Example 18

Solution

Example 19

Solution

Example 20

Solution

### Using Analytic Approach to Prove Results Relating to Rectilinear Figures

y

P

Solution

AO = BO given

∠AOP = ∠BOP = 90 given

PO = PO common side

∴ △AOP △BOP SAS

∴ AP = BP corr. sides,  △s

x

O

B

A

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.

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

y

x

P

P

O

O

B

B

A

A

O

B

A

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

P

Solution

Introduce a rectangular coordinate system to the figure.

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.

y

x

O

B

A

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

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

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.

(a, 0),

(a, 0) and

(0, b) respectively.

y

x

2

2

=

-

-

+

-

AP

[0

(

a

)]

(

b

0)

2

2

=

+

a

b

2

2

=

-

+

-

BP

(0

a

)

(

b

0)

2

2

=

-

+

(

a

)

b

2

2

=

+

a

b

O

B

A

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

P

(0, b)

Solution

(a, 0)

(a, 0)

∴ AB = BP

C

E

A

B

y

D

E

x

0

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.

Solution

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

Introduce a rectangular coordinate system as shown in the figure.

C(0, b)

B(a, 0)

A(a, 0)

The coordinates of A, D, B and C are

A(a, 0),

C(0, b) and

D(0, b)

B(a, 0),

D(0, b)

respectively.

C

=

-

-

+

-

-

2

2

a

0

0

b

(

)

[

(

)]

E

A

B

=

-

+

2

2

a

b

(

)

C(0, b)

y

=

+

2

2

a

b

D

B(a, 0)

A(a, 0)

E

x

0

=

-

+

-

-

2

2

BD

a

0

0

b

(

)

[

(

)]

D(0, b)

=

+

2

2

a

b

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.

Solution

C

=

-

-

+

-

2

2

AC

0

a

b

0

[

(

)]

(

)

E

A

B

=

+

2

2

a

b

C(0, b)

y

D

=

-

+

-

2

2

BC

0

a

b

0

(

)

(

)

B(a, 0)

A(a, 0)

E

x

0

=

-

+

2

2

a

b

(

)

D(0, b)

=

+

2

2

a

b

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.

Solution

C

E

A

B

C(0, b)

y

D

B(a, 0)

A(a, 0)

E

x

0

D(0, b)

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.

Solution

∵AD = BD = BC = AC