Chapter 1 - Orienting yourself: The use of coordinates

NCERT Solutions for Class 9 Maths Chapter 1: Coordinate Geometry

Strengthen your understanding of Class 9 Maths Chapter 1: Coordinate Geometry with chapter-wise NCERT solutions designed for effective learning and exam preparation. Aligned with the latest NCERT syllabus (2026–27) and CBSE guidelines, these expert-created solutions feature step-by-step answers to every textbook question and include important questions for focused practice. Curated by ALLEN Experts, the solutions are available for free PDF download, making revision and offline study convenient anytime, anywhere.

1.0Download NCERT Solutions for Class 9 Maths Chapter 1 – Coordinate Geometry PDF

Class 9 Maths Chapter 1: Coordinate Geometry introduces students to the Cartesian coordinate system and the basics of locating points on a plane. It helps build a strong foundation for understanding graphs and coordinate-based concepts. Students can download the free PDF of NCERT Solutions for Class 9 Maths Chapter 1: Coordinate Geometry, featuring step-by-step solutions to all exercises, prepared according to the latest NCERT syllabus and CBSE-aligned guidelines.

NCERT Solutions Class 9 Maths Chapter 1

2.0Learning Outcomes : NCERT Class 9 Maths Chapter 1 Solutions

Understand the Cartesian System: Explain the concept of a plane and how the Cartesian plane is formed by two perpendicular number lines.
Identify Axes and Origin: Define the x-axis, y-axis, and origin, and describe their role in locating points on a plane.
Determine Coordinates of a Point: Find the x-coordinate (abscissa) and y-coordinate (ordinate) of any given point in the Cartesian plane.
Plot Points Accurately: Plot points on graph paper using ordered pairs (x, y) with correct scale and sign conventions.
Recognize the Four Quadrants: Identify the four quadrants (I, II, III, IV) and determine the sign conventions of coordinates in each quadrant.
Interpret Ordered Pairs: Understand that the order of x and y in an ordered pair (x, y) matters and affects the point's location.
Locate Points on Axes: Identify points lying on the x-axis (y = 0) and y-axis (x = 0) and understand why the origin has coordinates (0, 0).
Apply Graphical Representation: Use coordinate geometry to represent real-life data and relationships graphically.
Solve NCERT Exercise Problems: Confidently attempt and solve all questions from Exercises 1.1, 1.2, and 1.3 based on plotting and identifying coordinates.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports future chapters on linear equations in two variables, graphs, and higher-level coordinate geometry in Class 10 and beyond.

3.0Detailed Class 9 Maths Chapter 1 Coordinate Geometry Solutions

Exercise 1.1

1. Figure shows Reiaan's room with points OABC marking its corners. The x and y axis are marked in the figure. Point 0 is the origin.

Referring to Figure, answer the following questions: (i) If $D_{1} R_{1}$ 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 $D_{1}$ ? (iii) If $R_{1}$ 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? (iv) If $B_{1} (0, 1.5)$ and $B_{2} (0, 4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Sol. (i) Distance of door from $y$ -axis (left wall) $= 8$ units Distance of door from x -axis $= 0$ units [Because door is lying on x-axis.] (ii) Coordinate of $D_{1} = (8, 0)$ (iii) Width of room door $= 11.5 - 8$ $= 3.5$ units Yes, this is a comfortable width for a room door. Also, a person in a wheelchair would likely be able to enter easily. A width of 3.5 units is much more generous than the standard requirement. (iv) Width of bathroom door ( $B_{1} B_{2}$ ) $= 4 - 1.5 = 2.5$ units Hence, bathroom door is narrower than the room door.

Exercise 1.2

On a graph sheet, mark the x -axis and y -axis and the origin 0 . 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 Figure, answer the given questions.

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?

Sol.

Because study table is rectangular in shape that's why distance of opposite sides are equal. So, fourth foot of table is at (8,7). (ii) No, this is likely not a good spot, because (8, 7) and (11, 7) fall directly in front of bed.

Placing a table there would block the path to the bed. (iii) Width $= 11 - 8 = 3$ units

Length $= 9 - 7 = 2$ units No, we cannot make out the height of the table from this image.

2. If the bathroom door has a hinge at $B_{1}$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Sol. Door $= B_{1} (0, 1.5)$ to $B_{2} (0, 4)$ Width of door $= 4 - 1.5 = 2.5$ feets

But the wardrobe starts at $W_{1} (3, 0)$ So, the door will not hit the wardrobe. If the door is made 3 feet or wider, it will smash into the wardrobe. So, make some changes:

Reversing the hinge so the door swings outward into the bathroom instead.
Moving the wardrobe further to the right.

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 $3 ft \times 2 ft$ space for the washbasin and a $2 ft \times 3 ft$ space for the toilet. Write the coordinates of the corners of these spaces.

Sol. (i) The coordinate of $O (0, 0), F (0, 9), R (- 6, 9)$ and $P (- 6, 0)$ (ii) SWHR has 4 sides. So, it is a quadrilateral and coordinates of $S (- 6, 6), H (- 3, 6), W (- 2, 9)$ and $R (- 6, 9)$ . (iii) Washbasin ( $3ft \times 2ft$ )

So, corners are $(- 6, 0), (- 4, 0), (- 4, 3)$ and $(- 6, 3)$ . Toilet ( $2ft \times 3ft$ ) So, corners are $(- 6, 3), (- 3, 3), (- 3, 5)$ and $(- 6, 5)$ .

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 $5 ft \times 3 ft$ dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Sol. (i)

So, the coordinates are $P (- 6, 0), A (12, 0), B (12, - 15)$ and $C (- 6, 15)$ . (ii) Center point of dinning room: $x$ midpoint $(- 6, 12) = \frac{- 6 + 12}{2} = 3$ y midpoint $(0, - 15) = \frac{0 - 15}{2} = - 7.5$ Now, center of room ( $3, - 7.5$ ) → The length ( 5 ft .) along x -axis:

$3 + 2.5 = 5.5 3 - 2.5 = 0.5$

→ The width ( 3 ft .) along $y$ -axis:

$- 7.5 + 1.5 = - 6 - 7.5 - 1.5 = - 9$

So, the coordinate of table $(5.5, - 6), (5.5, - 9), (0.5, - 6), (0.5, - 9)$ .

End of Chapter Exercise

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

Sol. $(0, 0)$ origin $\to$ where both axes intersect.

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? Sol. $W (- 5, y)$ then x coordinate of H will be always -5 but y coordinate will be any real number and lie in II or III quadrant.

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. Sol. (i) $AM ⊥ MP$ (ii) MP (Parallel to $y$ -axis)

Or AM (Parallel to x -axis) (iii) $M (- 5, - 2)$ and $P (- 5, 2)$ will be minor images, if we consider x -axis as mirror.

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.) Sol.

Use formula $($ distance $) = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ $IZ = (5 - 5)^{2} + (- 6 - 0)^{2} = 6$ units $ZN = (5 - 0)^{2} + (- 6 + 6)^{2} = 5$ units $NI = (5 - 0)^{2} + (0 + 6)^{2} = 61$ 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?

Sol. Without negative numbers, we would be restricted to the first quadrant. No, to locate every point, we would have to use all quadrant.

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.

Sol.

MA = [0 - (- 3)]^{2} + [0 - (- 4)]^{2} = 3^{2} + 4^{2}

= 9 + 16 = 5

units

AG = (6 - 0)^{2} + (8 - 0)^{2} = 6^{2} + 8^{2}

= 36 + 64 = 10

units

MG = [6 - (- 3)]^{2} + [8 - (- 4)]^{2} = 9^{2} + 1 2^{2}

= 81 + 144 = 15

units So,

MA + AG = MG

$5 + 10 = 15 units$

Hence, points are collinear.

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.

Sol.

RB = [- 2 - (- 5)]^{2} + [- 5 - (- 1)]^{2}

= 3^{2} + (- 4)^{2} = 9 + 16 = 5

units

BC = [4 - (- 2)]^{2} + [- 12 - (- 5)]^{2}

= 6^{2} + (- 7)^{2} = 36 + 49 = 85

units

RC = [4 - (- 5)]^{2} + [- 12 - (- 1)]^{2}

= 9^{2} + (- 11)^{2} = 81 + 121 = 202

units So,

RB + BC \neq = RC

Hence 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.

Sol. (i) $OA = (4 - 0)^{2} + (0 - 0)^{2} = 4$ units $OB = (0 - 0)^{2} + (4 - 0)^{2} = 4$ units $AB = (4 - 0)^{2} + (4 - 0)^{2} = 32$ units

For right angle $OA^{2} + OB^{2} = AB^{2}$ $(4)^{2} + (4)^{2} = (32)^{2}$ $32 = 32$ Hence proved (ii)

$OC = [0 - (- 3)]^{2} + [0 - (- 4)]^{2} = 3^{2} + 4^{2} = 9 + 16 = 5 units OD = (0 - 3)^{2} + [0 - (- 4)]^{2} = (- 3)^{2} + 4^{2} = 9 + 16 = 5 units$

So, $OC = OD$ , hence $△ OCD$ is an isosceles triangle.

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.

S	M	T	Is M the midpoint of ST? Yes or No	Reason 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 ?

Sol.

SM = [0 - (- 3)]^{2} + (0 - 0)^{2} = 3^{2} = 3

units

MT = (3 - 0)^{2} + (0 - 0)^{2} = 3^{2} = 3

units

ST = [3 - (- 3)]^{2} + (0 - 0)^{2} = 6^{2} = 6

units

SM = MT

So,

SM + MT = ST

Hence, M is mid point of ST .

S	M	T
$(2, 3)$	$(3, 4)$	$(4, 5)$

$SM = (3 - 2)^{2} + (4 - 3)^{2} = 1 + 1 = 2$ units MT $= (4 - 3)^{2} + (5 - 4)^{2} = 1 + 1 = 2$ units ST $= (4 - 2)^{2} + (5 - 3)^{2} = (2)^{2} + (2)^{2} = 8$ units So, $SM + MT = ST$ $2 + 2 = 22$ Hence, M is mid point of ST .

SM = (0 - 0)^{2} + (5 - 0)^{2} = 5^{2} = 5

units MT

= (0 - 0)^{2} + (- 10 - 5)^{2} = (- 15)^{2} = 15

units

ST = (0 - 0)^{2} + (- 10 - 0)^{2} = (- 10)^{2} = 10

units

SM \neq = MT

So,

SM + MT \neq = ST

Hence, M is not a mid point of ST.

SM = [(0 - (- 8)]^{2} + (- 2 - 7)^{2} = 8^{2} + (- 9)^{2} = 145

units MT

= (6 - 0)^{2} + [- 3 - (- 2)]^{2} = 6^{2} + (- 1)^{2} = 37

units ST

= [6 - (- 8)]^{2} + (- 3 - 7)^{2} = 1 4^{2} + (- 10)^{2} = 296

units So,

SM \neq = MT

Also,

SM + MT \neq = ST

Hence, M is mid point of ST.

For midpoint we have connection $x = \frac{x _{1} + x _{2}}{2}, y = \frac{y _{1} + y _{2}}{2}$

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 $B (x, y)$ . Sol.

Midpoint formula $x = \frac{x _{1} + x _{2}}{2}, y = \frac{y _{1} + y _{2}}{2}$ $- 7 = \frac{3 + x}{2}, 1 = \frac{- 4 + y}{2}$ $- 14 - 3 = x, 2 + 4 = y$ $x = - 17, y = 6$ 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 $B (16, - 2)$ . Sol.

Given $AP = PQ = QB$ x-step: $16 - 4 = 12$ Now divide 12 by 3 $12 \div 3 = 4$ y-step: $- 2 - 7 = - 9$ Now divide -9 by 3 $- 9 \div 3 = - 3$ Coordinates of $P = [4 + 4, 7 + (- 3)]$ $(8, 4)$ Coordinates of $Q = [8 + 4, 4 + (- 3)]$ $(12, 1)$

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. Sol. (i) If $A, B$ and C lie on circle then distance $OA, OB$ and OC are equal.

$OA = (0 - 1)^{2} + [0 - (- 8)]^{2} = (- 1)^{2} + (8)^{2} = 65 units OB = [0 - (- 4)]^{2} + (0 - 7)^{2} = (4)^{2} + (- 7)^{2} = 16 + 49 = 65 units OC = [0 - (- 7)]^{2} + [0 - (- 4)]^{2} = 7^{2} + 4^{2} = 49 + 16 = 65 units$

Hence the radius of circle is $65$ units. (ii) For $D (- 5, 6)$

$OD = (- 5 - 0)^{2} + (6 - 0)^{2} = 25 + 36 = 61$

Since $61 < 65$ , point D lies with in the circle. For E ( 0,9 ):

$OE = 0^{2} + 9^{2} = 81 = 9$

Since $9 > 65$ , point E lies outside the circle.

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.

Sol.

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ Now by using midpoint formula $\frac{x _{1} + x _{2}}{2} = 5$ and $\frac{y _{1} + y _{2}}{2} = 1$ $x_{1} + x_{2} = 10$ $y_{1} + y_{2} = 2$

Again $\frac{x _{2} + x _{3}}{2} = 6$ and $\frac{y _{2} + y _{3}}{2} = 5$ $x_{2} + x_{3} = 12$ $y_{2} + y_{3} = 10$

Also, $\frac{x _{1} + x _{3}}{2} = 0$ and $\frac{y _{1} + y _{3}}{2} = 3$ $x_{1} + x_{3} = 0$ $y_{1} + y_{3} = 6$

Now add (i), (iii) and (v) we get $x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3} = 10 + 12 + 0$ $2 (x_{1} + x_{2} + x_{3}) = 22$ $x_{1} + x_{2} + x_{2} = 11$ $x_{1} = (x_{1} + x_{2} + x_{3}) - (x_{2} + x_{3}) = 11 - 12 = - 1$ $x_{2} = (x_{1} + x_{2} + x_{3}) - (x_{1} + x_{3}) = 11 - 0 = 11$ $x_{3} = (x_{1} + x_{2} + x_{3}) - (x_{1} + x_{2}) = 11 - 10 = 1$ Now add (ii), (iv) and (vi) we get $2 (y_{1} + y_{2} + y_{3}) = 18$ $y_{1} + y_{2} + y_{3} = 9$ $y_{1} = (y_{1} + y_{2} + y_{3}) - (y_{2} + y_{3}) = 9 - 10 = - 1$ $y_{2} = (y_{1} + y_{2} + y_{3}) - (y_{1} + y_{3}) = 9 - 6 = 3$ $y_{3} = (y_{1} + y_{2} + y_{3}) - (y_{1} + y_{2}) = 9 - 2 = 7$ The coordinates of $A (- 1, - 1), B (11, 3)$ and $C (1, 7)$ .

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 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 ).

Sol. (i)

(ii) (a) There is only one cross street at

(4, 3)

. (b) There is only one cross street at

(3, 4)

.

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.

Sol. (i)

Circle A: Centre $(100, 150)$ and radius $= 80$ So, $100 - 80 = 20$

$150 - 80 = 70$

Since both are greater than 0 , it doesn't cross the left or bottom boundaries. Similarly, for circle B: Centre $(250, 230)$ and radius $= 100$ So, $250 - 100 = 150$

$230 - 100 = 130$

No, neither circle goes outside the screen boundaries. (ii) Distance between centers $A (100, 150)$ and $B (250, 230)$ :

$AB = (250 - 100)^{2} + (230 - 150)^{2} = (150)^{2} + (80)^{2} = 22500 + 6400 = 28900 = 170$

Sum of radii $= 80 + 100 = 180$ Thus, the two circles interest.

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? Sol.

$A B = (- 1 - 2)^{2} + (2 - 1)^{2} = 10 units BC = [(- 2 - (- 1)]^{2} + (- 1 - 2)^{2} = 10 units C D = [1 - (- 2)]^{2} + [- 2 - (- 1)]^{2} = 10 units A D = (1 - 2)^{2} + (- 2 - 1)^{2} = 10 units A C = (- 2 - 2)^{2} + (- 1 - 1)^{2} = 20 units B D = [1 - (- 1)]^{2} + (- 2 - 2)^{2} = 20 units$

So, all sides and both diagonals are equal, hence the given figure is a square. Now, Area of square $=$ Side × Side $= 10 \times 10$ $= 10$ sq. units

4.0Key Topics in NCERT Class 9 Maths Chapter 1 Coordinate Geometry

Topic	What Students Learn
Coordinate System	Understand how coordinates are used to locate objects and positions in everyday life.
Cartesian Plane	Learn the Cartesian coordinate system with the x-axis, y-axis, and origin.
Coordinates of a Point	Represent the position of a point using an ordered pair (x, y).
Quadrants and Sign Convention	Identify the four quadrants and understand the signs of coordinates in each quadrant.
Points on the Axes	Recognise points lying on the x-axis and y-axis and determine their coordinates.
Plotting Points	Plot and locate points accurately on the Cartesian plane.
Distance Between Two Points	Calculate the distance between two points on a coordinate plane using the distance formula.
Midpoint of a Line Segment	Find the midpoint of a line segment joining two given points.
Applications of Coordinates	Apply coordinate geometry concepts to solve map-based, navigation, and real-life problems.

5.0Exercise-wise NCERT Class 9 Maths Chapter 1 – Coordinate Geometry

Explore the key concepts covered in each exercise of NCERT Class 9 Maths Chapter 1 – Coordinate Geometry. The exercise-wise breakdown helps students revise systematically and strengthen key concepts.

Exercise	Key Concepts Covered
Exercise 1.1	Introduction to the Cartesian coordinate system, x-axis, y-axis, origin, coordinates of a point, and locating points on the Cartesian plane.
Exercise 1.2	Plotting points, identifying quadrants, understanding sign conventions, and solving coordinate-based problems.
End-of-Chapter Exercise	Comprehensive revision of coordinate geometry concepts through mixed conceptual and application-based questions.

6.0Quick Revision on Class 9 Maths Chapter 1 - Coordinate Geometry

7.0Related Study Materials Class 9 Maths

Enhance your knowledge with ALLEN Class 9th Maths study materials for all chapters as per the latest NCERT syllabus. Along with NCERT Solutions, you can refer to the latest syllabus, NCERT books, revision notes, sample papers and previous years’ question papers to revise well and prepare confidently for your school and CBSE exams.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 1 Maths Class 9 NCERT Solutions

Develops Coordinate Geometry Basics: Builds a clear understanding of the Cartesian plane, axes, origin and quadrants.
Enhances Point Representation: Assists students with correctly identifying and writing the coordinates of points.
Helps in Plotting Skills: It teaches the right way of plotting points on the coordinate plane.
Explains Point Positions: Discusses how the signs of coordinates determine the location of points in quadrants.
Develops Graphical Interpretation Skills: improves the ability to read and interpret points on the Cartesian plane.
Builds Strong Foundation: Equips students to handle the higher level concepts that involve graphs, linear equations and coordinate geometry in higher classes.

Frequently Asked Questions

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Frequently Asked Questions

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How many quadrants are there in the Cartesian plane?

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NCERT Solutions for Class 9 Maths Chapter 1: Coordinate Geometry

1.0Download NCERT Solutions for Class 9 Maths Chapter 1 – Coordinate Geometry PDF

2.0Learning Outcomes : NCERT Class 9 Maths Chapter 1 Solutions

3.0Detailed Class 9 Maths Chapter 1 Coordinate Geometry Solutions

Exercise 1.1

Exercise 1.2

End of Chapter Exercise

4.0Key Topics in NCERT Class 9 Maths Chapter 1 Coordinate Geometry

5.0Exercise-wise NCERT Class 9 Maths Chapter 1 – Coordinate Geometry

6.0Quick Revision on Class 9 Maths Chapter 1 - Coordinate Geometry

7.0Related Study Materials Class 9 Maths

8.0Advantages of Chapter 1 Maths Class 9 NCERT Solutions