Class 9 Maths Ganita Manjari Chapter 1 Solutions (NCERT 2026–27) – Orienting Yourself: The Use of Coordinates

These Class 9 Maths Ganita Manjari Chapter 1 solutions cover Orienting Yourself: The Use of Coordinates from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step so you can understand each concept and revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 1 Exercises: 1.1, 1.2, End-of-Chapter Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Ganita Manjari, Orienting Yourself: The Use of Coordinates, builds the idea of locating any point in a plane using an ordered pair of numbers. Through the story of Reiaan setting up his new room, the chapter introduces the two-dimensional Cartesian coordinate system — the x-axis, the y-axis, the origin and the four quadrants — and how to plot and read points as (x, y). It then shows how to find the distance between two points using the Baudhāyana–Pythagoras theorem, and leads you towards the idea of the midpoint of a segment. The Class 9 Maths Ganita Manjari Chapter 1 solutions below work through every part of the chapter step by step.

Key Concepts & Definitions

Cartesian plane: the plane formed by two perpendicular number lines (also called the coordinate plane or xy-plane).

Coordinate axes: the horizontal line is the x-axis; the vertical line is the y-axis.

Origin: the point where the axes meet, with coordinates (0, 0).

Coordinates (x, y): x (the abscissa) is the distance from the y-axis; y (the ordinate) is the distance from the x-axis.

Quadrants: the axes divide the plane into four parts — Q I (+, +), Q II (−, +), Q III (−, −), Q IV (+, −).

Points on axes: a point on the x-axis is (x, 0); a point on the y-axis is (0, y).

Important Formulas (Chapter 1)

Distance between two points (x1, y1) and (x2, y2):
d = √[(x2 − x1)2 + (y2 − y1)2]

Distance from the origin to (x, y): d = √(x2 + y2)

Horizontal segment (same y): distance = |x2 − x1|  •  Vertical segment (same x): distance = |y2 − y1|

Midpoint of (x1, y1) and (x2, y2): M = ( (x1 + x2)/2 , (y1 + y2)/2 )

Exercise Set 1.1

Fig. 1.3 shows Reiaan’s room with points OABC marking its corners — O(0, 0), C(0, 10), B(12, 10), A(12, 0). The x- and y-axes are marked in the figure. Referring to Fig. 1.3, answer the following questions:

(i) If D1R1 represents the door to Reiaan’s room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

SOLUTION The door D1R1 lies on the floor along the x-axis, with D1(8, 0) and R1(11.5, 0). Distance of the door from the x-axis = 0 (it lies on the x-axis). Distance of the door from the left wall (y-axis) = x-coordinate of the nearer end D1 = 8 units (ft).

(ii) What are the coordinates of D1?

SOLUTIONFrom Fig. 1.3, D1 = (8, 0).

(iii) If R1 is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

SOLUTION Width of door = D1R1 = |11.5 − 8| = 3.5 units (ft). A standard room door is about 3 ft, so 3.5 ft is a comfortable width. A wheelchair needs about 2.5–2.75 ft of clear width, so a person in a wheelchair can enter easily.

(iv) If B1 (0, 1.5) and B2 (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

SOLUTION Width of bathroom door = |4 − 1.5| = 2.5 ft. Since 2.5 ft < 3.5 ft, the bathroom door is narrower than the room door.
Think and Reflect 1. What are the standard widths for a room door? Look around your home and in school. 2. Are the doors in your school suitable for people in wheelchairs? Answer. 1. Standard room doors are about 2.5–3 ft (75–90 cm) wide; main entrance doors are wider. 2. Many older school doors are narrow, so they may not be wheelchair-friendly; a clear width of at least about 2.75 ft (≈ 85 cm) is recommended for easy wheelchair access.
Think and Reflect 1. What is the x-coordinate of a point on the y-axis? 2. Is there a similar generalisation for a point on the x-axis? 3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer. 4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true? Answer. 1. The x-coordinate of any point on the y-axis is 0. 2. Yes — the y-coordinate of any point on the x-axis is 0. 3. Q(y, x) coincides with P(x, y) only when x = y. 4. The claim is true: swapping the coordinates gives the same point exactly when x = y.

Exercise Set 1.2

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (−7, 0) to (13, 0) on the x-axis and from (0, −15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using Fig. 1.5, answer the given questions. Parts that depend on reading Fig. 1.5 or are design/observation tasks are answered with the method and the values that follow from the coordinates.

1. Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7). (i) Where will the fourth foot of the table be? (ii) Is this a good spot for the table? (iii) What is the width of the table? The length? Can you make out the height of the table?

SOLUTION (i) For a rectangle, the fourth foot shares the x of (8, 9) and the y of (11, 7): fourth foot = (8, 7). (ii) It is a good spot only if the table does not block the door or window and gets enough light — place it so the chair faces the window. (Observation; accept reasonable answers.) (iii) Width = |11 − 8| = 3 ft; length = |9 − 7| = 2 ft. The height cannot be found from the plan, because a 2-D coordinate map shows only the floor footprint, not vertical height.

2. If the bathroom door has a hinge at B1 and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

SOLUTION The open door sweeps a quarter-circle about the hinge B1, of radius equal to the door’s width. Check whether any corner of the wardrobe lies inside this arc on your plan; if it does, the door will hit the wardrobe. If the door is made wider, the sweep radius increases, so it is more likely to hit the wardrobe. A practical change is to make the door open outwards or use a sliding door.

3. Look at Reiaan’s bathroom. (i) What are the coordinates of the four corners O, F, R, and P of the bathroom? (ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners. (iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

SOLUTION (i) Read the coordinates of O, F, R, P directly from Fig. 1.5 (they are the four corners of the bathroom rectangle). (ii) The showering area SHWR is a rectangle; list its four corner coordinates from the figure. (iii) Draw rectangles of the given size inside the bathroom: a 3 ft × 2 ft basin with one corner at (x, y) has its opposite corner at (x + 3, y + 2); a 2 ft × 3 ft toilet at (a, b) has opposite corner (a + 2, b + 3). Write the four corners of each. Exact values depend on Fig. 1.5 / your plan.

4. Other rooms in the house: (i) Reiaan’s room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners. (ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

SOLUTION (i) Draw the dining room as an 18 ft × 15 ft rectangle with its length along PA (read P and A from Fig. 1.5) and label the four corners. (ii) If the room corners are (a, b) and (a + 18, b + 15), the centre is (a + 9, b + 7.5). The 5 ft × 3 ft table’s feet are at (a + 9 ± 2.5, b + 7.5 ± 1.5), i.e. (a + 6.5, b + 6), (a + 11.5, b + 6), (a + 11.5, b + 9), (a + 6.5, b + 9).

1.4 Distance Between Two Points in the 2-D Plane

When a segment is not parallel to either axis, its length is found using the Baudhāyana–Pythagoras theorem: the horizontal shift and the vertical shift are the two legs of a right triangle, and the segment is the hypotenuse. This gives d = √[(x2 − x1)2 + (y2 − y1)2].

Think and Reflect 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis? 2. Can these distances help you find the distance AD? Answer. 1. Distance along the x-axis = 7 − 3 = 4; distance along the y-axis = 4 − 1 = 3. 2. Yes — these are the two legs of a right triangle, so AD = √(42 + 32) = √25 = 5 units.
Think and Reflect 1. What has remained the same and what has changed with this reflection? 2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)? Answer. 1. The side lengths stay the same; only the signs of the x-coordinates change (each point moves to the opposite side of the y-axis). 2. Yes — reflecting in the x-axis also preserves all lengths; there the signs of the y-coordinates change instead.

Class 9 Maths Ganita Manjari Chapter 1 Solutions — End-of-Chapter Exercises

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

SOLUTIONThe two axes meet at the origin, so the x-coordinate = 0 and the y-coordinate = 0. The point is (0, 0).

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 through W parallel to the y-axis is the vertical line x = −5, so every point on it has x = −5. Hence H = (−5, y) for some y. As x is negative: H lies in Quadrant II (if y > 0) or Quadrant III (if y < 0); if y = 0, H is on the x-axis.

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. (ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

SOLUTION (i) AM (from A(0,−2) to M(−5,−2)) is horizontal and MP (from M(−5,−2) to P(−5,2)) is vertical, so AM ⊥ MP. (ii) AM lies along y = −2, so AM is parallel to the x-axis (also MP is parallel to the y-axis). (iii) M(−5, −2) and P(−5, 2) have the same x and opposite y, so they are mirror images in the x-axis.

4. Plot point Z (5, −6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

SOLUTION One valid construction: take I(5, 0) and N(0, 0), with the right angle at I. IZ = |0 − (−6)| = 6 units; NI = |5 − 0| = 5 units. NZ = √(52 + 62) = √61 ≈ 7.81 units. So the three sides are 5, 6 and √61 units.

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 Without negative numbers we could use only x ≥ 0 and y ≥ 0 — that is, only the first quadrant (and the positive axes). This would not let us locate every point of the plane, because points in the other three quadrants need negative coordinates.

*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 Method: find the three distances; the points are collinear if the sum of the two shorter distances equals the longest. MA = √[(0+3)2 + (0+4)2] = √25 = 5. AG = √[62 + 82] = √100 = 10. MG = √[92 + 122] = √225 = 15. Since MA + AG = 5 + 10 = 15 = MG, the points are collinear (A lies between M and G).

*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 RB = √[32 + (−4)2] = √25 = 5. BC = √[62 + (−7)2] = √85 ≈ 9.22. RC = √[92 + (−11)2] = √202 ≈ 14.21. RB + BC ≈ 14.22 ≠ RC ≈ 14.21, so the points are not collinear.

*8. Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

SOLUTION (i) O(0, 0), A(4, 0), B(0, 4): the right angle is at O and OA = OB = 4, so it is right-angled and isosceles. (ii) O(0, 0), P(−3, −4) in Q III and Q(3, −4) in Q IV: OP = OQ = √(32 + 42) = 5, so the triangle is isosceles. (Other correct answers are possible.)

*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
SMTMidpoint of S, TIs M the midpoint? — Reason
(−3, 0)(0, 0)(3, 0)(0, 0)Yes — M equals the average of S and T
(2, 3)(3, 4)(4, 5)(3, 4)Yes — averages match M
(0, 0)(0, 5)(0, −10)(0, −5)No — average is (0, −5), not (0, 5)
(−8, 7)(0, −2)(6, −3)(−1, 2)No — average is (−1, 2), not (0, −2)
Connection: M is the midpoint exactly when its coordinates are the averages of S and T, i.e. M = ( (xS + xT)/2 , (yS + yT)/2 ).

*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 −7 = (3 + x)/2 ⇒ 3 + x = −14 ⇒ x = −17. 1 = (−4 + y)/2 ⇒ −4 + y = 2 ⇒ y = 6. ∴ B = (−17, 6).

*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 the case when the points are A (4, 7) and B (16, −2).

SOLUTION P divides AB so that AP : PB = 1 : 2, and Q so that AQ : QB = 2 : 1. P = ( 4 + (16 − 4)/3 , 7 + (−2 − 7)/3 ) = (8, 4). Q = ( 4 + 2(16 − 4)/3 , 7 + 2(−2 − 7)/3 ) = (12, 1). Check: P is the midpoint of A and Q, and Q is the midpoint of P and B — both confirm the answer.

*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? (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 (i) OA = √(12 + 82) = √65; OB = √(42 + 72) = √65; OC = √(72 + 42) = √65. All equal, so A, B, C lie on a circle centred at O with radius √65 ≈ 8.06 units. (ii) OD = √(52 + 62) = √61 ≈ 7.81 < √65, so D is inside K. OE = √(0 + 92) = 9 > √65, so E is outside K.

*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 Taking D, E, F as midpoints of BC, CA, AB respectively, each vertex = sum of its two adjacent midpoints − the opposite midpoint. A = F + E − D = (0 + 6 − 5, 3 + 5 − 1) = (1, 7). B = F + D − E = (0 + 5 − 6, 3 + 1 − 5) = (−1, −1). C = D + E − F = (5 + 6 − 0, 1 + 5 − 3) = (11, 3). Check: midpoint of BC = (5, 1) = D, of CA = (6, 5) = E, of AB = (0, 3) = F. ✓

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 (N–S) direction and East–West (E–W) 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. A street intersection where the i-th N–S street meets the j-th E–W street is written as (i, j). Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).

SOLUTION Each ordered pair names one unique crossing, so (a) exactly 1 intersection is (4, 3) and (b) exactly 1 intersection is (3, 4). They are different points: (4, 3) is the 4th N–S street with the 3rd E–W street, while (3, 4) is the 3rd N–S street with the 4th E–W street — order matters.

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 the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.

SOLUTION (i) Circle 1 spans x ∈ [20, 180], y ∈ [70, 230]; Circle 2 spans x ∈ [150, 350], y ∈ [130, 330]. Both lie inside [0, 800] × [0, 600], so no part of either circle is outside the screen. (ii) AB = √[(250 − 100)2 + (230 − 150)2] = √[1502 + 802] = √28900 = 170. Sum of radii = 80 + 100 = 180. Since 170 < 180, the two circles intersect.

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 AB = BC = CD = DA = √[32 + 12] = √10, so all four sides are equal. Diagonals AC = BD = √[42 + 22] = √20 are also equal. Equal sides and equal diagonals ⇒ ABCD is a square. Area = side2 = (√10)2 = 10 square units.

Common Mistakes to Avoid

Watch out for these

  • Writing a point as (y, x) instead of (x, y) — the x-coordinate (abscissa) always comes first.
  • Mixing up the axes: distance from the x-axis is |y|, and distance from the y-axis is |x|.
  • Sign errors in quadrants — recheck (+, +), (−, +), (−, −), (+, −) for Q I–IV.
  • In the distance formula, forgetting to square the differences, or taking the square root of each term separately.
  • For the midpoint, averaging (dividing by 2) — not just adding the coordinates.
  • Assuming three points are collinear because they look close; always verify with distances or equal slopes.

Practice MCQs & Assertion–Reason

1. The point where the x-axis and y-axis meet is called the:

(a) quadrant    (b) origin    (c) abscissa    (d) ordinate

2. The coordinates of the origin are:

(a) (1, 1)    (b) (0, 1)    (c) (0, 0)    (d) (1, 0)

3. The point (−5, 3) lies in:

(a) Quadrant I    (b) Quadrant II    (c) Quadrant III    (d) Quadrant IV

4. The x-coordinate of every point on the y-axis is:

(a) 1    (b) 0    (c) the y-value    (d) −1

5. The distance of the point (x, y) from the x-axis is:

(a) x    (b) |x|    (c) |y|    (d) √(x2 + y2)

6. The distance between (0, 0) and (6, 8) is:

(a) 10    (b) 14    (c) √14    (d) 48

7. The point (3, −5) lies in:

(a) Quadrant I    (b) Quadrant II    (c) Quadrant III    (d) Quadrant IV

8. The midpoint of (−3, 0) and (3, 0) is:

(a) (3, 0)    (b) (0, 0)    (c) (−3, 0)    (d) (0, 3)

9. A point on the x-axis has coordinates of the form:

(a) (0, y)    (b) (x, 0)    (c) (x, x)    (d) (0, 0)

10. The distance between (2, 3) and (2, −1) is:

(a) 2    (b) 3    (c) 4    (d) 5

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(c), 6-(a), 7-(d), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The point (0, −4) lies on the y-axis.

Reason: A point whose x-coordinate is 0 lies on the y-axis.

A-R 2. Assertion: The distance between (−3, −4) and (0, 0) is 5.

Reason: The distance of (x, y) from the origin is √(x2 + y2).

A-R 3. Assertion: (2, 3) and (3, 2) are the same point.

Reason: (x, y) = (y, x) only when x = y.

A-R 4. Assertion: The points (1, 1), (2, 2), (3, 3) are collinear.

Reason: They all satisfy y = x.

A-R 5. Assertion: The point (−2, −5) lies in Quadrant III.

Reason: In Quadrant III both coordinates are negative.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Two perpendicular axes locate any point in a plane; they meet at the origin (0, 0).
  • For (x, y): x is the distance from the y-axis, y is the distance from the x-axis.
  • Quadrant signs: Q I (+, +), Q II (−, +), Q III (−, −), Q IV (+, −).
  • x-axis points are (x, 0); y-axis points are (0, y).
  • (x, y) = (y, x) only when x = y.
  • Distance = √[(x2 − x1)2 + (y2 − y1)2] (Baudhāyana–Pythagoras theorem).
  • Midpoint = averages of the coordinates; points are collinear if the part-distances add up (or slopes are equal).

How to score full marks in this chapter

Always write coordinates as (x, y), show each step of the distance formula, and keep answers in surd form (e.g. √65) unless a decimal is asked. Draw a neat, labelled graph for plotting questions, and for “collinear / square / circle” questions state the test you are using (distances or slopes) before the calculation.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 1 about?

It introduces the 2-D Cartesian coordinate system — axes, origin, quadrants and plotting points (x, y) — and the distance between two points using the Baudhāyana–Pythagoras theorem, leading to the midpoint idea.

What is the distance formula used in this chapter?

The distance between (x1, y1) and (x2, y2) is √[(x2 − x1)2 + (y2 − y1)2].

How many exercises does Chapter 1 have?

Exercise Set 1.1, Exercise Set 1.2, several “Think and Reflect” boxes and 16 End-of-Chapter Exercises — all solved on this page.

Are these Class 9 Maths Ganita Manjari Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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