UNIQUE STUDY POINT ← Class IX
Home›Class IX›Maths›Ch 1

NCERT Solutions Class 9 Maths Chapter 1: Orienting Yourself

Class IX Maths 📘 NCERT Solutions Ch 1: Orienting Yourself: The Use of Coordinates NCERT book: Ganita Manjari
Updated 09 Nov 2026

NCERT Solutions · Chapter 1

Orienting Yourself: The Use of Coordinates

  • 27Questions solved
  • 17Figures and graphs
  • 4Sections

Prepared by Sumeet Sahu, Unique Study Point, Indore · Every calculation re-done

These are the solutions for Class 9 Maths Chapter 1, Orienting Yourself: The Use of Coordinates, from the new NCERT textbook Ganita Manjari (Part 1, session 2026-27). All 27 questions are solved step by step, with the figures and graphs you need for each answer.

★ Key takeaways
  1. A point is written as (x, y): x is the distance from the y-axis and y is the distance from the x-axis. The order matters: (x, y) and (y, x) are different points unless x = y.
  2. Every point on the y-axis has x = 0, and every point on the x-axis has y = 0.
  3. Distance formula: d = √(x2 − x1)² + (y2 − y1)². Midpoint: take the average of the x-coordinates and the average of the y-coordinates.
  4. Three points A, B, C lie on one straight line if the two shorter distances add up to the longest one (for example AB + BC = AC).
🎯 Jump to a question (27 questions)

Reiaan’s room: reading coordinates from a plan (Q1 to Q5)

Question 1: Door of Reiaan’s room

Q1

Figure shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

  • 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?
  • ii. What are the coordinates of D₁?
  • 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?
  • OR 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?
Reiaan’s room drawn on a grid. The corners are O (0, 0), A (12, 0), B (12, 10) and C (0, 10). The wardrobe, the bed, the room door D₁ and its end R₁ (11.5, 0), and the bathroom door ends B₁ and B₂ are marked.
Answer

(i) 8 units from the y-axis and 0 units from the x-axis. (ii) D₁ = (8, 0). (iii) The door is 3.5 units wide, which is comfortable. OR: the bathroom door (2.5 units) is narrower.

  1. (i) Distances of the doorThe door lies on the x-axis, so its distance from the x-axis is 0 units. D₁ = (8, 0), so it starts 8 units from the y-axis.
  2. (ii) Coordinates of D₁From the figure, D₁ = (8, 0).
  3. (iii) Width of the doorD₁ = (8, 0) and R₁ = (11.5, 0).Width = 11.5 − 8 = 3.5 unitsIf one unit is 1 ft, then 3.5 ft = 42 inches. Standard room doors are about 30 to 36 inches wide, so this is slightly wider than average. A wheelchair needs a clear width of at least about 32 inches (≈ 2.7 ft), and 3.5 ft is more than that. So the width is comfortable and a wheelchair user can enter easily.
  4. OR: the bathroom doorB₁ = (0, 1.5) and B₂ = (0, 4).Width = 4 − 1.5 = 2.5 unitsSince 2.5 < 3.5, the bathroom door is narrower than the room door.

Question 2: Placing the study table

Q2

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 the figure, answer the given question.

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?
Reiaan’s room with the bathroom on the left of the y-axis. The bathroom has the showering area SHWR, the wardrobe W₁ W₂ W₃ W₄, the bed S₁ S₂ S₃ S₄ and the door end points D₁ and R₁ marked.
Answer

(i) The fourth foot is at (8, 7). (ii) Yes, it is a good spot. (iii) Width 3 units, length 2 units; the height cannot be found from the plan.

  1. (i) The fourth footThe three given points are three corners of a rectangle: A = (8, 9), B = (11, 9), C = (11, 7). The fourth corner has the same x-coordinate as A (8) and the same y-coordinate as C (7).Fourth foot = (8, 7)
  2. (ii) Is it a good spot?Yes. The table sits neatly inside the room, does not block the door or any pathway, and is near a wall, which is practical for study.
  3. (iii) Width, length and heightWidth (along the x-axis): distance between (8, 9) and (11, 9) = 11 − 8 = 3 units.
    Length (along the y-axis): distance between (11, 9) and (11, 7) = 9 − 7 = 2 units.
    Height: it cannot be found, because the plan is a top view (2-D) and shows no vertical dimension.
Graph with the four table feet plotted at (8, 9), (11, 9), (8, 7) and (11, 7)

Question 3: Will the bathroom door hit the wardrobe?

Q3

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 the figure, answer the given question.

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?

Plan of Reiaan’s home: the bedroom, the bathroom with the showering area, and the dining room, with the axes and the corner points marked.
Answer

No, the door will not hit the wardrobe. If the door is made wider than 3 units it would hit it.

  1. Width of the doorB₁ = (0, 1.5) and B₂ = (0, 4), so the width is 4 − 1.5 = 2.5 units. With the hinge at B₁, the door sweeps a quarter circle of radius 2.5 units about B₁.
  2. Position of the wardrobeThe wardrobe has corners W₁ (3, 0), W₂ (7, 0), W₃ (7, 2) and W₄ (3, 2). Its point nearest to B₁ is (3, 1.5), anddistance from B₁ = 3 − 0 = 3 units
  3. ConclusionSince 3 > 2.5, the swinging door stops 0.5 unit short of the wardrobe. It will not hit.
  4. If the door is widerA door wider than 3 units would reach the wardrobe. Then the door can be made to open inward into the bathroom, or the wardrobe can be shifted a little to the right, or the door width can be kept at about 3 units or less.

Question 4: Reiaan’s bathroom

Q4

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 the figure, answer the given question.

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.
Plan of Reiaan’s home: the bedroom, the bathroom with the showering area, and the dining room, with the axes and the corner points marked.
Answer

(i) O (0, 0), F (0, 9), R (−6, 9), P (−6, 0). (ii) SHWR is a trapezium with S (−6, 6), H (−3, 6), W (−2, 9), R (−6, 9). (iii) One possible placement is given below.

  1. (i) Corners of the bathroomFrom the figure: O = (0, 0), F = (0, 9), R = (−6, 9), P = (−6, 0).
  2. (ii) Showering area SHWRS = (−6, 6), H = (−3, 6), W = (−2, 9), R = (−6, 9). SH and RW are the two parallel sides (SH is horizontal and RW is horizontal), while SR and HW are not parallel. Exactly one pair of opposite sides is parallel, so SHWR is a trapezium.
  3. (iii) Washbasin and toiletWashbasin (3 ft × 2 ft), in the bottom-left corner of the bathroom: (−6, 0), (−3, 0), (−3, 2), (−6, 2).
    Toilet (2 ft × 3 ft), just above it: (−6, 2), (−4, 2), (−4, 5), (−6, 5).

    Answers may differ: any non-overlapping placement inside the bathroom and below the showering area is correct.

Question 5: The dining room

Q5

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 the figure, answer the given question.

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.
Plan of Reiaan’s home: the bedroom, the bathroom with the showering area, and the dining room, with the axes and the corner points marked.
Answer

(i) The corners are (−6, 0), (12, 0), (12, −15), (−6, −15). (ii) The feet of the table are (0.5, −9), (5.5, −9), (5.5, −6), (0.5, −6).

  1. (i) Corners of the dining roomP = (−6, 0) and A = (12, 0), so PA = 12 − (−6) = 18 ft, as given. The room is 15 ft wide and lies below PA, so it extends down to y = −15.Corners: (−6, 0), (12, 0), (12, −15), (−6, −15)
  2. (ii) Centre of the roomThe room extends from x = −6 to x = 12 and from y = 0 to y = −15.Centre = ((−6 + 12) ÷ 2, (0 + (−15)) ÷ 2) = (3, −7.5)
  3. Feet of the 5 ft × 3 ft tableLength 5 along the x-axis and width 3 along the y-axis, so half-length = 2.5 and half-width = 1.5. The corners are (3 − 2.5, −7.5 − 1.5), (3 + 2.5, −7.5 − 1.5), (3 + 2.5, −7.5 + 1.5) and (3 − 2.5, −7.5 + 1.5).(0.5, −9), (5.5, −9), (5.5, −6), (0.5, −6)

Points, axes and quadrants (Q6 to Q10)

Question 6: Intersection of the two axes

Q6

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

Answer

The axes meet at the origin, (0, 0): both coordinates are 0.

The x-axis and the y-axis intersect at the origin, so the x-coordinate is 0 and the y-coordinate is 0. The point is (0, 0).

Question 7: Points on a line parallel to the y-axis

Q7

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?

Answer

H = (−5, y), where y can be any real number. H lies in Quadrant II or Quadrant III, or on the x-axis when y = 0.

Every point on a line parallel to the y-axis has the same x-coordinate as W, so the x-coordinate of H is −5.

Value of yPosition of H = (−5, y)
y > 0Quadrant II
y < 0Quadrant III
y = 0On the x-axis, at (−5, 0)

Question 8: Quadrilateral RAMP

Q8

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?
Answer

(i) AM and MP. (ii) AM is parallel to the x-axis (and MP to the y-axis). (iii) M and P, in the x-axis.

  1. (i) Perpendicular sidesA(0, −2) and M(−5, −2) have the same y-coordinate, so AM is horizontal. M(−5, −2) and P(−5, 2) have the same x-coordinate, so MP is vertical. A horizontal line and a vertical line are perpendicular, so AM ⊥ MP.
  2. (ii) Side parallel to an axisAM has the same y-coordinate (−2) at both ends, so AM ∥ x-axis. (Also MP ∥ y-axis, because both ends have x = −5.)
  3. (iii) Mirror imagesM(−5, −2) and P(−5, 2) have the same x-coordinate and opposite y-coordinates. So M and P are mirror images in the x-axis.
Graph with the points R (3, 0), A (0, −2), M (−5, −2) and P (−5, 2) joined to form RAMP

Question 9: Right-angled triangle IZN

Q9

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.)

Answer

Take I = (5, 0) and N = (0, −6). Then IZ = 6 units, ZN = 5 units and IN = √61 units.

  1. Choose the pointsI = (5, 0) is on the x-axis, directly above Z, and N = (0, −6) is on the y-axis, directly to the left of Z. IZ is vertical and ZN is horizontal, so the triangle IZN is right-angled at Z.
  2. IZSame x-coordinate, so IZ = 0 − (−6) = 6 units.
  3. ZNSame y-coordinate, so ZN = 5 − 0 = 5 units.
  4. ININ = √(5 − 0)² + (0 − (−6))² = √25 + 36 = √61 units (≈ 7.81).
  5. CheckIZ² + ZN² = 36 + 25 = 61 = IN², as the Pythagoras theorem requires.
Graph with the right-angled triangle I (5, 0), Z (5, −6), N (0, −6)

Question 10: A coordinate system without negative numbers

Q10

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?

Answer

No. Without negative numbers we could locate points only in Quadrant I and on the positive parts of the axes.

Without negative numbers the coordinates could be only zero or positive. Then on the x-axis we could mark only points to the right of the origin, and on the y-axis only points above the origin. We could locate only:

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

We could not locate points in Quadrants II, III and IV, or on the negative parts of the axes. So such a system could not locate all the points on a plane.

Distance, midpoint and shapes (Q11 to Q21)

Question 11: Are M, A and G on the same straight line?

Q11

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.

Answer

Yes. MA + AG = 5 + 10 = 15 = MG, so the three points lie on one straight line.

Method: find the three distances with the distance formula. If the two shorter distances add up to the longest, the points are on a line.

d = √(x2 − x1)² + (y2 − y1)²

  1. MAMA = √(0 + 3)² + (0 + 4)² = √9 + 16 = 5
  2. AGAG = √(6 − 0)² + (8 − 0)² = √36 + 64 = 10
  3. MGMG = √(6 + 3)² + (8 + 4)² = √81 + 144 = 15
  4. CheckMA + AG = 5 + 10 = 15 = MG.

Question 12: Are R, B and C on the same straight line?

Q12

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.

Answer

No. RB + BC ≠ RC, so R, B and C are not on one straight line.

  1. RBRB = √(−2 + 5)² + (−5 + 1)² = √9 + 16 = 5
  2. BCBC = √(4 + 2)² + (−12 + 5)² = √36 + 49 = √85
  3. RCRC = √(4 + 5)² + (−12 + 1)² = √81 + 121 = √202
  4. CheckRB + BC = 5 + √85 ≈ 14.22, but RC = √202 ≈ 14.21. The two are not equal.

When plotted, R, B and C do not lie on one line, as the graph shows.

Graph with the points R (−5, −1), B (−2, −5) and C (4, −12) plotted and joined

Question 13: Triangles with the origin as a vertex

Q13

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.
Answer

(i) O (0, 0), A (4, 0), B (0, 4). (ii) O (0, 0), P (−3, −4), Q (3, −4). (These are possible answers.)

  1. (i) Right-angled isosceles triangleTake O = (0, 0), A = (4, 0), B = (0, 4). OA = 4 and OB = 4, and OA ⊥ OB (one is on the x-axis, the other on the y-axis). So triangle OAB is a right-angled isosceles triangle.
  2. (ii) Isosceles triangleTake O = (0, 0), P = (−3, −4), Q = (3, −4). P lies in Quadrant III and Q lies in Quadrant IV.OP = √9 + 16 = 5   OQ = √9 + 16 = 5Since OP = OQ, triangle OPQ is isosceles.

Answers may differ: many other triangles are correct.

Graph with the right-angled isosceles triangle O (0, 0), A (4, 0), B (0, 4)
Graph with the isosceles triangle O (0, 0), P (−3, −4), Q (3, −4)

Question 14: Is M the midpoint of ST?

Q14

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.

SMTIs M the midpoint of ST? Yes or NoReason for your answer
(−3, 0)(0, 0)(3, 0)
(2, 3)(3, 4)(4, 5)
(0, 0)(0, 5)(0, −10)
(−8, 7)(0, −2)(6, −3)

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Answer

Rows 1 and 2: Yes. Rows 3 and 4: No. Connection: the coordinates of M are the averages of the coordinates of S and T.

  1. Row 1: S(−3, 0), M(0, 0), T(3, 0)SM = √(0 + 3)² + 0² = 3   MT = √3² + 0² = 3   ST = 6SM = MT and SM + MT = ST, so M is on ST and halves it. Yes.
  2. Row 2: S(2, 3), M(3, 4), T(4, 5)SM = √1² + 1² = √2   MT = √1² + 1² = √2   ST = √8 = 2√2SM = MT and SM + MT = ST. Yes.
  3. Row 3: S(0, 0), M(0, 5), T(0, −10)SM = 5   MT = 15SM ≠ MT (the midpoint of ST would be (0, −5)). No.
  4. Row 4: S(−8, 7), M(0, −2), T(6, −3)SM = √8² + 9² = √145   MT = √6² + 1² = √37SM ≠ MT (the midpoint of ST would be (−1, 2)). No.
  5. The connectionIn rows 1 and 2, where M is the midpoint:x-coordinate of M = (x of S + x of T) ÷ 2   y-coordinate of M = (y of S + y of T) ÷ 2For example, in row 2: (2 + 4) ÷ 2 = 3 and (3 + 5) ÷ 2 = 4. So M = ((xS + xT) ÷ 2, (yS + yT) ÷ 2).
Four small graphs, one for each row of the table of points S, M and T

Question 15: Finding B when M is the midpoint of AB

Q15

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).

Answer

B = (−17, 6).

  1. Use the midpoint connection

    M(−7, 1) is the midpoint of A(3, −4) and B(x, y), so

    −7 = (3 + x) ÷ 2   and   1 = (−4 + y) ÷ 2
  2. Solve for x3 + x = −14  ⇒  x = −17
  3. Solve for y−4 + y = 2  ⇒  y = 6
  4. CheckAM = √(−7 − 3)² + (1 + 4)² = √125 and MB = √(−17 + 7)² + (6 − 1)² = √125. They are equal, and M is on AB, so M is the midpoint.

B = (−17, 6)

Question 16: Points of trisection

Q16

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).

Answer

P = (8, 4) and Q = (12, 1).

Idea: P is the midpoint of AQ, and Q is the midpoint of PB. Let P = (x₁, y₁) and Q = (x₂, y₂).

  1. Equations for the x-coordinatesx₁ = (4 + x₂) ÷ 2  …(1)    x₂ = (x₁ + 16) ÷ 2  …(2)Put (1) in (2): x₂ = ((4 + x₂) ÷ 2 + 16) ÷ 2 = (x₂ + 36) ÷ 4, so 4x₂ = x₂ + 36, 3x₂ = 36.x₂ = 12  ⇒  x₁ = (4 + 12) ÷ 2 = 8
  2. Equations for the y-coordinatesy₁ = (7 + y₂) ÷ 2  …(3)    y₂ = (y₁ − 2) ÷ 2  …(4)Put (3) in (4): y₂ = ((7 + y₂) ÷ 2 − 2) ÷ 2 = (y₂ + 3) ÷ 4, so 4y₂ = y₂ + 3, 3y₂ = 3.y₂ = 1  ⇒  y₁ = (7 + 1) ÷ 2 = 4
  3. CheckP is one-third of the way from A to B: (4 + 12 ÷ 3, 7 − 9 ÷ 3) = (8, 4), and Q is two-thirds of the way: (12, 1).

P = (8, 4) and Q = (12, 1)

Line segment AB with the trisection points P and Q marked, A (4, 7) and B (16, −2)

Question 17: Points on a circle with centre at the origin

Q17

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.

Answer

(i) OA = OB = OC = √65, so A, B, C lie on circle K of radius √65 (≈ 8.06). (ii) D is inside the circle and E is outside it.

  1. (i) Distances from the originOA = √1² + (−8)² = √65OB = √(−4)² + 7² = √65OC = √(−7)² + (−4)² = √65All three points are at the same distance from O, so they lie on a circle with centre O. Radius = √65 units.
  2. (ii) Compare with the radiusOD = √(−5)² + 6² = √61 ≈ 7.81 < √65 ≈ 8.06OE = √0² + 9² = 9 > √65 ≈ 8.06D lies inside the circle and E lies outside the circle.
Circle K with centre O and radius √65 through A (1, −8), B (−4, 7) and C (−7, −4), with points D (−5, 6) inside and E (0, 9) outside

Question 18: Vertices of a triangle from the midpoints of its sides

Q18

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.

Answer

A = (1, 7), B = (−1, −1) and C = (11, 3).

Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). D is the midpoint of BC, E of CA and F of AB.

  1. Equations for the x-coordinatesx₂ + x₃ = 10 …(1)   x₃ + x₁ = 12 …(2)   x₁ + x₂ = 0 …(3)From (3), x₂ = −x₁. Put this in (1): x₃ = 10 + x₁. Put this in (2): (10 + x₁) + x₁ = 12, so 2x₁ = 2.x₁ = 1   x₂ = −1   x₃ = 11
  2. Equations for the y-coordinatesy₂ + y₃ = 2 …(4)   y₃ + y₁ = 10 …(5)   y₁ + y₂ = 6 …(6)From (6), y₂ = 6 − y₁. Put this in (4): y₃ = y₁ − 4. Put this in (5): (y₁ − 4) + y₁ = 10, so 2y₁ = 14.y₁ = 7   y₂ = −1   y₃ = 3
  3. CheckMidpoint of BC = ((−1 + 11) ÷ 2, (−1 + 3) ÷ 2) = (5, 1) = D. Midpoint of CA = ((11 + 1) ÷ 2, (3 + 7) ÷ 2) = (6, 5) = E. Midpoint of AB = ((1 − 1) ÷ 2, (7 − 1) ÷ 2) = (0, 3) = F.

A = (1, 7), B = (−1, −1), C = (11, 3)

Triangle ABC with the midpoints D (5, 1), E (6, 5) and F (0, 3) of its sides

Question 19: Model of a city

Q19

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.

  • i. Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
  • ii. There are street intersections in the model. Each street intersection is formed by two streets, one running in the N-S direction and another in the E-W direction. Each street intersection is referred to in the following manner: if the second street running in the N-S direction and 5th street in the E-W direction meet at some crossing, then we call this street intersection (2, 5). 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).
Answer

(i) Draw 10 vertical and 10 horizontal lines, 1 cm apart. (ii) (a) Only one intersection is called (4, 3). (b) Only one intersection is called (3, 4).

  1. (i) Drawing the modelThe scale is 1 cm = 200 m, and adjacent streets are 200 m apart, so adjacent lines are 1 cm apart. Draw 10 vertical parallel lines (N-S streets) and 10 horizontal parallel lines (E-W streets). They form a square grid.
  2. (ii) Naming an intersectionAn intersection is named (N-S street number, E-W street number). So (4, 3) is where the 4th N-S street meets the 3rd E-W street, and (3, 4) is where the 3rd N-S street meets the 4th E-W street.
  3. (a) and (b)One N-S street and one E-W street meet at only one point. So (a) one intersection is called (4, 3), and (b) one intersection is called (3, 4). They are different points.

The answers to (a) and (b) are one each, but the two points are different, because the order of the numbers matters.

Model of the city: 10 north-south streets and 10 east-west streets forming a square grid
City model with the street intersections (2, 5), (3, 4) and (4, 3) marked

Question 20: Two circular icons on a screen

Q20

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.
Answer

(i) No part of either circle is outside the screen. (ii) Yes, the two circles intersect.

  1. (i) Limits of each circle
    Circlex from … toy from … toInside 0 to 800 and 0 to 600?
    Centre (100, 150), r = 8020 to 18070 to 230Yes
    Centre (250, 230), r = 100150 to 350130 to 330Yes
  2. (ii) Distance between the centresAB = √(250 − 100)² + (230 − 150)² = √22500 + 6400 = √28900 = 170Sum of the radii = 80 + 100 = 180 and difference of the radii = 100 − 80 = 20. Since 20 < 170 < 180, the circles intersect (at two points).
The two circular icons on the 800 by 600 pixel screen: centre (100, 150), radius 80 and centre (250, 230), radius 100

Question 21: Is ABCD a square?

Q21

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?

Answer

Yes, ABCD is a square, with area 10 square units.

A quadrilateral is a square if all four sides are equal and the two diagonals are equal.

  1. SidesAB = √(−1 − 2)² + (2 − 1)² = √9 + 1 = √10BC = √(−2 + 1)² + (−1 − 2)² = √1 + 9 = √10CD = √(1 + 2)² + (−2 + 1)² = √9 + 1 = √10DA = √(2 − 1)² + (1 + 2)² = √1 + 9 = √10All four sides are equal.
  2. DiagonalsAC = √(−2 − 2)² + (−1 − 1)² = √16 + 4 = √20BD = √(1 + 1)² + (−2 − 2)² = √4 + 16 = √20The diagonals are equal.
  3. AreaAll sides are equal and the diagonals are equal, so ABCD is a square.Area = (side)² = (√10)² = 10 square units
Graph with the points A (2, 1), B (−1, 2), C (−2, −1) and D (1, −2) joined to form the square ABCD

Axes, ordered pairs and reflections (Q22 to Q27)

Question 22: x-coordinate of a point on the y-axis

Q22

What is the x-coordinate of a point on the y-axis?

Answer

The x-coordinate of any point on the y-axis is 0.

The x-coordinate is the horizontal distance from the y-axis. A point on the y-axis has no horizontal distance from it, so x = 0. Every point on the y-axis has the form (0, y), where y can be any real number.

We use this to find the y-intercept of a line or a curve: put x = 0 in its equation to see where it crosses the y-axis.

Question 23: y-coordinate of a point on the x-axis

Q23

Is there a similar generalization for a point on the x-axis?

Answer

Yes. The y-coordinate of any point on the x-axis is 0.

A point on the x-axis has no vertical distance from it, so y = 0. Every point on the x-axis has the form (x, 0), where x can be any real number.

To find the x-intercept of a line or a curve, put y = 0 in its equation.

Question 24: Can Q(y, x) coincide with P(x, y)?

Q24

Does point Q(y, x) ever coincide with point P(x, y)? Justify your answer.

Answer

Yes, but only when x = y, that is, for points on the line through the origin at 45°.

Two points coincide only if their corresponding coordinates are equal. So P(x, y) and Q(y, x) coincide only if x = y (first coordinates) and y = x (second coordinates). Both conditions say the same thing: x = y.

If x ≠ y, the points are different. For example, (2, 3) and (3, 2) are distinct points, while (4, 4) stays the same when the coordinates are swapped.

Question 25: Is (x, y) = (y, x) only when x = y?

Q25

If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?

Answer

Yes, the claim is true.

An ordered pair is defined by the position of each number. Two ordered pairs are equal only if their first numbers are equal and their second numbers are equal. So (x, y) = (y, x) needs x = y and y = x, that is, x = y. Hence x = y is both necessary and sufficient.

If x ≠ y, the first numbers differ, so the points are at different places. For example, (1, 5) and (5, 1) are different points.

Question 26: Distance along the axes from A to D

Q26

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?

Answer

4 units along the x-axis and 3 units along the y-axis.

Along the x-axis: 7 − 3 = 4 unitsAlong the y-axis: |1 − 4| = 3 units (downward)

Question 27: Reflecting the triangle in the x-axis

Q27

Would these observations be the same if △ADM is reflected in the x-axis (instead of the y-axis)?

Answer

Yes, the distances along the axes are the same (4 units and 3 units); only the signs and the quadrant change.

Reflection in the x-axis changes (x, y) to (x, −y). Take A(3, 4) and D(7, 1).

OriginalReflected in the x-axisReflected in the y-axis
A(3, 4)(3, −4)(−3, 4)
D(7, 1)(7, −1)(−7, 1)
Distance along the x-axis444
Distance along the y-axis333

A reflection keeps all distances unchanged, so the side lengths and the distances along the axes stay the same. In the x-axis reflection the y-coordinates change sign, so the points move to the other side of the x-axis (from Quadrant I to Quadrant IV) and the vertical direction of travel from A to D is reversed.

Important formulas of Chapter 1

ConceptFormula or rule
Distance between (x₁, y₁) and (x₂, y₂)d = √(x2 − x1)² + (y2 − y1)²
Midpoint of the segment joining (x₁, y₁) and (x₂, y₂)((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
Points on the y-axis / x-axis(0, y) and (x, 0)
Reflection in the x-axis / y-axis(x, y) → (x, −y) and (x, y) → (−x, y)
Three points on one lineThe sum of the two shorter distances equals the longest distance
Points on a circle with centre OAll are at the same distance (the radius) from O
Signs in the quadrantsI: (+, +)   II: (−, +)   III: (−, −)   IV: (+, −)

Practice: multiple choice questions

Tap an option to check it at once. Your score updates as you answer.

Your score0 / 12
1

Which quadrant contains the point (−4, 7)?

x is negative and y is positive, so the point is in Quadrant II.
2

Every point on the y-axis has:

Points on the y-axis have no horizontal distance from it, so x = 0.
3

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

√(6² + 8²) = √100 = 10.
4

The midpoint of the segment joining (2, −4) and (6, 8) is:

((2 + 6) ÷ 2, (−4 + 8) ÷ 2) = (4, 2).
5

The reflection of (3, −2) in the x-axis is:

Reflection in the x-axis changes (x, y) to (x, −y).
6

The reflection of (5, 4) in the y-axis is:

Reflection in the y-axis changes (x, y) to (−x, y).
7

The point Q(y, x) coincides with P(x, y) when:

Both coordinates must match, which needs x = y.
8

The points A(0, 0), B(3, 4) and C(6, 8) are:

AB = 5, BC = 5 and AC = 10, so AB + BC = AC and the points are collinear.
9

The point (−3, −1) lies in:

Both coordinates are negative, which is Quadrant III.
10

The distance between (1, 2) and (4, 6) is:

√((4 − 1)² + (6 − 2)²) = √(9 + 16) = 5.
11

Assertion (A): The points (3, 4) and (4, 3) are different points.
Reason (R): In an ordered pair the order of the coordinates matters.

Both are true, and the order of the coordinates is exactly why the two points differ.
12

Assertion (A): The point (0, −5) lies in the fourth quadrant.
Reason (R): Its y-coordinate is negative.

A is false: (0, −5) lies on the y-axis, not in any quadrant. R is true.

Frequently asked questions

How many questions are solved in Class 9 Maths Chapter 1?

All 27 questions of Orienting Yourself: The Use of Coordinates from the new NCERT Ganita Manjari textbook (2026-27).

What do the coordinates (x, y) of a point mean?

x is the distance of the point from the y-axis and y is its distance from the x-axis, with signs that show the side. The order matters.

What is the distance formula?

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

How do I check whether three points lie on one straight line?

Find the three distances. If the sum of the two shorter distances equals the longest one, the points are on one line.

How do I find the midpoint of two points?

Take the average of the two x-coordinates and the average of the two y-coordinates.

Where can I get the PPT and worksheet for this chapter?

Use the links below: the free PPT and the worksheet with answers for Coordinates are on this site.

More free CBSE study material

Notes, worksheets, PPTs and practice papers for Classes 6 to 10 are available at uniquestudyonline.com.

Solutions prepared by Unique Study Point by Sumeet Sahu. The questions are from the NCERT Mathematics textbook Ganita Manjari (Class 9, Part 1).

More study material: Class IX Maths

Sumeet Sahu (14 years of teaching experience), Unique Study Point, Amitesh Nagar, Indore. Free NCERT/CBSE study material for Class VI to X.