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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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.
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
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
Followup 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
2
2
(a)
+
units
2
=


+


CD
(
7
1)
[6
(
2)]
units
2
2
=

+
(
8)
8
units
=
+
64
64
units
=
128
units
(or
8
2
units)
Followup 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)
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.
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
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 x1x2.
Coordinates of C =
(x2, y1)
y
L
y2
x2
B( , )
vertical change
x
0
y1
x1
A(, )
y1
x2
C
(, )
horizontal change
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:
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
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
Followup 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?
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
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 xaxis,
2. The slope of a vertical line is .
undefined
For a line that is parallel to the yaxis,
It is meaningless to divide a number by 0.
x
0
Followup 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.
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 xaxis (measured anticlockwise from the xaxis to L)
Straight line L
x
positive xaxis
0
is called the inclination of L.
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?
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.
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
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
.
700
(cor.
to
3
sig.
fig.)
0
Followup 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
=
63
Followup 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)
Solution
Solution
Solution
Solution
Solution
Solution
Solution
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.
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
Followup 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
Solution
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 anticlockwise about O through 90 to A.
Let me show you.
Then, the coordinates of A
are (1, 2).
If L1L2, then m1m2 = –1.
Proof
The converse of the above result is also true:
If m1m2 = –1, thenL1L2.
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

=
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)
Followup question 7
In the figure, AB ⊥ CD. Find x.
Solution
∵ ABCD
∴ Slope of AB slope of CD = –1
B
(x2, y2)
M
(x1, y1)
A
x
0
Midpoint
If M is a point on the line segment AB such that M bisects AB
then M is called the midpoint of AB.
(i.e. AM = MB),
Let’s try to find the coordinates of the midpoint M of AB.
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 =
x2x
y2y
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
B(x2, y2)
M(x, y)
A(x1, y1)
x
0
If M(x, y) is the midpoint of the line segment joining A(x1, y1) and B(x2, y2), then
and
This is known as the midpoint formula.
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 midpoint M of AB in the figure.
Let (x, y) be the coordinates of M.
By the midpoint formula, we have
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)
Followup question 8
In the figure, if M is the midpoint of AB, find the coordinates of M.
Solution
Let (x, y) be the coordinates of M.
By the midpoint formula, we have
Solution
Solution
Solution
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’.
PF and BF.
Similar to the case for the midpoint, 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.
∴ 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.
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
x
0
B(3, –2)
P(x, y)
A(–13, –5)
\
Coordinates of P = (–9.8, –4.4)
Followup 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
Solution
Solution
Solution
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.
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.
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.
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
E
A
B
y
D
E
x
0
Followup 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.
=


+


2
2
AD
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
Followup 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
=


+

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
Followup 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
E
A
B
C(0, b)
y
D
B(a, 0)
A(a, 0)
E
x
0
D(0, b)
Followup 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
∴ADBC is a rhombus.
Solution
Solution
Solution
Solution