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

GSEB SolutionsMathsScience

**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:**

- 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:**