GSEB Solutions for Class 9 Mathematics – Coordinate Geometry

GSEB Solutions for Class 9 Mathematics – Coordinate Geometry (English Medium)

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Exercise 4:

Solution 1:

Solution 2:

Solution 3(1):

Given, A = {-3, 2, 4} and B = {-1, 1}
A × B = {-3, 2, 4} × {-1, 1}
A × B = {(-3, -1), (-3, 1), (2, -1), (2, 1), (4, -1), (4, 1)}
Graph of A × B:

Solution 3(2):

Given A = -3, 2, 4}
A × A = {-3, 2, 4} × {-3, 2, 4}
A × A =  {(-3, -3), (-3, 2), (-3, 4), (2, -3), (2, 2), (2, 4) (4, -3) (4, 2) (4, 4)}
Graph of A × A:

Solution 3(3):

Given,  B = {-1, 1}
B × B = {-1, 1} × {-1, 1}
B × B =  {(-1, -1), (-1, 1), (1, -1), (1, 1)}

Solution 3(4):

Given, A = {-3, 2, 4} and B = {-1, 1}
B × A = {-1, 1} × {-3, 2, 4}
B × A = {(-1, -3), (-1, 2), (-1, 4),(1, -3), (1, 2) (1, 4)}

Solution 4:

Solution 5:

Given is x = -1, y = 5, z = 3 and w =-4.
Now,
(x + y, z + w) = (-1 + 5, 3 – 4) = (4, -1).
Point (4, -1) lies in the fourth quadrant.
So, point (x + y, z + w) lies in the fourth quadrant.
Then,
(y – z, w + x) = (5 – 3, -4 – 1) = (2, -5).
Point (2, -5) lies in the fourth quadrant.
So, point (y – z, w + x) lies in the fourth quadrant.
Then, (x – w, y + z) = (-1 + 4, 5 + 3) = (3, 8).
Point (3, 8) lies in the first quadrant.
So, point (x – w, y + z) lies in the first quadrant.

Solution 6(1):

Solution 6(2):

b. II
If x < 0 and y > 0 then the points lies in the II quadrant.
For a point, if the abscissa is -3 and the ordinate is 5, then it lies in the II quadrant.

Solution 6(3):

c.(0, 0)
Both the axes intersect at the origin and its coordinates are (0, 0).

Solution 6(4):

Solution 6(5):

d. IV
If x > 0 and y < 0, the point lies in the IV quadrant.
∴ Point (5, -2) lies in the IV quadrant.

Solution 6(6):

d.7
The abscissa is the x-coordinate.
∴ For the point (7, -4), the abscissa is 7.

Solution 6(7):

d.-5
The ordinate is the y-coordinate.
For the point (3, -5), the ordinate is -5 .

Solution 6(8):

c. 0
For the origin O, abscissa and ordinate are both zero.

Solution 6(9):

b. X’OY’
The 3^rd quadrant is the interior of ∠X’OY’.

Solution 6(10):

b. x-axis
The co-ordinates of any point on the Y-axis are of the form (0, b), where |b| is the distance of the point from the X-axis.

Solution 6(11):

Solution 6(12):

d. I
For x = 3, y = 2, u= -9, v = 13 the point (x + y, u + v) lies in the I quadrant.

Solution 6(13):

a.x = 3, y = 3
In the plane, (x, y) = (y, x), if x = 3, y = 3.

Solution 6(14):

b. I and III
If the co-ordinates of the points are of the same sign (both positive or both negative), then points lie in the I or III quadrants.

Solution 6(15):

d. II and IV
The points having co-ordinates with opposite signs lie in II and IV quadrant.

Solution 6(16):

d. (a, 0)
Any point on the x-axis is of the type (a, 0).

Solution 6(17):

c. four
The coordinate axes divide the plane into four parts called quadrants.

Solution 6(18):

b. origin
The x-axis is a horizontal line passing through the origin.

Solution 6(19):

c. y-axis
The vertical line through the origin is called the Y-axis.

Solution 6(20):

c. 2^nd
The 2^nd quadrant is bounded by the negative x-axis and the positive y-axis.

Solution 6(21):

d. X-axis and Y-axis both
In the plane origin O (0, 0) lies on the X-axis and Y-axis both.

Solution 6(22):

Solution 6(23):

Solution 6(24):

a. y-axis
The point (0, -2) lies on the y-axis.

Solution 6(25):

c. interior of ∠YOX’
The point (-3, 4) lies in the interior of ∠YOX’.

Exercise 4.1:

Solution 1:

1. The horizontal axis is called the X-axis and the vertical axis is called the Y-axis.
2. The axes partition the Cartesian Plane into four parts. These parts are named as Quadrant I, Quadrant II, Quadrant III and Quadrant IV.
3. The axes intersect in the Cartesian plane and their point of intersection is called the Origin and co-ordinates of the origin are (0, 0).

Solution 2:

1. The co-ordinates of P are (4, 1) and the co-ordinates of Q are (-2, 4).
2. Point identified as (-2, 4) is point Q.
3. The abscissa of point A is -3.
4. The ordinate of point R is -2.
5. The co-ordinates of the points A, B, C, D, E, F, G, H, I, J, R, S and T are given below.

Point	Co-ordinates
A B C D E F G H I J R S T	(-3, -4) (2, -1) (2, 3) (-4, 3) (-3, 1) (-2, -1) (2, -3) (3, -4) (3, 2) (5, 4) (-4, -2) (-1, -3) (4, -2)

Exercise 4.2:

Solution 1:

Solution 2:

For x = -3, y = 3(-3) – 2 = -9 – 2 = -11
∴ The point to be plotted is (-3, -11).
For x = -2, y = 3(-2) – 2  = -6 – 2 = -8
∴ The point to be plotted is (-2, -8)
Similarly, all the points to be plotted are given in the table below :

Solution 3:

Given,
P = {0, 1, -1} and Q {-3, 2}
P × Q = {0, 1, -1} × {-3, 2}
P × Q = {(0, -3), (0, 2), (1, -3), (1, 2), (-1, -3), (-1, 2)}
Graph of P × Q

Q × P = {-3, 2} × {0, 1, -1}
∴Q × P = {(-3, 0), (-3, 1), (-3, -1), (2, 0), (2, 1), (2, -1)}
Graph of Q × P:

Solution 4(1):

Given,
A = {-2, 3} and B = {-1, 1, 4}
A × B = {-2, 3} × {-1, 1, 4}
A × B = = {(-2, -1), (-2, 1), (-2, 4), (3, -1), (3, 1), (3, 4)}

Solution 4(2):

Given,
A = {-2, 3} and B = {-1, 1, 4}
B × A = {-1, 1, 4} × {-2, 3}
B × A = {(-1, -2), (-1, 3), (1, -2), (1, 3), (4, -2), (4, 3)}

Solution 4(3):

Given,
A = {-2, 3}
A × A = {-2, 3} × {-2, 3}
A × A = {(-2, -2), (-2, 3), (3, -2), (3, 3)}

Solution 4(4):

Given,
B = {-1, 1, 4}
B × B = {-1, 1, 4} × {-1, 1, 4}
B × B =  {(-1, -1), (-1, 1), (-1, 4), (1, -1), (1, 1), (1, 4),(4,-1), (4, 1), (4, 4)}

Solution 5:

Solution 6:

Solution 7: