Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

COORDINATE GEOMETRY

99

COORDINATE GEOMETRY

7

7
...
The distance of a point from the y-axis is called its
x-coordinate, or abscissa
...
The coordinates of a point on the x-axis are of the form
(x, 0), and of a point on the y-axis are of the form (0, y)
...
Draw a set of a pair of perpendicular axes on a graph
paper
...
Then join the points P(3
...
Also join the points X(5
...

Now join S(4, 5), T(4
...
Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6)
...
Further, in Chapter 2, you have seen the graph of
y = ax2 + bx + c (a ¹ 0), is a parabola
...
It helps us to study geometry
using algebra, and understand algebra with the help of geometry
...
You will also study how to find the coordinates of the
point which divides a line segment joining two given points in a given ratio
...
2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15
km north of the town A
...
Let us see
...
7
...
You may use the Pythagoras Theorem
to calculate this distance
...

Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig
...
2
...


Fig
...
1

From the figure you can see that OA = 4
units and OB = 6 units
...
e
...

So, if two points lie on the x-axis, we can
easily find the distance between them
...
Can you find the distance between
them
...
7
...


Fig
...
2

Next, can you find the distance of A from C (in Fig
...
2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i
...
, AC = 32  42 = 5 units
...

Now, if we consider two points not lying on coordinate axis, can we find the
distance between them? Yes! We shall use Pythagoras theorem to do so
...

In Fig
...
3, the points P(4, 6) and Q(6, 8) lie in the first quadrant
...
Also, draw a perpendicular
from P on QS to meet QS at T
...
So, RS = 2 units
...


2024-25

COORDINATE GEOMETRY

101

Therefore, QT = 2 units and PT = RS = 2 units
...
7
...
Draw QS perpendicular to the
x-axis
...


Fig
...
3

Fig
...
4
Then PT = 11 units and QT = 7 units
...


2024-25

102

MATHEMATICS

Let us now find the distance between any two
points P(x 1, y 1) and Q(x 2, y 2 )
...
A perpendicular from the
point P on QS is drawn to meet it at the point
T (see Fig
...
5)
...


So, RS = x2 – x1 = PT
...
So, QT = y2 – y1
...
7
...
So, the distance between the points P(x1, y1) and Q(x2, y2) is
PQ =

 x2 –

x1  +  y2 – y1  ,
2

2

which is called the distance formula
...
In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
OP =
2
...


 x1  x2 2   y1  y2 2


...

Solution : Let us apply the distance formula to find the distances PQ, QR and PR,
where P(3, 2), Q(–2, –3) and R(2, 3) are the given points
...
07 (approx
...
21 (approx
...
41 (approx
...

2024-25

COORDINATE GEOMETRY

103

Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have  P = 90°
...

Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices
of a square
...
One way
of showing that ABCD is a square is to use the property that all its sides should be
equal and both its digonals should also be equal
...
Thereore, ABCD is a
square
...
Here AD2 + DC2 =
34 + 34 = 68 = AC2
...
A quadrilateral
with all four sides equal and one
angle 90° is a square
...

Example 3 : Fig
...
6 shows the
arrangement of desks in a
classroom
...

Do you think they are seated in a
line? Give reasons for your
answer
...
7
...
Therefore, they are seated in a line
...

Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5)
...
So, AP2 = BP2
i
...
,

(x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2

i
...
,

x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25

i
...
,

x–y= 2

which is the required relation
...
From your earlier studies,
you know that a point which is equidistant
from A and B lies on the perpendicular
bisector of AB
...
7
...

Example 5 : Find a point on the y-axis which
is equidistant from the points A(6, 5) and
B(– 4, 3)
...
So, let the point
Fig
...
7
P(0, y) be equidistant from A and B
...
e
...
e
...
e
...

Let us check our solution : AP =
BP =

(6 – 0) 2  (5 – 9) 2  36  16  52
(– 4 – 0) 2  (3 – 9) 2  16  36  52

Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and
the perpendicular bisector of AB
...
1
1
...
Find the distance between the points (0, 0) and (36, 15)
...
2
...
Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear
...
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle
...
In a classroom, 4 friends are
seated at the points A, B, C and
D as shown in Fig
...
8
...

Using distance formula, find
which of them is correct
...
Name the type of quadrilateral
formed, if any, by the following
points, and give reasons for
your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)

Fig
...
8

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

7
...

8
...


2024-25

106

MATHEMATICS

9
...
Also find the
distances QR and PR
...
Find a relation between x and y such that the point (x, y) is equidistant from the point
(3, 6) and (– 3, 4)
...
3 Section Formula
Let us recall the situation in Section 7
...

Suppose a telephone company wants to
position a relay tower at P between A and B
is such a way that the distance of the tower
from B is twice its distance from A
...
7
...
If we take A as the origin O,
and 1 km as one unit on both the axis, the
coordinates of B will be (36, 15)
...
How do we find these
coordinates?

Fig
...
9

Let the coordinates of P be (x, y)
...
Draw PC perpendicular to BE
...


OD OP 1
PD OP 1

 , and


PC PB 2
BC PB 2
y
1
x
1
 
So,
and

15  y 2
36  x 2
These equations give x = 12 and y = 5
...

Now let us use the understanding that
you may have developed through this
example to obtain the general formula
...
e
...
7
...


PB m2

Fig
...
10

2024-25

COORDINATE GEOMETRY

107

Draw AR, PS and BT perpendicular to the x-axis
...
Then, by the AA similarity criterion,
 PAQ ~  BPC
Therefore,
Now,

PQ
PA AQ

=
(1)
BC
BP
PC
AQ = RS = OS – OR = x – x1
PC = ST = OT – OS = x2 – x
PQ = PS – QS = PS – AR = y – y1
BC = BT– CT = BT – PS = y2 – y

Substituting these values in (1), we get
m1
x  x1
y  y1

=
x2  x y2  y
m2
m1 x2  m2 x1
m1
x  x1
Taking
=
, we get x =
m1  m2
x2  x
m2
Similarly, taking

m1
y  y1
m y  m2 y1
=
, we get y = 1 2
m2
y2  y
m1  m2

So, the coordinates of the point P(x, y) which divides the line segment joining the
points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 are

 m1 x2  m2 x1 , m1 y2  m2 y1 


m1  m2 
 m1  m2
This is known as the section formula
...

If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be

 kx2  x1 , ky2  y1 


k1 
 k1
Special Case : The mid-point of a line segment divides the line segment in the ratio
1 : 1
...

 11
  2
Let us solve a few examples based on the section formula
...

Solution : Let P(x, y) be the required point
...

Example 7 : In what ratio does the point (– 4, 6) divide the line segment joining the
points A(– 6, 10) and B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio m1 : m2
...

So,

–4=

3m1  6m2
 8m1  10m2
and 6 
m1  m2
m1  m2

Now,

–4=

3m1  6m2
m1  m2

gives us

– 4m1 – 4m2 = 3m1 – 6m2
i
...
,

7m1 = 2m2

i
...
,

m1 : m2 = 2 : 7

You should verify that the ratio satisfies the y-coordinate also
...

Alternatively : The ratio m1 : m2 can also be written as

m1
:1, or k : 1
...
Using the section formula, we get

 3k  6 , 8k  10 
(– 4, 6) = 

k 1 
 k 1
So,

–4=

(2)

3k  6
k 1

i
...
,

– 4k – 4 = 3k – 6

i
...
,

7k = 2

i
...
,

k:1= 2:7

You can check for the y-coordinate also
...

Note : You can also find this ratio by calculating the distances PA and PB and taking
their ratios provided you know that A, P and B are collinear
...
e
...

Solution : Let P and Q be the points of
trisection of AB i
...
, AP = PQ = QB
(see Fig
...
11)
...
7
...
Therefore, the coordinates of P, by
applying the section formula, are

 1( 7)  2(2) , 1(4)  2( 2) 

 , i
...
, (–1, 0)
1 2
1 2


Now, Q also divides AB internally in the ratio 2 : 1
...
e
...

Note : We could also have obtained Q by noting that it is the mid-point of PB
...

Example 9 : Find the ratio in which the y-axis divides the line segment joining the
points (5, – 6) and (–1, – 4)
...

Solution : Let the ratio be k : 1
...

Therefore,
So,

k  5
= 0
k 1
k=5

That is, the ratio is 5 : 1
...

3 

Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p
...

So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
i
...
,

 6  9 , 1  4   8  p , 2  3

= 

2   2
2 
 2

i
...
,

 15 , 5 
8  p, 5

 = 

2
 2 2
 2

so,

15
8 p
=
2
2

i
...
,

p= 7

2024-25

COORDINATE GEOMETRY

111

EXERCISE 7
...
Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3
...
Find the coordinates of the points of trisection of the line segment joining (4, –1)
and (–2, –3)
...
To conduct Sports Day activities, in
your rectangular shaped school
ground ABCD, lines have been
drawn with chalk powder at a
distance of 1m each
...
7
...
Niharika runs
th the
4
distance AD on the 2nd line and
1
th
5
the distance AD on the eighth line
and posts a red flag
...
Preet runs

Fig
...
12

4
...

5
...
Also find the coordinates of the point of division
...
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y
...
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is
(2, – 3) and B is (1, 4)
...
If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that
3
AP = AB and P lies on the line segment AB
...
Find the coordinates of the points which divide the line segment joining A(– 2, 2) and
B(2, 8) into four equal parts
...
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in
1
order
...
4 Summary
In this chapter, you have studied the following points :

1
...


2
...

3
...
The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is
 x1  x2 , y1  y2 


...
3 discusses the Section Formula for the coordinates (x, y) of a
point P which divides internally the line segment joining the points
A(x1, y 1) and B(x2, y2) in the ratio m1 : m2 as follows :
x=

m1 x2  m2 x1 ,
m1  m2

y=

m1 y2  m2 y1
m1  m2

Note that, here, PA : PB = m1 : m2
...
You will study
Section Formula for such case in higher classes

---