Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises: Orienting Yourself — The Use of Coordinates

Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises cover the main ideas from Orienting Yourself: The Use of Coordinates. These exercises help students revise the x-axis, y-axis, origin, quadrants, coordinates of points, midpoint, distance between two points and real-life use of coordinate geometry.

Chapter 1 introduces the Cartesian coordinate system Class 9 through maps, room layouts and points on a plane. The end-of-chapter questions take this learning further by asking students to predict positions, plot points, check collinearity, use midpoint ideas and solve real-life coordinate geometry problems. These Class 9 Maths Ganita Manjari Chapter 1 Solutions are written step-by-step so students can practise the complete chapter in one place.

Key Takeaways

Origin and Axes: The coordinate plane begins from the origin where the axes meet.
Quadrants: Signs of x and y decide the quadrant of a point.
Collinearity: Equal slopes help check whether points lie on the same line.
Midpoint and Distance: Coordinate formulas help find missing points, circle positions and shapes.

Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises Structure 2026

Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises

The Ganita Manjari Class 9 Chapter 1 End of Chapter Exercises include 16 questions. Some questions are direct, while starred questions need deeper reasoning. These Class 9 Maths coordinates exercise answers follow the textbook sequence and use clear coordinate geometry methods.

End of Chapter Exercises Question 1

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution:

The x-axis and y-axis intersect at the origin.

The coordinates of the origin are:

O = (0, 0)

So:

• x-coordinate = 0
• y-coordinate = 0

Answer: The x-coordinate is 0 and the y-coordinate is 0.

End of Chapter Exercises Question 2

Point W has x-coordinate equal to –5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Solution:

A line parallel to the y-axis has the same x-coordinate for all points on it.

Since point W has x-coordinate –5, any point H on the line through W parallel to the y-axis will also have x-coordinate –5.

So, the coordinates of H will be:

H = (–5, y)

Now:

• If y > 0, H lies in Quadrant II.
• If y < 0, H lies in Quadrant III.
• If y = 0, H lies on the x-axis, not in any quadrant.

Answer: The coordinates of H are (–5, y). H can lie in Quadrant II or Quadrant III, depending on the value of y.

End of Chapter Exercises Question 3

Consider the points R (3, 0), A (0, –2), M (–5, –2) and P (–5, 2). If they are joined in the same order, predict:

(i) Two sides of RAMP that are perpendicular to each other.

Solution:

Points:

• R (3, 0)
• A (0, –2)
• M (–5, –2)
• P (–5, 2)

Side AM goes from A (0, –2) to M (–5, –2). Since the y-coordinate remains the same, AM is horizontal.

Side MP goes from M (–5, –2) to P (–5, 2). Since the x-coordinate remains the same, MP is vertical.

A horizontal side and a vertical side are perpendicular.

Answer: AM and MP are perpendicular to each other.

(ii) One side of RAMP that is parallel to one of the axes.

Solution:

Side AM is horizontal because both A and M have y-coordinate –2.

So, AM is parallel to the x-axis.

Answer: AM is parallel to the x-axis.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be?

Solution:

Points M (–5, –2) and P (–5, 2) have the same x-coordinate and opposite y-coordinates.

So, they are mirror images of each other in the x-axis.

Answer: M (–5, –2) and P (–5, 2) are mirror images in the x-axis.

End of Chapter Exercises Question 4

Plot point Z (5, –6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.

Solution:

Answers may differ because students can choose different points. One possible right-angled triangle is:

• Z (5, –6)
• I (5, 0)
• N (0, –6)

Here:

• ZI is vertical.
• ZN is horizontal.
• The right angle is at Z.

Now find the lengths:

ZI = 0 − (−6) = 6 units

ZN = 5 − 0 = 5 units

Using the Baudhāyana–Pythagoras Theorem:

IN² = ZI² + ZN²

IN² = 6² + 5²

IN² = 36 + 25

IN² = 61

IN = √61 units

Answer: One possible right-angled triangle has vertices Z (5, –6), I (5, 0), N (0, –6). Its side lengths are 6 units, 5 units and √61 units.

End of Chapter Exercises Question 5

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Solution:

If there were no negative numbers, we could mark only points with non-negative coordinates.

That means we could locate points:

• on the positive x-axis,
• on the positive y-axis,
• and in Quadrant I.

But we would not be able to locate points to the left of the y-axis or below the x-axis.

Answer: Without negative numbers, the coordinate system would locate only points in Quadrant I and on the positive axes. It would not allow us to locate all points on a 2-D plane.

End of Chapter Exercises Question 6

Are the points M (–3, –4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

Solution:

We can check by comparing slopes.

Slope of MA:

Slope of MA = [0 − (−4)] / [0 − (−3)]

Slope of MA = 4 / 3

Slope of AG:

Slope of AG = [8 − 0] / [6 − 0]

Slope of AG = 8 / 6 = 4 / 3

Since both slopes are equal, the points lie on the same straight line.

Answer: Yes, M (–3, –4), A (0, 0) and G (6, 8) are on the same straight line. This can be checked by comparing slopes.

End of Chapter Exercises Question 7

Use your method from Problem 6 to check if the points R (–5, –1), B (–2, –5) and C (4, –12) are on the same straight line. Now plot both sets of points and check your answers.

Solution:

Use the slope method.

Slope of RB:

Slope of RB = [−5 − (−1)] / [−2 − (−5)]

Slope of RB = (−4) / 3

Slope of BC:

Slope of BC = [−12 − (−5)] / [4 − (−2)]

Slope of BC = (−7) / 6

Since:

−4 / 3 ≠ −7 / 6

the three points are not on the same straight line.

Answer: No, R (–5, –1), B (–2, –5) and C (4, –12) are not on the same straight line.

End of Chapter Exercises Question 8

Using the origin as one vertex, plot the vertices of:

(i) A right-angled isosceles triangle.

Solution:

One possible right-angled isosceles triangle is:

• O (0, 0)
• A (3, 0)
• B (0, 3)

Here:

OA = 3 units

OB = 3 units

OA and OB are perpendicular because one lies on the x-axis and the other lies on the y-axis.

Answer: One possible triangle has vertices O (0, 0), A (3, 0), B (0, 3).

(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Solution:

Use the origin as one vertex.

Choose:

• O (0, 0)
• A (–3, –4), in Quadrant III
• B (3, –4), in Quadrant IV

Now:

OA = √[(-3)² + (-4)²] = √25 = 5 units

OB = √[(3)² + (-4)²] = √25 = 5 units

So, OA = OB.

Answer: One possible triangle has vertices O (0, 0), A (–3, –4), B (3, –4).

End of Chapter Exercises Question 9

The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

Solution:

The midpoint of S (x₁, y₁) and T (x₂, y₂) is:

M = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]

Answer: M is the midpoint in the first two cases only.

Connection between the coordinates of M, S and T

If M is the midpoint of ST, then:

x-coordinate of M = [x-coordinate of S + x-coordinate of T] / 2

y-coordinate of M = [y-coordinate of S + y-coordinate of T] / 2

End of Chapter Exercises Question 10

Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, –4) and B (x, y).

Solution:

Using the midpoint formula:

M = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]

Given:

• M = (–7, 1)
• A = (3, –4)
• B = (x, y)

For the x-coordinate:

−7 = (3 + x) / 2

−14 = 3 + x

x = −17

For the y-coordinate:

1 = (−4 + y) / 2

2 = −4 + y

y = 6

Answer: The coordinates of B are (–17, 6).

End of Chapter Exercises Question 11

Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for A (4, 7) and B (16, –2).

Solution:

Given:

• A (4, 7)
• B (16, –2)

Find the difference between the coordinates of B and A.

Change in x = 16 − 4 = 12

Change in y = −2 − 7 = −9

Since P and Q trisect AB, each part is one-third of the total change.

One-third change in x = 12 / 3 = 4

One-third change in y = −9 / 3 = −3

P is closer to A:

P = (4 + 4, 7 + (−3))

P = (8, 4)

Q is closer to B, or two-thirds from A:

Q = (4 + 8, 7 + (−6))

Q = (12, 1)

Answer: The points of trisection are P (8, 4) and Q (12, 1).

End of Chapter Exercises Question 12

(i) Given the points A (1, –8), B (–4, 7) and C (–7, –4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?

Solution:

Find the distance of each point from the origin.

For A (1, –8):

OA² = 1² + (−8)²

OA² = 1 + 64 = 65

OA = √65

For B (–4, 7):

OB² = (−4)² + 7²

OB² = 16 + 49 = 65

OB = √65

For C (–7, –4):

OC² = (−7)² + (−4)²

OC² = 49 + 16 = 65

OC = √65

Since OA = OB = OC, all three points lie on a circle with centre O.

Answer: The points lie on circle K with centre O (0, 0). The radius is √65 units.

(ii) Given the points D (–5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

Solution:

Circle K has radius:

√65

So:

r² = 65

For D (–5, 6):

OD² = (−5)² + 6²

OD² = 25 + 36 = 61

Since:

61 < 65

D lies inside the circle.

For E (0, 9):

OE² = 0² + 9²

OE² = 81

Since:

81 > 65

E lies outside the circle.

Answer: D (–5, 6) lies inside the circle, and E (0, 9) lies outside the circle.

End of Chapter Exercises Question 13

The midpoints of the sides of triangle ABC are the points D, E and F. Given that the coordinates of D, E and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.

Solution:

Assume:

• D is the midpoint of BC
• E is the midpoint of CA
• F is the midpoint of AB

Given:

• D = (5, 1)
• E = (6, 5)
• F = (0, 3)

Using midpoint relations:

A = E + F − D

B = F + D − E

C = D + E − F

Now calculate A:

A = (6, 5) + (0, 3) − (5, 1)

A = (1, 7)

Calculate B:

B = (0, 3) + (5, 1) − (6, 5)

B = (−1, −1)

Calculate C:

C = (5, 1) + (6, 5) − (0, 3)

C = (11, 3)

Answer: The coordinates are A (1, 7), B (–1, –1), C (11, 3).

End of Chapter Exercises Question 14

A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South direction and East–West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(i) Using 1 cm = 200 m, draw a model of the city in your notebook.

Solution:

Draw two perpendicular main roads crossing at the centre. Then draw 10 equally spaced streets parallel to the North–South road and 10 equally spaced streets parallel to the East–West road.

Since:

1 cm = 200 m

each street should be drawn 1 cm apart.

Answer: Draw a square grid of streets with each parallel street 1 cm apart.

(ii) Using the given convention, find:

(a) How many street intersections can be referred to as (4, 3)?

Solution:

The ordered pair (4, 3) means the 4th street running in the North–South direction and the 3rd street running in the East–West direction.

These two streets meet at exactly one point.

Answer: 1 street intersection can be referred to as (4, 3).

(b) How many street intersections can be referred to as (3, 4)?

Solution:

The ordered pair (3, 4) means the 3rd street running in the North–South direction and the 4th street running in the East–West direction.

These two streets also meet at exactly one point.

Answer: 1 street intersection can be referred to as (3, 4).

End of Chapter Exercises Question 15

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at B (250, 230). Determine:

(i) Whether any part of either circle lies outside the screen.

Solution:

The screen extends from:

• x = 0 to x = 800
• y = 0 to y = 600

For circle A:

• Centre A = (100, 150)
• Radius = 80

Horizontal range:

100 − 80 = 20

100 + 80 = 180

Vertical range:

150 − 80 = 70

150 + 80 = 230

So, circle A lies fully inside the screen.

For circle B:

• Centre B = (250, 230)
• Radius = 100

Horizontal range:

250 − 100 = 150

250 + 100 = 350

Vertical range:

230 − 100 = 130

230 + 100 = 330

So, circle B also lies fully inside the screen.

Answer: No part of either circle lies outside the screen.

(ii) Whether the two circles intersect each other.

Solution:

Find the distance between centres A and B.

A = (100, 150)
B = (250, 230)

AB² = (250 − 100)² + (230 − 150)²

AB² = 150² + 80²

AB² = 22500 + 6400

AB² = 28900

AB = 170 pixels

Sum of radii:

80 + 100 = 180 pixels

Difference of radii:

100 − 80 = 20 pixels

Since:

20 < 170 < 180

the two circles intersect.

Answer: Yes, the two circles intersect each other.

End of Chapter Exercises Question 16

Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Solution:

Find the lengths of the sides.

AB:

AB² = (−1 − 2)² + (2 − 1)²

AB² = (−3)² + 1²

AB² = 9 + 1 = 10

AB = √10

BC:

BC² = (−2 − (−1))² + (−1 − 2)²

BC² = (−1)² + (−3)²

BC² = 1 + 9 = 10

BC = √10

CD:

CD² = (1 − (−2))² + (−2 − (−1))²

CD² = 3² + (−1)²

CD² = 9 + 1 = 10

CD = √10

DA:

DA² = (2 − 1)² + (1 − (−2))²

DA² = 1² + 3²

DA² = 1 + 9 = 10

DA = √10

All four sides are equal.

Now check one right angle using slopes.

Slope of AB:

Slope of AB = (2 − 1) / (−1 − 2)

Slope of AB = 1 / (−3) = −1/3

Slope of BC:

Slope of BC = (−1 − 2) / (−2 − (−1))

Slope of BC = −3 / −1 = 3

Since:

(−1/3) × 3 = −1

AB is perpendicular to BC.

So, ABCD is a square.

Area of square:

Area = side²

Area = (√10)²

Area = 10 square units

Answer: Yes, ABCD is a square. Its area is 10 square units.

Final Answers for Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises

Concept Used in Orienting Yourself The Use of Coordinates End of Chapter Exercises

Orienting Yourself The Use of Coordinates End of Chapter Exercises revise the full coordinate geometry chapter. Students practise the coordinate plane Class 9 topic through both direct and application-based questions.

The main concepts are:

• origin and coordinate axes,
• points on the x-axis and y-axis,
• quadrants,
• ordered pairs,
• mirror images in axes,
• midpoint of a segment,
• collinearity using slopes,
• distance between two points,
• coordinates in real-life maps and screens.

These ideas build a strong base for Class 9 Ganita Manjari coordinate geometry solutions and later coordinate geometry chapters.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1