 Lesson: Coordinate Geometry
Exercise 3.1 (24 Multiple choice Questions and
Question: 1
Point (–3, 5) lies in the:
Solution:
b
The x-coordinate of the point (-3, 5) is negative and
the sign of the y-coordinate is positive, so, it lies in the
Question: 2
Which of the following option shows the signs of the
abscissa and ordinate of a point in the second
(a) (+, +)
(b) (–, –)
(c) (–, +)
(d) (+, –)
Solution: c
Question: 3
Point (0, –7) lies:
(a) on the x –axis.
(c) on the y-axis.
Solution:
c
The x - coordinate of the point (0, −7) is 0. Therefore,
it the y – axis.
Question: 4
Point (– 10, 0) lies:
(a) on the negative direction of the x-axis.
(b) on the negative direction of the y-axis.
Solution:
a
The x - coordinate of the point (10, 0) is negative
and the y - coordinate is 0.So, it lies on the negative
direction of the x - axis. Question: 5
Abscissa of all the points on the x-axis is:
(a) 0
(b) 1
(c) 2
(d) any number
Solution:
d
Abscissa of all the points on the x-axis can be any
number.
Question: 6
Ordinate of all points on the x-axis is:
(a) 0
(b) 1
(c) –1
(d) Any number
Solution:
a
On the x-axis, ordinate of all the points is 0.
Question: 7
The point at which the two coordinate axes meet is
called the:
(a) abscissa (b) ordinate
(c) origin
Solution:
c
The point at which the two coordinate axes meet is
called the origin.
Question: 8
A point both of whose coordinates are negative will lie
in:
Solution:
c
In the third quadrant, the coordinates of all the points
will be negative. Hence, option c is correct.
Question: 9
Points (1, – 1), (2, – 2), (4, – 5), (– 3, – 4) (d) do not lie in the same quadrant
Solution:
d
The point (1, −1), (2, −2), and (4, −5) lie in the
Hence, option (d) is correct.
Question: 10
If y-coordinate of a point is zero, then this point
always lies:
(c) on x - axis
(d) on y – axis
Solution:
c
The y-coordinate of all the points on the x–axis is
always zero. Hence, option (c) is correct.
Question: 11
The points (–5, 2) and (2, – 5) lie in the:
(b) II and III quadrants, respectively
(c) II and IV quadrants, respectively
(d) IV and II quadrants, respectively Solution:
c
The y-coordinate of the point
(–5, 2) is negative and
the coordinate is positive.
Hence, it lies in the II quadrant.
The x-coordinate of the point
(2, –5) is positive and
the y-coordinate is negative.
Hence, it lies in the IV quadrant.
Question: 12
If the perpendicular distance of a point P from the
x - axis is 5 units and the foot of the perpendicular lies
on the negative direction of x – axis, then the point P
has
(a) x-coordinate = 5
(b) y -coordinate = 5 only
(c) y -coordinate = 5 only
(d) y-coordinate = 5 or 5
Solution:
d
Since, the perpendicular distance of a point P from the
x - axis is 5 units and the foot of the perpendicular lies
on the negative direction of x - axis, then the point P can lie either on the II
y - coordinate can be either 5 or 5.
Question: 13
On plotting the points O(0, 0), A(3, 0), B(3, 4), C (0,
4) and joining OA, AB, BC and CO, which one of the
following figures is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
b
Here, the point O(0, 0), is the origin. The point
A(3, 0) lies on the positive direction of the x - axis, the
point B(3, 4),lies in the first quadrant and the point
C(0, 4), lies on the positive direction of y - axis. On
joining OA, AB, BC and CO, the figure obtained is a
rectangle. Question: 14
If P(1, 1), Q(3, 4), R(1, 1), S(2, 3), and
T(4, 4), are plotted on a graph paper, then the point
(s) in the fourth quadrant are:
(a) R and T
(b) Q and R
(c) Only S
(d) P and R
Solution:
b The x-coordinate of the point P(1, 1) is 1 and the y
- coordinate is 1.
So, it lies in the 2
th
Similarly, we can plot all the points;
Q(3, 4), R(1, 1), S(2, 3) and T(4, 4).
It is clear from the graph that the points and lie in the
Question: 15
If the coordinates of two points are P(2, 3) and
Q(3, 5), then (abscissa of P) (abscissa of Q) is:
(a) 5
(b) 1 (c) 1
(d) 2
Solution:
b
Abscissa of the point P is 2 and for the point Q
is 3.
Hence, (abscissa of P) – (abscissa of Q)
= (2) (3)
= 2 + 3
= 1
Question: 16
If P(5, 1), Q(8, 0), R(0, 4), S(0, 5), and O(0, 0), are
plotted on a graph paper, then the point (s) on the
x - axis are:
(a) P and R
(b) R and S
(c) Only Q
(d) Q and O
Solution:
d
A point lies on the x-axis, if its y-coordinate is zero.
Hence, on plotting the given points on a graph paper,
we can see that Q and O lie on the x-axis. Question: 17
Abscissa of a point is positive in:
Solution:
b
Abscissa of any point in the I quadrant or the IV
quadrant is always positive. Hence, option (b) is
correct.
Question: 18 The points whose abscissa and ordinate have different
signs will lie in:
Solution:
d
In the I quadrant, the abscissa and the ordinate of any
point are positive.
In the II quadrant, the abscissa of any point is negative
and the ordinate is positive.
In the III quadrant, the abscissa of any point is
negative and the ordinate is also negative.
In the IV quadrant, the abscissa of any point is
positive and the ordinate is negative.
Hence, option (d) is correct.
Question: 19
In the given figure, coordinates of P are: (a) (−4, 2)
(b) (−2, 4)
(c) (4, 2)
(d) (2, 4)
Solution:
b
The perpendicular distance of the point P is 2 units
from y - axis. So, the value of the x - coordinate will
be 2.
Also, the perpendicular distance of the point from the
x-axis is 4 units. So, the value of the
y- coordinate will be 4. Therefore, the coordinates of the point P will be.
Question: 20
In the following figure, the point identified by the
coordinates (–5, 3) is:
(a) T
(b) R
(c) L
(d) S
Solution:
c
Question: 21 The point whose ordinate is and which lies on
y - axis is:
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Solution:
b
Since, the point lies on y – axis, the x - coordinate will
be 0.Also, the ordinate or value of y - coordinate is 4.
Therefore, the coordinates of the point will be
(0, 4).
Question: 22
Which of the points P(0, 3), Q(1, 0), R(0, 1), S(5,
0),
T(1, 2), do not lie on the x – axis?
(a) P and P only
(b) Q and S only
(c) P, R and T
(d) Q, S and T
Solution:
c The y - coordinate will be zero for all the points that
lie on x - axis. The points P,R and T have
y coordinate value greater than zero. Hence, these
points do not lie on the x - axis.
Question: 23
The point which lies on y - axis at a distance of
5 units, in the negative direction from x - axis is 5:
(a) (0, 5)
(b) (5, 0)
(c) (0, 5)
(d) (5, 0)
Solution:
c
Since, the point lies on the y - axis, the x - coordinate
will be zero. Also, this point lies on the y - axis at a
distance of 5 units, in the negative direction of
y - axis.
Hence, the y - coordinate will be 5.
Therefore, option (c) is correct.
Question: 24
The perpendicular distance of the point P (3, 4) from
the y-axis is: (a) 3
(b) 4
(c) 5
(d) 7
Solution:
a
The perpendicular distance of the point P (3, 4) from
the y-axis is 3.
Exercise 3.2
Question: 1
Write whether the following statements are
(i) Point (3, 0) lies in the first quadrant.
(ii) Point (1, 1) and (1, 1) lies in the first quadrant.
(iii) The coordinates of a point whose ordinate is and
abscissa is 1 are
(iv) A point lies on y - axis at a distance of 2 units
from the x - axis. Its coordinates are (2, 0).
(v) (3, 0) is a point in the quadrant.
Write whether the following statements are True or
Solution: (i) False, because if the ordinate of a point is zero, the
point lies on the x-axis.
(ii) False.
(1, –1 ) lies in the IV quadrant and (–1, 1) lies in the II
(iii) False, because in the coordinates of a point, the
abscissa is written first and then the ordinate.
(iv) False, because a point on the y-axis is of the form
(0, y).
(v) True, because in the II quadrant, the signs of the
abscissa and the ordinate are – and + respectively.
Exercise 3.3
Question: 1
Write the coordinates of each of the points P, Q, R, S,
T and O from the given below figure. Solution:
Here, the points P and S lie in the quadrant. So, their
coordinates will be positive.
Now,
the perpendicular distance of P from both axes is1.
Therefore, the coordinates of Pare (1, 1).
The perpendicular distance of S from the x - axis
is 1 unit and from the y - axis is 2 units. So, the
coordinates of are (2, 1).
The point Q lies on the x - axis in the negative
direction. So, its y – coordinate is zero and
x - coordinate is 3.
So, the coordinates of Q are (3, 0). The point R lies in the III quadrant.
Hence, both coordinates will be negative.
Now, its perpendicular distance from the x - axis is 3
units and from the y - axis is 2 units. So, the
x - coordinates of the point R are (2, 3).
The point lies in the III quadrant.
Hence, both coordinates will be negative.
Now, its perpendicular distance from the x - axis is 2
units and from the y - axis is 4 units. So, the
coordinates of T are (4, 2).
The point O is the intersection of both the axes. So, it
is the origin and its coordinates are O(0, 0).
Question: 2
Plot the following points and write the name of the
figure obtained by joining them in order:
P(– 3, 2), Q (– 7, – 3), R (6, – 3), S (2, 2)
Solution:
Let X’OX and Y’OY be the coordinate axes and lets
mark points on it. Here point P(3, 2) lies in the II
Q(–7, –3) lies in the III quadrant, R( 6, -3) lies in the these points on the graph paper, the figure obtained is
the trapezium PQRS.
Question: 3
Plot the points (x, y) given by the following table:
x
2
4
3
2
3
0
y
4
2
0
5
3
0
Solution:
On plotting the given points on a graph paper, we get
the points P(2, 4), Q(4,2) R(-3, 0), S(-2, 5), T(3, -3)
and O(0, 0) as shown in the graph. Question: 4
Plot the following points and check whether they are
collinear or not:
(i) (1, 3), (– 1, – 1), (– 2, – 3)
(ii) (1, 1), (2, – 3), (– 1, – 2)
(iii) (0, 0), (2, 2), (5, 5)
Solution: (i) Plotting the points P(1,3) , Q(–1, –1) and R (–2, –3)
on a graph paper and joining these points, we get a
straight line. Hence, these points are collinear.
(ii) Plotting the points P(1, 1), Q(2, –3) and R (–1, –2)
on a graph paper and joining these points, we get three
lines i.e., the given points do not lie on the same line.
Therefore, the given points are not collinear. (iii) Plotting the points O(0, 0) , A(2, 2) and B (5,5) on
a graph paper and joining these points, we get a
straight line. Hence, the given points are collinear. Question: 5
Without plotting the points indicate the quadrant in
which they will lie, if:
(i) ordinate is 5 and abscissa is – 3
(ii) abscissa is – 5 and ordinate is – 3
(iii) abscissa is – 5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Solution:
(i) The given point is (–3, 5). Here, the abscissa is
negative and ordinate is positive. Therefore, it lies in (ii) The given point is (–5, –3). Here, both the abscissa
and the ordinate are negative. Therefore, it lies in the
(iii) The given point is (–5, 3). Here, the abscissa is
negative and the ordinate is positive. Therefore, it lies
(iv) The given point is (3, 5). Here, both the abscissa
and the ordinate are positive. Therefore, it lies in the I
Question: 6
In the given figure, LM is a line parallel to the y-axis
at a distance of 3 units.
(i) What are the coordinates of the points P, R and Q?
(ii) What is the difference between the abscissa of the
points L and M? Solution:
Given LM is a line parallel to the y-axis and its
perpendicular distance from y axis is 3 units.
(i)The coordinates of the point P= (3, 2) [Since, its
perpendicular distance from the x- axis is 2 units] The
coordinate of the point Q= (3, –1) [Since, its
perpendicular distance from the x- axis is 1 unit and in
the negative direction of y-axis].
The coordinates of point R= (3, 0) [Since, it lies on the
x-axis, its y-coordinate is zero].
(ii) The abscissa of the point L= 3 and the abscissa of
the point M=3 The difference between the abscissas of the points L
and M = 3 – 3 = 0
Question: 7
In which quadrant or on which axis each of the
following points lie?
(– 3, 5), (4, – 1), (2, 0), (2, 2), (– 3, – 6)
Solution:
(i) The x-coordinate of the point (-3, 5) is negative and
the y-coordinate is positive. Therefore, it lies
(ii) The x-coordinate of the point (4, -1) is positive and
the y-coordinate is negative. Therefore, it lies (iii) The x-coordinate of the point (2, 0) is positive and
the y-coordinate is zero. So, it lies on the x-axis.
(iv) The x-coordinate and the y-coordinate of the point
(2, 2) are positive. So, it lies in the I quadrant.
(v) The x-coordinate and the y-coordinate of the point
(-3, -6) are negative. So, it lies in the III quadrant.
Question: 8
Which of the following points lie on y-axis?
A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (0, – 1),
F (– 1, 0), G (0, 5), H (– 7, 0), I (3, 3).
Solution:
A point lies on the y-axis, if its x-coordinate is zero.
Here, x coordinates of the points C(0, 1), D(0, 0), E(0,
-1) and G(0, 5) are zero. So, these points lie on the y-
axis.
Also, D(0, 0) is the intersection point of both the axes.
So it lies on the y- axis as well as on the x- axis.
Question: 9
Plot the points (x, y) given by the following table. Use
scale 1 cm = 0.25 units: X
1.25
0.25
1.5
-
1.75
Y
-0.5
1
1.5
-
0.25
Solution:
Let X’OX and Y’OY be the coordinate axes.
Let’s plot the given points (1.25, -0.5), (0.25, 1), (1.5,
1.5) and
(–1.75, – 0.25) on a graph paper. After plotting, we get
the points P (1.25, –0.5), Q(0.25, 1), R(1.5, 1.5) and
S(–1.75, –0.25). Question: 10
A point lies on the x-axis at a distance of 7 units from
the y-axis. What are its coordinates? What will be the
coordinates if it lies on y-axis at a distance of –7 units
from x-axis?
Solution:
The given point lies on the positive direction of x-axis.
So, its y-coordinate will be zero. It is at a distance of 7
units from the y-axis. So, its coordinates are (7, 0).
If it lies on the negative direction of y-axis, then its x-
coordinate will be zero. Its distance from x-axis is 7
units. Its coordinates are (0, –7).
Question: 11
Find the coordinates of the point:
(i) which lies on x and y axes both.
(ii) whose ordinate is – 4 and which lies on y-axis.
(iii) whose abscissa is 5 and which lies on x-axis.
Solution:
(i) The point which lies on the x and the y-axis both is
the origin whose coordinates are (0, 0).
(ii) The point whose ordinate is –4 and which lies on
y-axis, i.e., whose x-coordinate is zero, is (0, –4).
(iii) The point whose abscissa is 5 and which lies on x-
axis, i.e., whose y-coordinate is zero, is (5, 0). Question: 12
Taking 0.5 cm as 1 unit, plot the following points on
the graph paper:
A (1, 3), B (– 3, – 1), C (1, – 4), D (– 2, 3), E (0, – 8),
F (1, 0)
Solution:
The x and y–coordinates of the point A(1, 3) are
positive. Therefore, it lies in the I quadrant.
The x and y–coordinates of point B(–3, –1) are
negative. Therefore, it lies in III quadrant.
The x-coordinate of the point C(1, –4) is positive and
the y-coordinative is negative.
Therefore, it lies in the IV quadrant.
The x-coordinate of the point D(–2, 3) is negative and
the y-coordinate is positive. So, it lies in the II
The x-coordinate of the point E( 0, –8 ) is zero.
Therefore, it lies on the y-axis.
The x and the y-coordinate of the point F(1, 0) is zero.
So, it lies on the x-axis.
On plotting the given points, we can draw follwing
graph. Exercise 3.4
Question: 1
Points A (5, 3), B (– 2, 3) and D (5, – 4) are three
vertices of a square ABCD. Plot these points on a
graph paper and hence find the coordinates of the
vertex C.
Solution: The graph obtained by plotting the points A, B and D
is given below.We take a point C on the graph such
that ABCD is a square i.e., all sides AB, BC, CD and
So, the abscissa of C should be equal to the abscissa of
B i.e., –2, and the ordinate of C should be equal to the
ordinate of D i.e., –4.
Hence, the coordinates of C are (–2, –4).
Question: 2
Write the coordinates of the vertices of a rectangle
whose length and breadth are 5 and 3 units
respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the
Solution:
Given length of a rectangle = 5 units
Given breadth of a rectangle = 3 units
One vertex is at origin i.e., D(0, 0) and one of the
other vertices lies in the III quadrant.
The length of the rectangle is 5 units in the negative
direction of the x–axis.
The vertex of A is (–5, 0). Also, the breadth of the
rectangle is 3 units in the negative direction of the y-
axis. The vertex of C is (0, –3). The fourth vertex B is
(–5, –3). Question: 3
Plot the points P (1, 0), Q (4, 0) and S (1, 3). Find the
coordinates of the point R such that PQRS is a square.
Solution:
The y-coordinate of the point P(1, 0) is zero.
Therefore, it lies on the x-axis .
The y- coordinate of the point Q(4, 0) is zero.
Therefore, it lies on the x-axis.
Both coordinates of the point S(1, 3) are positive.
Therefore, it lies on the I quadrant.
On plotting these points, we get the
Graph as shown.
Now, let’s take a point R on the graph such that PQRS
is a square. Then, all
sides will be equal i.e., PQ = QR = RS = PS.
So, the abscissa of R should be equal to
the abscissa of Q i.e., 4 and the ordinate of R should
be equal to the ordinate of S i.e., 3.
Hence, the coordinates of R are (4, 3). Question: 4
From the given figure, answer the following:
(i) Write the points whose abscissa is 0.
(ii) Write the points whose ordinate is 0.
(iii) Write the points whose abscissa is – 5. Solution:
(i) The point whose abscissa is 0 will always lie on the
y-axis. So, the required points whose abscissa is 0 are
A, L and O.
(ii) The point whose ordinate is 0 will always lie on
the x-axis. So, the required points whose ordinate is 0
are G, I and O.
(iii) Here, the abscissa ‘– 5’ is negative. The points
with abscissa –5 will lie in II and III quadrants. So, the
required points whose abscissa is –5, are D and H.
Question: 5
Plot the points A (1, – 1) and B (4, 5) (i) Draw a line segment joining these points.
Write the coordinates of a point on this line segment
between the points A and B.
(ii) Extend this line segment and write the coordinates
of a point on this line which lies outside the line
segment AB.
Solution:
In the point A(1, –1), the x-coordinate is positive and
the y-coordinative is negative. So, it lies in the IV
In the point B(4, 5), both the coordinates are positive.
So, it lies in the I quadrant.
On plotting these point, we get the graph as shown.
(i) On joining the points A and B, we get the line
segment AB.
Now, to find the coordinates of a point on this line
segment between A and B, we draw a perpendicular
line to the x – axis from x = 2.
Since, x = 2 lies between A and B, it intersect the line
segment AB at S. Now, we draw a perpendicular from
S to the y–axis such that it intersects the y–axis at y =
1. Thus, we get a point (2, 1) which lie on the line
segment AB. (ii) Now, we extend the line segment AB such that it
intersects the y-axis at point Q. Thus, we get
the point Q(0, – 3) which lies outside the line segment
AB.