Class 9 · MathsNCERT 2026 Revised Syllabus
Chapter 1: Orienting Yourself: The Use of Coordinates (0904CH01)
Ganita Manjari Part I · 21 questions · 3 exercises
Chapter code: 0904ch01 · Book: Ganita Manjari Part I · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 1. Every exercise has a clear, step-by-step answer for chapter code 0904CH01.
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Exercise Set 1.1(1 questions)
1Fig. 1.3 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. Fig. 1.3 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?(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 heshe be able to do so 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?
School Level Solution
Fig. 1.3
Exercise Set 1.2(4 questions)
1Place 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?
School Level Solution
Fig. 1.5
2If 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?
School Level Solution
From Fig. 1.5, the bathroom door is hinged at B₁ and opens into the bedroom.The wardrobe starts at W₁ and lies in the path through which the door swings.Therefore, the door will hit the wardrobe.If the door is made wider, it would swing even more into the bedroom.Therefore, the wardrobe should be moved a little farther away from the door or the door should open towards the bathroom. Ans.
3Look 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.
School Level Solution
i) The four corners of the bathroom are O = (0, 0), F = (0, 10), R = (−5, 10), and P = (−5, 0).
ii) From Fig. 1.5, S = (0, 8), H = (−3, 5), W = (−5, 10), and R = (−5, 8). Thus SHWR is a quadrilateral, not a rectangle.
iii) One valid marking on the 5 ft × 10 ft bathroom grid is: washbasin (3 ft × 2 ft) with corners (−4, 2), (−1, 2), (−1, 4), (−4, 4); toilet (2 ft × 3 ft) with corners (−5, 2), (−3, 2), (−3, 5), (−5, 5). These spaces have the required dimensions and do not overlap.
4Other 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.
School Level Solution
The dining room has length 18 ft from P = (−5, 0) to A = (13, 0), and width 15 ft, so its four corners are P = (−5, 0), A = (13, 0), and the corresponding upper corners (13, 15) and (−5, 15).
The centre of the dining room is the midpoint of its diagonals: ((−5 + 13)/2, (0 + 15)/2) = (4, 7.5).
For a 5 ft × 3 ft table centred at (4, 7.5), take half-length 2.5 ft and half-width 1.5 ft. Its four corner points are (1.5, 6), (6.5, 6), (6.5, 9), and (1.5, 9).
End-of-Chapter Exercises(16 questions)
1
What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
School Level Solution
The point of intersection of the x-axis and y-axis is the origin O.
At the origin, x-coordinate = 0 and y-coordinate = 0.
(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?
School Level Solution
x-coordinate of W = −5
A line through W parallel to the y-axis will have the same x-coordinate.
Therefore, the coordinates of H can be (−5, y), where y can be any real number.
If y is positive, H will lie in Quadrant II and if y is negative, H will lie in Quadrant III.
Therefore, H can lie in Quadrant II or Quadrant III.
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.
School Level Solution
(i) In RAMP, AM is horizontal and MP is vertical.
Therefore, AM ⟂ MP.
(ii) Since AM is horizontal, AM is parallel to the x-axis.
(iii) M (−5, −2) and P (−5, 2) have the same x-coordinate and opposite y-coordinates.
Therefore, they are mirror images of each other in the x-axis.
Hence, the predictions are verified.
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.)
School Level Solution
One possible construction is to take I (5, 0) and N (0, −6).
IZ = 6 units
ZN = 5 units
IN = √((5 − 0)² + (0 − (−6))²)
= √(25 + 36)
= √61 units
Therefore, the three sides are 5 units, 6 units 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?
School Level Solution
If we did not have negative numbers, the x- and y-coordinates could only be zero or positive.
Therefore, we could locate only the points in the first quadrant and on the positive x-axis, positive y-axis and the origin.
No, this system would not allow us to locate all the points on a 2-D plane.
★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.
School Level Solution
We can check this by comparing the slopes of the line segments AM and AG.
Slope of AM = −4−3 = 43
= 43
Slope of AG = 86 = 43
Both slopes are equal.
Therefore, M, A and G are on the same straight line.
★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.
School Level Solution
Slope of RB = −5 − (−1)−2 − (−5) = −43
Slope of BC = −12 − (−5)4 − (−2) = −76
The slopes are not equal.
Therefore, R, B and C are not on the same straight line.
The plot also verifies the result.
★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.
School Level Solution
(i) One possible set of vertices is O (0, 0), A (2, 0) and B (0, 2).
OA = OB = 2 units and OA ⟂ OB.
Hence, it is a right-angled isosceles triangle.
(ii) One possible set of vertices is O (0, 0), A (−2, −2) and B (2, −2).
OA = OB = √8 = 2√2 units.
Hence, it is an isosceles triangle with A in Quadrant III and B in Quadrant IV.
★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.
School Level Solution
| S | M | T | Is M the midpoint of ST? Yes or No | Reason for your answer |
|---|---|---|---|---|
| (−3, 0) | (0, 0) | (3, 0) | Yes | Midpoint = (0, 0) |
| (2, 3) | (3, 4) | (4, 5) | Yes | Midpoint = (3, 4) |
| (0, 0) | (0, 5) | (0, −10) | No | Midpoint = (0, −5) |
| (−8, 7) | (0, −2) | (6, −3) | No | Midpoint = (−1, 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).
School Level Solution
Using midpoint formula,
−7 = 3 + x2
−14 = 3 + x
x = −17
1 = −4 + y2
2 = −4 + y
y = 6
Therefore, the coordinates of B are (−17, 6). Ans.
★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).
School Level Solution
First, find the midpoint M of AB.
M = (4 + 162, 7 − 22)
= (10, 52)
P is the midpoint of AM.
P = (4 + 102, 7 + 5/22)
= (7, 194)
Q is the midpoint of MB.
Q = (10 + 162, 5/2 − 22)
= (13, 14)
Therefore, P = (7, 194) and Q = (13, 14). Ans.
★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.
School Level Solution
(i) Since the centre is O (0, 0),
OA² = 1² + (−8)² = 1 + 64 = 65
OB² = (−4)² + 7² = 16 + 49 = 65
OC² = (−7)² + (−4)² = 49 + 16 = 65
Thus, OA = OB = OC = √65
Therefore, A, B and C lie on the same circle K with radius = √65 units.
(ii) OD² = (−5)² + 6² = 25 + 36 = 61
OE² = 0² + 9² = 81
Since 61 < 65, D lies inside the circle.
Since 81 > 65, E lies outside the circle.
Therefore, D is inside and E is outside circle 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.
School Level Solution
D, E and F are the midpoints of AB, BC and CA respectively.
Since D is the midpoint of AB, E of BC and F of CA,
A = D + F − E
= (5, 1) + (0, 3) − (6, 5)
= (−1, −1)
B = D + E − F
= (5, 1) + (6, 5) − (0, 3)
= (11, 3)
C = E + F − D
= (6, 5) + (0, 3) − (5, 1)
= (1, 7)
Therefore, A = (−1, −1), B = (11, 3), C = (1, 7). Ans.
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.(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roadsstreets 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).
School Level Solution
(i) Each adjacent pair of streets is 200 m apart.
Using the scale 1 cm = 200 m, adjacent parallel streets will be 1 cm apart in the model.
So, we draw 10 parallel N–S streets and 10 parallel E–W streets.
This gives the required model of the city.
(ii)(a) (4, 3) means the 4th N–S street and the 3rd E–W street.
These two streets meet at exactly one point.
No. of such intersections = 1
(ii)(b) (3, 4) means the 3rd N–S street and the 4th E–W street.
These two streets also meet at exactly one point.
No. of such intersections = 1
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.
School Level Solution
(i) For circle A, centre = (100, 150), radius = 80.
Its leftmost x = 100 − 80 = 20 and rightmost x = 100 + 80 = 180.
Its bottom y = 150 − 80 = 70 and top y = 150 + 80 = 230.
All these values lie within the screen from x = 0 to 800 and y = 0 to 600.
For circle B, centre = (250, 230), radius = 100.
Its leftmost x = 150, rightmost x = 350, bottom y = 130 and top y = 330.
These also lie within the screen.
Therefore, no part of either circle lies outside the screen.
(ii) Distance between centres AB = √((250 − 100)² + (230 − 150)²)
= √(150² + 80²)
= √28900 = 170 pixels
Sum of radii = 80 + 100 = 180 pixels.
Difference of radii = 100 − 80 = 20 pixels.
Since 20 < 170 < 180, the two circles intersect at two points.
Therefore, the circles intersect each other. Ans.
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?
School Level Solution