Coordinates are ordered pairs of numbers used to describe the exact position of a point in a plane. In Class 9 Maths Chapter 1, students learn how the x-axis, y-axis, origin and quadrants form the Cartesian plane and how these ideas help in plotting points and finding distances.
NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 1 help students solve textbook questions from Orienting Yourself: The Use of Coordinates in the new Ganita Manjari Class 9 Maths textbook. This chapter introduces coordinates, the Cartesian plane, x-axis, y-axis, origin, quadrants, plotting points and distance between two points in a 2-D plane.
The chapter uses practical examples such as room layouts, doors, tables, bathrooms, dining rooms, city roads and computer graphics to show how coordinates help locate objects accurately. These exercise-wise NCERT Solutions for Class 9 Maths Chapter 1 are useful for homework, school exam preparation, CBSE 2026-27 revision and concept clarity.
NCERT Solutions for Class 9 Maths Chapter 1 PDF Download
NCERT Solutions for Class 9 Maths Chapter 1 PDF gives students a complete set of exercise-wise answers for Orienting Yourself: The Use of Coordinates. The PDF is useful for offline study, quick revision, classroom assignments and exam practice.
Students can use the PDF to revise:
- Coordinate axes and origin
- Points on the x-axis and y-axis
- Quadrants in the Cartesian plane
- Plotting points on graph paper
- Distance between two points
- Baudhāyana-Pythagoras Theorem applications
- Room-layout and city-grid questions
- End-of-chapter coordinate geometry problems
Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 1
| Exercise |
Main Focus |
Solution Type |
| Exercise Set 1.1 |
Reiaan’s room layout |
Diagram-based coordinate questions |
| Exercise Set 1.2 |
Graph plotting and room planning |
Coordinate plotting and rectangle questions |
| End-of-Chapter Exercises |
Mixed coordinate geometry |
Conceptual, numerical and application questions |
NCERT Solutions for Class 9 Maths Chapter 1 Exercise Set 1.1
Exercise Set 1.1 introduces students to coordinate-based room layout questions. Students read points from a room map and answer questions about door position, distance from axes, door width and accessibility.
Question 1. If D₁R₁ represents the door to Reiaan’s room, how far is the door from the left wall, which is the y-axis, of the room? How far is the door from the x-axis?
Answer:
The left wall of the room is represented by the y-axis. The distance of the door from the left wall is given by the x-coordinate of point D₁ or R₁, depending on the point from which the distance is measured.
Since D₁R₁ lies on the x-axis in the room layout, the door is 0 units from the x-axis.
Final answer:
The door is on the x-axis, so its distance from the x-axis is 0 ft. Its distance from the left wall should be read from the x-coordinate of D₁ in Fig. 1.3.
Question 2. What are the coordinates of D₁?
Answer:
Point D₁ lies on the x-axis, so its y-coordinate is 0.
Therefore, the coordinates of D₁ are of the form:
D₁ = (x, 0)
The exact x-coordinate should be read from Fig. 1.3 in the textbook.
Final answer:
D₁ = (x, 0), where x is the position of D₁ on the x-axis in Fig. 1.3.
Question 3. 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?
Answer:
The width of the door is the distance between D₁ and R₁.
Given:
R₁ = (11.5, 0)
Let:
D₁ = (x, 0)
Door width = |11.5 - x|
Since both points lie on the x-axis, only the difference between their x-coordinates is needed.
Final answer:
Door width = |11.5 - x| ft, where x is the x-coordinate of D₁ from Fig. 1.3.
A comfortable room door should be wide enough for easy entry. If the calculated width is around 3 ft or more, it is generally more comfortable and more suitable for wheelchair access. If it is much less than that, it may be difficult for a wheelchair user to enter easily.
Question 4. 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?
Answer:
The bathroom door width is the distance between B₁ and B₂.
Given:
B₁ = (0, 1.5)
B₂ = (0, 4)
Since both points lie on the y-axis:
Bathroom door width = |4 - 1.5|
Bathroom door width = 2.5 ft
Now compare this with the room door width found in Question 3.
Final answer:
The bathroom door is 2.5 ft wide. It is narrower than the room door if the room door width is more than 2.5 ft.
NCERT Solutions for Class 9 Maths Chapter 1 Exercise Set 1.2
Exercise Set 1.2 asks students to use graph paper and plot coordinates. It includes questions on placing a study table, identifying bathroom corners, marking spaces for a washbasin and toilet, sketching a dining room and placing a dining table at the centre.
Question 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?
Answer:
The three given points form three corners of a rectangle:
A = (8, 9)
B = (11, 9)
C = (11, 7)
The missing fourth point must have:
- x-coordinate same as A, which is 8
- y-coordinate same as C, which is 7
Therefore, the fourth foot is:
(8, 7)
Final answer:
The fourth foot of the table will be at (8, 7).
(ii) Is this a good spot for the table?
Answer:
The suitability of the table position depends on the room layout in Fig. 1.5. Students should check whether the table blocks the door, wardrobe or walking space.
Final answer:
The table is a good spot if it does not block the door, wardrobe or movement area in the room layout.
(iii) What is the width of the table? The length? Can you make out the height of the table?
Answer:
The horizontal distance between (8, 9) and (11, 9) is:
11 - 8 = 3 ft
The vertical distance between (11, 9) and (11, 7) is:
9 - 7 = 2 ft
So, the table has dimensions:
3 ft × 2 ft
The height of the table cannot be found from the floor map because the map shows only the top view or floor layout.
Final answer:
The table is 3 ft long and 2 ft wide. Its height cannot be found from this 2-D floor map.
Question 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?
Answer:
This is a diagram-based question. Students should observe Fig. 1.5 and check the arc of the bathroom door when it opens into the bedroom.
If the opening path of the door overlaps with the wardrobe, then the door will hit the wardrobe. If it does not overlap, the door will open safely.
Final answer:
The answer depends on the door movement shown in Fig. 1.5. If the door is made wider, it may need to open outward, slide sideways or be repositioned so that it does not hit the wardrobe.
Question 3. Look at Reiaan’s bathroom.
(i) What are the coordinates of the four corners O, F, R and P of the bathroom?
Answer:
The exact coordinates of O, F, R and P should be read from Fig. 1.5.
Since O is usually the origin:
O = (0, 0)
The remaining points F, R and P should be written according to their positions on the graph.
Final answer:
O = (0, 0). The coordinates of F, R and P should be read directly from Fig. 1.5.
(ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.
Answer:
The shape of SHWR can be identified by observing the four points S, H, W and R in Fig. 1.5.
If opposite sides are equal and parallel and all angles are right angles, then SHWR is a rectangle. If all four sides are equal, then it is a square.
Final answer:
The shape of SHWR should be identified from Fig. 1.5. The coordinates of S, H, W and R should be read directly from the graph.
(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.
Answer:
Students can choose suitable positions inside the bathroom so that the spaces do not overlap with the door or showering area.
For a 3 ft × 2 ft washbasin space, if one corner is taken as (a, b), then the other corners may be:
(a, b), (a + 3, b), (a + 3, b + 2), (a, b + 2)
For a 2 ft × 3 ft toilet space, if one corner is taken as (p, q), then the other corners may be:
(p, q), (p + 2, q), (p + 2, q + 3), (p, q + 3)
Final answer:
Answers may vary depending on where students place the washbasin and toilet. The chosen coordinates should form rectangles of dimensions 3 ft × 2 ft and 2 ft × 3 ft.
Question 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.
Answer:
The dining room is rectangular.
Given:
Length = 18 ft
Width = 15 ft
If one corner is taken as P = (a, b), and the length extends horizontally from P to A, then:
A = (a + 18, b)
The other two corners will be:
(a, b + 15) and (a + 18, b + 15)
Final answer:
If P = (a, b), then the dining room corners can be:
P = (a, b)
A = (a + 18, b)
Third corner = (a + 18, b + 15)
Fourth corner = (a, b + 15)
(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.
Answer:
The dining room has length 18 ft and width 15 ft.
The centre of the room is located halfway along both dimensions.
Half of 18 ft = 9 ft
Half of 15 ft = 7.5 ft
The table has dimensions:
5 ft × 3 ft
Half of 5 ft = 2.5 ft
Half of 3 ft = 1.5 ft
If the centre of the dining room is (a + 9, b + 7.5), then the four feet of the table are:
(a + 9 - 2.5, b + 7.5 - 1.5)
(a + 9 + 2.5, b + 7.5 - 1.5)
(a + 9 + 2.5, b + 7.5 + 1.5)
(a + 9 - 2.5, b + 7.5 + 1.5)
Simplifying:
(a + 6.5, b + 6)
(a + 11.5, b + 6)
(a + 11.5, b + 9)
(a + 6.5, b + 9)
Final answer:
If P = (a, b), the table feet are:
(a + 6.5, b + 6), (a + 11.5, b + 6), (a + 11.5, b + 9), (a + 6.5, b + 9)
NCERT Solutions for Class 9 Maths Chapter 1 End-of-Chapter Exercises
The end-of-chapter exercises cover the origin, coordinates of points, quadrants, right-angled triangles, collinearity, midpoint, trisection, circles, city-grid coordinates, computer graphics and checking whether a quadrilateral is a square.
Question 1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Answer:
The point of intersection of the x-axis and y-axis is called the origin.
Coordinates of origin = (0, 0)
Final answer:
x-coordinate = 0, y-coordinate = 0
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?
Answer:
A line parallel to the y-axis has the same x-coordinate for all its points.
Since W has x-coordinate -5, every point on the line through W parallel to the y-axis will have x-coordinate -5.
So, H can be written as:
H = (-5, y)
If y is positive, H lies in Quadrant II.
If y is negative, H lies in Quadrant III.
If y = 0, H lies on the x-axis.
Final answer:
H = (-5, y). H can lie in Quadrant II or Quadrant III, depending on the value of y.
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 the following.
(i) Two sides of RAMP that are perpendicular to each other.
Answer:
A (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.
Final answer:
AM and MP are perpendicular to each other.
(ii) One side of RAMP that is parallel to one of the axes.
Answer:
AM is parallel to the x-axis because A and M have the same y-coordinate.
Final 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?
Answer:
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.
Final answer:
M and P are mirror images in the x-axis.
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.
Answer:
Point Z is (5, -6). It lies in Quadrant IV.
One possible right-angled triangle can be formed by taking:
I = (5, 0)
N = (0, -6)
Z = (5, -6)
Now:
ZI = 6 units
ZN = 5 units
IN = √(5² + 6²)
IN = √(25 + 36)
IN = √61 units
Final answer:
One possible triangle has side lengths 5 units, 6 units and √61 units.
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?
Answer:
Without negative numbers, we could mark only points to the right of the origin and above the origin. This would cover only Quadrant I and the positive parts of the axes.
We would not be able to locate points in Quadrants II, III and IV.
Final answer:
No, without negative numbers, the coordinate system would not allow us to locate all points on a 2-D plane.
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.
Answer:
Check whether the slopes between the points are equal.
Slope of MA:
= [0 - (-4)] / [0 - (-3)]
= 4 / 3
Slope of AG:
= (8 - 0) / (6 - 0)
= 8 / 6
= 4 / 3
Since both slopes are equal, the points are collinear.
Final answer:
Yes, M (-3, -4), A (0, 0) and G (6, 8) lie on the same straight line.
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.
Answer:
Check slopes.
Slope of RB:
= [-5 - (-1)] / [-2 - (-5)]
= -4 / 3
Slope of BC:
= [-12 - (-5)] / [4 - (-2)]
= -7 / 6
Since:
-4/3 ≠ -7/6
The points are not collinear.
Final answer:
No, R, B and C do not lie on the same straight line.
Question 8. Using the origin as one vertex, plot the vertices of the following.
(i) A right-angled isosceles triangle.
Answer:
One possible right-angled isosceles triangle can have vertices:
O = (0, 0)
A = (4, 0)
B = (0, 4)
OA = 4 units
OB = 4 units
OA is perpendicular to OB
Final answer:
One possible set of vertices is (0, 0), (4, 0), (0, 4).
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Answer:
One possible isosceles triangle can have vertices:
O = (0, 0)
A = (-3, -4)
B = (3, -4)
Here, A lies in Quadrant III and B lies in Quadrant IV.
OA = √[(-3)² + (-4)²] = 5
OB = √[(3)² + (-4)²] = 5
Final answer:
One possible set of vertices is (0, 0), (-3, -4), (3, -4).
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.
Row 1: S (-3, 0), M (0, 0), T (3, 0)
Answer:
Midpoint of S and T:
= [(-3 + 3)/2, (0 + 0)/2]
= (0, 0)
This is M.
Final answer:
Yes, M is the midpoint.
Row 2: S (2, 3), M (3, 4), T (4, 5)
Answer:
Midpoint of S and T:
= [(2 + 4)/2, (3 + 5)/2]
= (3, 4)
This is M.
Final answer:
Yes, M is the midpoint.
Row 3: S (0, 0), M (0, 5), T (0, -10)
Answer:
Midpoint of S and T:
= [(0 + 0)/2, (0 + (-10))/2]
= (0, -5)
This is not M.
Final answer:
No, M is not the midpoint.
Row 4: S (-8, 7), M (0, -2), T (6, -3)
Answer:
Midpoint of S and T:
= [(-8 + 6)/2, (7 + (-3))/2]
= (-2/2, 4/2)
= (-1, 2)
This is not M.
Final answer:
No, M is not the midpoint.
Connection:
If M is the midpoint of S(x₁, y₁) and T(x₂, y₂), then:
M = [(x₁ + x₂)/2, (y₁ + y₂)/2]
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).
Answer:
Using midpoint formula:
M = [(x₁ + x₂)/2, (y₁ + y₂)/2]
Given:
A = (3, -4)
B = (x, y)
M = (-7, 1)
For x-coordinate:
(3 + x)/2 = -7
3 + x = -14
x = -17
For y-coordinate:
(-4 + y)/2 = 1
-4 + y = 2
y = 6
Final answer:
B = (-17, 6)
Question 11. Let P and Q be points of trisection of AB, with P closer to A and Q closer to B. Find P and Q when A (4, 7) and B (16, -2).
Answer:
P and Q divide AB into three equal parts.
Given:
A = (4, 7)
B = (16, -2)
Change in x:
16 - 4 = 12
One-third of x-change:
12/3 = 4
Change in y:
-2 - 7 = -9
One-third of y-change:
-9/3 = -3
P is one-third from A:
P = (4 + 4, 7 - 3)
P = (8, 4)
Q is two-thirds from A:
Q = (4 + 8, 7 - 6)
Q = (12, 1)
Final answer:
P = (8, 4), Q = (12, 1)
Question 12. Given the points A (1, -8), B (-4, 7) and C (-7, -4), show that they lie on a circle K whose centre is the origin O (0, 0). What is the radius of circle K?
Answer:
Find the distance of each point from origin.
OA = √[(1)² + (-8)²]
OA = √(1 + 64)
OA = √65
OB = √[(-4)² + (7)²]
OB = √(16 + 49)
OB = √65
OC = √[(-7)² + (-4)²]
OC = √(49 + 16)
OC = √65
Since OA = OB = OC, all three points lie on the same circle with centre O.
Final answer:
The radius of circle K is √65 units.
Given 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:
Radius of circle K = √65
OD = √[(-5)² + 6²]
OD = √(25 + 36)
OD = √61
Since √61 < √65, D lies inside the circle.
OE = √[0² + 9²]
OE = 9
OE = √81
Since √81 > √65, E lies outside the circle.
Final answer:
D lies inside the circle. E lies outside the circle.
Question 13. The midpoints of the sides of triangle ABC are D, E and F. Given that D, E and F are (5, 1), (6, 5) and (0, 3), respectively, find the coordinates of A, B and C.
Answer:
Let:
D = midpoint of AB = (5, 1)
E = midpoint of BC = (6, 5)
F = midpoint of CA = (0, 3)
Let:
A = (x₁, y₁)
B = (x₂, y₂)
C = (x₃, y₃)
Using midpoint formula:
(x₁ + x₂)/2 = 5, so x₁ + x₂ = 10
(y₁ + y₂)/2 = 1, so y₁ + y₂ = 2
(x₂ + x₃)/2 = 6, so x₂ + x₃ = 12
(y₂ + y₃)/2 = 5, so y₂ + y₃ = 10
(x₃ + x₁)/2 = 0, so x₃ + x₁ = 0
(y₃ + y₁)/2 = 3, so y₃ + y₁ = 6
Solving x-coordinates:
x₁ + x₂ = 10
x₂ + x₃ = 12
x₃ + x₁ = 0
Adding all:
2(x₁ + x₂ + x₃) = 22
x₁ + x₂ + x₃ = 11
Now:
x₃ = 11 - 10 = 1
x₁ = 11 - 12 = -1
x₂ = 11 - 0 = 11
Solving y-coordinates:
y₁ + y₂ = 2
y₂ + y₃ = 10
y₃ + y₁ = 6
Adding all:
2(y₁ + y₂ + y₃) = 18
y₁ + y₂ + y₃ = 9
Now:
y₃ = 9 - 2 = 7
y₁ = 9 - 10 = -1
y₂ = 9 - 6 = 3
Final answer:
A = (-1, -1), B = (11, 3), C = (1, 7)
Question 14. A city has two main roads crossing at the centre. All other streets 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.
Answer:
Students should draw two perpendicular main roads crossing at the centre. Then they should draw 10 parallel streets in the North-South direction and 10 parallel streets in the East-West direction, keeping 1 cm gap between consecutive streets.
Final answer:
A grid model should be drawn using the scale 1 cm = 200 m.
(ii)(a) How many street intersections can be referred to as (4, 3)?
Answer:
In a coordinate-style street grid, one pair of street numbers identifies one unique intersection.
Final answer:
Only one street intersection can be referred to as (4, 3).
(ii)(b) How many street intersections can be referred to as (3, 4)?
Answer:
Similarly, the pair (3, 4) identifies one unique intersection.
Final answer:
Only one street intersection can be referred to as (3, 4).
Note:
(4, 3) and (3, 4) are different intersections because the order of coordinates matters.
Question 15. A computer graphics program displays images on a rectangular screen whose coordinate system has origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels has centre A (100, 150). Another circular icon of radius 100 pixels has centre B (250, 230).
(i) Determine whether any part of either circle lies outside the screen.
Answer:
For circle A:
Centre = (100, 150)
Radius = 80
Left boundary = 100 - 80 = 20
Right boundary = 100 + 80 = 180
Bottom boundary = 150 - 80 = 70
Top boundary = 150 + 80 = 230
All values are within the screen limits:
0 ≤ x ≤ 800
0 ≤ y ≤ 600
So, circle A lies fully inside the screen.
For circle B:
Centre = (250, 230)
Radius = 100
Left boundary = 250 - 100 = 150
Right boundary = 250 + 100 = 350
Bottom boundary = 230 - 100 = 130
Top boundary = 230 + 100 = 330
All values are within screen limits.
Final answer:
No part of either circle lies outside the screen.
(ii) Determine whether the two circles intersect each other.
Answer:
Find the distance between centres A and B.
A = (100, 150)
B = (250, 230)
Distance AB = √[(250 - 100)² + (230 - 150)²]
= √[(150)² + (80)²]
= √(22500 + 6400)
= √28900
= 170 pixels
Sum of radii:
80 + 100 = 180 pixels
Since distance between centres is 170 pixels and the sum of radii is 180 pixels, the circles intersect.
Final answer:
Yes, the two circles intersect each other.
Question 16. Plot the points A (2, 1), B (-1, 2), C (-2, -1), and D (1, -2). Is ABCD a square? Explain why. What is the area of this square?
Answer:
Find the lengths of all sides.
AB = √[(-1 - 2)² + (2 - 1)²]
= √[(-3)² + 1²]
= √10
BC = √[(-2 - (-1))² + (-1 - 2)²]
= √[(-1)² + (-3)²]
= √10
CD = √[(1 - (-2))² + (-2 - (-1))²]
= √[3² + (-1)²]
= √10
DA = √[(2 - 1)² + (1 - (-2))²]
= √[1² + 3²]
= √10
All four sides are equal.
Now check diagonals:
AC = √[(-2 - 2)² + (-1 - 1)²]
= √[(-4)² + (-2)²]
= √20
BD = √[(1 - (-1))² + (-2 - 2)²]
= √[2² + (-4)²]
= √20
The diagonals are equal. Therefore, ABCD is a square.
Area of square = side²
= (√10)²
= 10 square units
Final answer:
ABCD is a square, and its area is 10 square units.
Topics Covered in NCERT Solutions for Class 9 Maths Chapter 1
- Coordinate system
- Cartesian plane
- x-axis and y-axis
- Origin
- Coordinates of a point
- Points on the x-axis
- Points on the y-axis
- Quadrants
- Plotting points on graph paper
- Distance between points on horizontal and vertical lines
- Distance between any two points
- Baudhāyana-Pythagoras Theorem
- Midpoint of a line segment
- Points of trisection
- Collinearity of points
- Circle-related coordinate questions
- City-grid coordinate questions
- Computer graphics coordinate questions
Important Formulas in NCERT Solutions for Class 9 Maths Chapter 1
| Concept |
Formula |
| Origin |
(0, 0) |
| Point on x-axis |
(x, 0) |
| Point on y-axis |
(0, y) |
| Distance on horizontal line |
|x₂ - x₁| |
| Distance on vertical line |
|y₂ - y₁| |
| Distance between two points |
√[(x₂ - x₁)² + (y₂ - y₁)²] |
| Midpoint of two points |
[(x₁ + x₂)/2, (y₁ + y₂)/2] |
NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions
Benefits of NCERT Solutions for Class 9 Maths Chapter 1
NCERT Solutions for Class 9 Maths Chapter 1 help students understand the new Ganita Manjari chapter in a structured and exam-focused way.
Key benefits include:
- Step-by-step answers to textbook exercises.
- Clear explanation of coordinate geometry basics.
- Better understanding of axes, origin and quadrants.
- Practice with plotting points on graph paper.
- Support for diagram-based questions.
- Easy revision of distance formula and midpoint formula.
- Preparation for CBSE 2026-27 school exams.
- Foundation for Class 10 coordinate geometry.