NCERT Textbook: Coordinate Geometry | Mathematics (Maths) Class 10 PDF Download

 Page 1


COORDINATE GEOMETRY 99
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. 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).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax
2
 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!
In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. 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.
COORDINA TE GEOMETR Y
2024-25
Page 2


COORDINATE GEOMETRY 99
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. 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).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax
2
 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!
In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. 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.
COORDINA TE GEOMETR Y
2024-25
100 MATHEMA TICS
7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without
actually measuring it. Let us see. This situation
can be represented graphically as shown in
Fig. 7.1. You may use the Pythagoras Theorem
to calculate this distance.
Now, suppose two points lie on the x-axis.
Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig. 7.2. The points A and B lie on the x-axis.
From the figure you can see that OA = 4
units and OB = 6 units.
Therefore, the distance of B from A, i.e.,
AB = OB – OA = 6 – 4 = 2 units.
So, if two points lie on the x-axis, we can
easily find the distance between them.
Now, suppose we take two points lying on
the y-axis. Can you find the distance between
them. If the points C(0, 3) and D(0, 8) lie on the
y-axis, similarly we find that CD = 8 – 3 = 5 units
(see Fig. 7.2).
Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i.e., AC = 
22
34 ? = 5 units. Similarly, you can
find the distance of B from D = BD = 10 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. Let us see
an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively.  So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.
Fig. 7.1
Fig. 7.2
2024-25
Page 3


COORDINATE GEOMETRY 99
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. 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).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax
2
 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!
In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. 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.
COORDINA TE GEOMETR Y
2024-25
100 MATHEMA TICS
7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without
actually measuring it. Let us see. This situation
can be represented graphically as shown in
Fig. 7.1. You may use the Pythagoras Theorem
to calculate this distance.
Now, suppose two points lie on the x-axis.
Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig. 7.2. The points A and B lie on the x-axis.
From the figure you can see that OA = 4
units and OB = 6 units.
Therefore, the distance of B from A, i.e.,
AB = OB – OA = 6 – 4 = 2 units.
So, if two points lie on the x-axis, we can
easily find the distance between them.
Now, suppose we take two points lying on
the y-axis. Can you find the distance between
them. If the points C(0, 3) and D(0, 8) lie on the
y-axis, similarly we find that CD = 8 – 3 = 5 units
(see Fig. 7.2).
Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i.e., AC = 
22
34 ? = 5 units. Similarly, you can
find the distance of B from D = BD = 10 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. Let us see
an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively.  So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.
Fig. 7.1
Fig. 7.2
2024-25
COORDINATE GEOMETRY 101
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we
have
 PQ
2
 =PT
2
 + QT
2
=2
2
 + 2
2
 = 8
So, PQ =
22
 units
How will we find the distance between two
points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3)
(see Fig. 7.4). Draw QS perpendicular to the
x-axis. Also draw a perpendicular PT from the
point P on QS (extended) to meet y-axis at the
point R.
Fig. 7.4
Then PT = 11 units and QT = 7 units. (Why?)
Using the Pythagoras Theorem to the right triangle PTQ, we get
PQ = 
22
11 7 ? = 
170
 units.
Fig. 7.3
2024-25
Page 4


COORDINATE GEOMETRY 99
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. 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).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax
2
 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!
In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. 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.
COORDINA TE GEOMETR Y
2024-25
100 MATHEMA TICS
7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without
actually measuring it. Let us see. This situation
can be represented graphically as shown in
Fig. 7.1. You may use the Pythagoras Theorem
to calculate this distance.
Now, suppose two points lie on the x-axis.
Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig. 7.2. The points A and B lie on the x-axis.
From the figure you can see that OA = 4
units and OB = 6 units.
Therefore, the distance of B from A, i.e.,
AB = OB – OA = 6 – 4 = 2 units.
So, if two points lie on the x-axis, we can
easily find the distance between them.
Now, suppose we take two points lying on
the y-axis. Can you find the distance between
them. If the points C(0, 3) and D(0, 8) lie on the
y-axis, similarly we find that CD = 8 – 3 = 5 units
(see Fig. 7.2).
Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i.e., AC = 
22
34 ? = 5 units. Similarly, you can
find the distance of B from D = BD = 10 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. Let us see
an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively.  So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.
Fig. 7.1
Fig. 7.2
2024-25
COORDINATE GEOMETRY 101
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we
have
 PQ
2
 =PT
2
 + QT
2
=2
2
 + 2
2
 = 8
So, PQ =
22
 units
How will we find the distance between two
points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3)
(see Fig. 7.4). Draw QS perpendicular to the
x-axis. Also draw a perpendicular PT from the
point P on QS (extended) to meet y-axis at the
point R.
Fig. 7.4
Then PT = 11 units and QT = 7 units. (Why?)
Using the Pythagoras Theorem to the right triangle PTQ, we get
PQ = 
22
11 7 ? = 
170
 units.
Fig. 7.3
2024-25
102 MATHEMA TICS
Let us now find the distance between any two
points P(x
1
, y
1
) and Q(x
2
, y
2
). Draw PR and QS
perpendicular to the x-axis. A perpendicular from the
point P on QS is drawn to meet it at the point
T (see Fig. 7.5).
Then, OR = x
1
, OS = x
2
. So, RS = x
2
 – x
1
 = PT.
Also, SQ = y
2
, ST = PR = y
1
. So, QT = y
2
 – y
1
.
Now, applying the Pythagoras theorem in ? PTQ, we get
PQ
2
 =PT
2
 + QT
2
=(x
2
 – x
1
)
2
 + (y
2
 – y
1
)
2
Therefore, PQ =
?? ? ?
22
21 2 1
xxyy ?? ?
Note that since distance is always non-negative, we take only the positive square
root. So, the distance between the points P(x
1
, y
1
) and Q(x
2
, y
2
) is
PQ =
?? ? ?
22
21 2 1
–+ – xxyy ,
which is called the distance formula.
Remarks :
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
OP = 
22
x y ? .
2. We can also write, PQ = 
?? ? ?
22
12 1 2
xx y y ?? ? . (Why?)
Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the
type of triangle formed.
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. We have
PQ = 
22 22
(3 2) (2 3) 5 5 50 ?? ? ? ? ? = 7.07 (approx.)
QR = 
22 2 2
(–2 – 2) (–3 – 3) (– 4) (– 6) 52 ?? ? ? = 7.21 (approx.)
PR = 
22 2 2
(3 –2) (2– 3) 1 ( 1) 2 ?? ??? = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore,
the points P, Q and R form a triangle.
Fig. 7.5
2024-25
Page 5


COORDINATE GEOMETRY 99
7
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. 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).
Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?
Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y = ax
2
 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!
In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. 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.
COORDINA TE GEOMETR Y
2024-25
100 MATHEMA TICS
7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without
actually measuring it. Let us see. This situation
can be represented graphically as shown in
Fig. 7.1. You may use the Pythagoras Theorem
to calculate this distance.
Now, suppose two points lie on the x-axis.
Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig. 7.2. The points A and B lie on the x-axis.
From the figure you can see that OA = 4
units and OB = 6 units.
Therefore, the distance of B from A, i.e.,
AB = OB – OA = 6 – 4 = 2 units.
So, if two points lie on the x-axis, we can
easily find the distance between them.
Now, suppose we take two points lying on
the y-axis. Can you find the distance between
them. If the points C(0, 3) and D(0, 8) lie on the
y-axis, similarly we find that CD = 8 – 3 = 5 units
(see Fig. 7.2).
Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i.e., AC = 
22
34 ? = 5 units. Similarly, you can
find the distance of B from D = BD = 10 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. Let us see
an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively.  So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.
Fig. 7.1
Fig. 7.2
2024-25
COORDINATE GEOMETRY 101
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we
have
 PQ
2
 =PT
2
 + QT
2
=2
2
 + 2
2
 = 8
So, PQ =
22
 units
How will we find the distance between two
points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3)
(see Fig. 7.4). Draw QS perpendicular to the
x-axis. Also draw a perpendicular PT from the
point P on QS (extended) to meet y-axis at the
point R.
Fig. 7.4
Then PT = 11 units and QT = 7 units. (Why?)
Using the Pythagoras Theorem to the right triangle PTQ, we get
PQ = 
22
11 7 ? = 
170
 units.
Fig. 7.3
2024-25
102 MATHEMA TICS
Let us now find the distance between any two
points P(x
1
, y
1
) and Q(x
2
, y
2
). Draw PR and QS
perpendicular to the x-axis. A perpendicular from the
point P on QS is drawn to meet it at the point
T (see Fig. 7.5).
Then, OR = x
1
, OS = x
2
. So, RS = x
2
 – x
1
 = PT.
Also, SQ = y
2
, ST = PR = y
1
. So, QT = y
2
 – y
1
.
Now, applying the Pythagoras theorem in ? PTQ, we get
PQ
2
 =PT
2
 + QT
2
=(x
2
 – x
1
)
2
 + (y
2
 – y
1
)
2
Therefore, PQ =
?? ? ?
22
21 2 1
xxyy ?? ?
Note that since distance is always non-negative, we take only the positive square
root. So, the distance between the points P(x
1
, y
1
) and Q(x
2
, y
2
) is
PQ =
?? ? ?
22
21 2 1
–+ – xxyy ,
which is called the distance formula.
Remarks :
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
OP = 
22
x y ? .
2. We can also write, PQ = 
?? ? ?
22
12 1 2
xx y y ?? ? . (Why?)
Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the
type of triangle formed.
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. We have
PQ = 
22 22
(3 2) (2 3) 5 5 50 ?? ? ? ? ? = 7.07 (approx.)
QR = 
22 2 2
(–2 – 2) (–3 – 3) (– 4) (– 6) 52 ?? ? ? = 7.21 (approx.)
PR = 
22 2 2
(3 –2) (2– 3) 1 ( 1) 2 ?? ??? = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore,
the points P, Q and R form a triangle.
Fig. 7.5
2024-25
COORDINATE GEOMETRY 103
Also, PQ
2
 + PR
2
 = QR
2
, by the converse of Pythagoras theorem, we have ? P = 90°.
Therefore, PQR is a right triangle.
Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices
of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. 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. Now,
AB =
22
(1 – 4) (7 2) 9 25 34 ?? ? ? ?
BC =
22
(4 1) (2 1) 25 9 34 ?? ? ? ? ?
CD =
22
(–1 4) (–1 – 4) 9 25 34 ?? ? ? ?
DA =
22
(1 4) (7 – 4) 25 9 34 ?? ? ? ?
AC =
22
( 1 1) (7 1) 4 64 68 ?? ? ? ? ?
BD =
22
(4 4) (2 4) 64 4 68 ?? ? ? ? ?
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral
ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a
square.
Alternative Solution : We find
the four sides and one diagonal, say,
AC as above. Here AD
2
 + DC
2
 =
34 + 34 = 68 = AC
2
. Therefore, by
the converse of Pythagoras
theorem, ? D = 90°. A quadrilateral
with all four sides equal and one
angle 90° is a square. So, ABCD
is a square.
Example 3 : Fig. 7.6 shows the
arrangement of desks in a
classroom. Ashima, Bharti and
Camella are seated at A(3, 1),
B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a
line? Give reasons for your
answer.
Fig. 7.6
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