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 3rd 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. 2nd
The 2nd 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:
- The horizontal axis is called the X-axis and the vertical axis is called the Y-axis.
- The axes partition the Cartesian Plane into four parts. These parts are named as Quadrant I, Quadrant II, Quadrant III and Quadrant IV.
- 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:
- The co-ordinates of P are (4, 1) and the co-ordinates of Q are (-2, 4).
- Point identified as (-2, 4) is point Q.
- The abscissa of point A is -3.
- The ordinate of point R is -2.
- 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: