What is Chapter 1 of Class 9 Maths Ganita Manjari about?

Chapter 1, Orienting Yourself: The Use of Coordinates, introduces the 2-D Cartesian coordinate system - axes, origin, quadrants, plotting points - and the distance between two points using the Baudhayana-Pythagoras theorem, along with the midpoint idea.

What is the distance formula in Ganita Manjari Chapter 1?

The distance between points (x1, y1) and (x2, y2) is the square root of (x2 - x1) squared plus (y2 - y1) squared.

Class 9 Maths Ganita Manjari Chapter 1 Solutions — Orienting Yourself: The Use of Coordinates

Q: How many exercises are there in Ganita Manjari Chapter 1?

Chapter 1 has Exercise Set 1.1, Exercise Set 1.2, several Think and Reflect boxes, and 16 End-of-Chapter Exercises. ClearStudy provides solutions to all of them.

Q: Are Class 9 Maths Ganita Manjari Chapter 1 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 are free and follow the official NCERT textbook.

These Class 9 Maths Ganita Manjari Chapter 1 solutions cover Orienting Yourself: The Use of Coordinates from the new NCF-2023 textbook (2026–27). The solutions follow Exercise Sets 1.1 and 1.2, Think and Reflect activities and the 16 end-of-chapter questions. Source checked: NCERT Ganita Manjari, Grade 9, Part I, first edition April 2026, Chapter 1, pages 1–15. Open-ended layouts are labelled as samples.

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

On this page

Chapter overview
Key concepts & definitions
Important formulas
Exercise Set 1.1
Exercise Set 1.2
1.4 Distance Between Two Points
End-of-Chapter Exercises (Q1–Q16)
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

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. Use the labelled plots alongside the calculations, then reproduce the constructions on graph paper.

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) and y (the ordinate) are signed coordinates. The perpendicular distances from the y-axis and x-axis are |x| and |y| respectively. Distances are never negative; the signs show direction.

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 $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ :
$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Distance from the origin to (x, y): $d=\sqrt{x^{2}+y^{2}}$

Horizontal segment (same y): distance = $|x_{2}-x_{1}|$ • Vertical segment (same x): distance = $|y_{2}-y_{1}|$

Midpoint of $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ : $M=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

In-text observations — pages 3–7

A floor plan records horizontal position, not height. A window’s wall location can be indicated by a plan symbol, but its height above the floor cannot be read without an elevation or a third coordinate.

On the axes: B(4.5, 0), E(−2.9, 0), G(0, −4.5), H(0, 4). For the quadrant activity, keep Q(−5, 3) and S(3, −5); sample choices are P(2, 2) in Quadrant I and R(−3, −2) in Quadrant III. Axes themselves belong to no quadrant.

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 $D_{1}R_{1}$ 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?

SOLUTION The door

D_{1}R_{1}

lies on the floor along the x-axis, with

D_{1}\left(8,0\right)

and

R_{1}\left(11.5,0\right)

. 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

D_{1}=8

units (ft).

(ii) What are the coordinates of $D_{1}$ ?

SOLUTIONFrom Fig. 1.3,

D_{1}=\left(8,0\right)

(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?

SOLUTION Width of door =

\begin{aligned}D_{1}R_{1}&=|11.5-8|\\&=3.5\end{aligned}

units (ft). Sample interpretation: a 3.5 ft (106.68 cm) opening appears reasonably spacious in this plan. Actual wheelchair access cannot be guaranteed from the plan alone: check the clear opening after the frame and door leaf, the wheelchair width, thresholds and turning space.

(iv) If $B_{1}\left(0,1.5\right)$ and $B_{2}\left(0,4\right)$ 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\,\mathrm{ft}

. Since

2.5\,\mathrm{ft}<3.5\,\mathrm{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. Observation activity: measure several actual clear door openings at home and school and record their widths in centimetres. There is no single answer for every building. For school accessibility, compare the measured clear opening with the wheelchair and check thresholds and space to approach and turn. Write your findings; do not assume that every door is suitable.

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\ne y

, then

\left(x,y\right)\ne \left(y,x\right)

; and

\left(x,y\right)=\left(y,x\right)

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 $1\,\mathrm{cm}=1$ unit.) Using Fig. 1.5, answer the given questions. In this room plan, one coordinate unit represents one foot. Suggested furniture layouts below are sample solutions; other placements with the correct dimensions and adequate clearance are possible.

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) The fourth foot is (8, 7), completing the rectangle with horizontal and vertical sides.

(ii) Sample assessment: the footprint $8\le x\le 11$ , $7\le y\le 9$ lies inside the bedroom and does not overlap the bed or wardrobe. It does not obstruct the room doorway at $y=0$ . This is a possible location, but chair space, lighting and window height must be checked in the actual room; the floor plan does not show them.

(iii) The horizontal side is $11-8=3\,\mathrm{ft}$ and the vertical side is $9-7=2\,\mathrm{ft}$ . Conventionally the longer side is the length (3 ft), and the shorter side the width (2 ft). Height cannot be obtained from a floor plan.

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?

SOLUTION

The hinge is $B_{1}\left(0,1.5\right)$ and the door width is $4-1.5=2.5\,\mathrm{ft}$ . When it swings into the bedroom, every point on the door stays within 2.5 ft of the hinge. The wardrobe occupies $3\le x\le 7$ and $0\le y\le 2$ . Its nearest point to the hinge is (3, 1.5), at distance 3 ft. Since $2.5<3$ , the door does not hit the wardrobe.

For the idealised 90° swing, a 3 ft door would just touch the wardrobe at (3, 1.5); a wider door can hit it. Sample suggestions: move the wardrobe farther away, use a sliding door, or redesign the opening direction after checking the bathroom fixtures and access.

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

SOLUTION

(i) The bathroom corners are O(0, 0), F(0, 9), R(−6, 9), P(−6, 0). It is 6 ft wide and 9 ft long.

(ii) S(−6, 6), H(−3, 6), W(−2, 9), R(−6, 9). SH and WR are horizontal and parallel, but SR is vertical and HW is sloping. Thus SHWR is a trapezium, not a rectangle. Its parallel sides have lengths 3 ft and 4 ft.

(iii) One sample layout: washbasin corners (−6, 0), (−3, 0), (−3, 2), (−6, 2); this gives $3\,\mathrm{ft}\times 2\,\mathrm{ft}$ . Toilet corners (−6, 2.5), (−4, 2.5), (−4, 5.5), (−6, 5.5); this gives $2\,\mathrm{ft}\times 3\,\mathrm{ft}$ . These footprints are within the bathroom, do not overlap each other or the shower, and leave the doorway at $x=0$ clear. The rectangles represent allocated spaces, not detailed fixture drawings.

Coordinate reconstruction of NCERT Fig. 1.5, with the proposed study table in green. Each grid interval represents 1 ft; equal scales on both axes. Letters match the textbook. The shower SHWR is a trapezium.

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

SOLUTION

(i) P(−6, 0) and A(12, 0) are $12-\left(-6\right)=18\,\mathrm{ft}$ apart. The dining room lies below PA; its four corners are (−6, 0), (12, 0), (12, −15), (−6, −15). The perpendicular width is 15 ft.

(ii) The room centre is $\left(\frac{-6+12}{2},\frac{0-15}{2}\right)=\left(3,-7.5\right)\text{.}$ Taking the table’s 5 ft side parallel to the x-axis and its 3 ft side parallel to the y-axis gives half-length 2.5 and half-width 1.5. Its feet are (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9). The left and right gaps are both 6.5 ft; the top and bottom gaps are both 6 ft. A 90° rotation is another valid centred placement.

Exercise 1.2 Q4: dining room below the x-axis and a centred

5\,\mathrm{ft}\times 3\,\mathrm{ft}

table. Equal coordinate scales; dimensions are in feet.

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 $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\text{.}$

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

\begin{aligned}\mathrm{AD}&=\sqrt{4^{2}+3^{2}}\\&=\sqrt{25}\\&=5\end{aligned}

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.

Worked triangle ADM and its reflection

For A(3, 4), D(7, 1), M(9, 6): $\begin{aligned}\mathrm{AD}&=\sqrt{4^{2}+\left(-3\right)^{2}}\\&=5\text{;}\end{aligned}$ $\mathrm{DM}=\sqrt{2^{2}+5^{2}}=\sqrt{29}$ ; $\begin{aligned}\mathrm{MA}&=\sqrt{\left(-6\right)^{2}+\left(-2\right)^{2}}\\&=\sqrt{40}\\&=2\sqrt{10}\end{aligned}$ units.

Reflection in the y-axis gives A′(−3, 4), D′(−7, 1), M′(−9, 6). Reflection in the x-axis gives (3, −4), (7, −1), (9, −6). Side lengths and angles are preserved; orientation is reversed.

End-of-Chapter Exercises — Questions 1–16

Stars mark the textbook’s challenge questions.

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=\left(-5,y\right)

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.

Q3: AM is horizontal and MP is vertical; M and P reflect across 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.

\mathrm{IZ}=|0-\left(-6\right)|=6

units;

\mathrm{NI}=|5-0|=5

units.

\begin{aligned}\mathrm{NZ}&=\sqrt{5^{2}+6^{2}}\\&=\sqrt{61}\\&\approx 7.81\end{aligned}

units. So the three sides are 5, 6 and $\sqrt{61}$ units.

Q4: one valid construction, with right angle at I. The legs are 5 and 6 units; hypotenuse

\sqrt{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\ge 0

and

y\ge 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.

\begin{aligned}\mathrm{MA}&=\sqrt{\left(0+3\right)^{2}+\left(0+4\right)^{2}}\\&=\sqrt{25}\\&=5\text{.}\end{aligned}

\begin{aligned}\mathrm{AG}&=\sqrt{6^{2}+8^{2}}\\&=\sqrt{100}\\&=10\text{.}\end{aligned}

\begin{aligned}\mathrm{MG}&=\sqrt{9^{2}+12^{2}}\\&=\sqrt{225}\\&=15\text{.}\end{aligned}

Since

\mathrm{MA}+\mathrm{AG}=5+10=15=\mathrm{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

\begin{aligned}\mathrm{RB}&=\sqrt{3^{2}+\left(-4\right)^{2}}\\&=\sqrt{25}\\&=5\text{.}\end{aligned}

\begin{aligned}\mathrm{BC}&=\sqrt{6^{2}+\left(-7\right)^{2}}\\&=\sqrt{85}\\&\approx 9.22\text{.}\end{aligned}

\begin{aligned}\mathrm{RC}&=\sqrt{9^{2}+\left(-11\right)^{2}}\\&=\sqrt{202}\\&\approx 14.21\text{.}\end{aligned}

Exact check: if

\mathrm{RB}+\mathrm{BC}=\mathrm{RC}

, then

5+\sqrt{85}=\sqrt{202}

. Squaring gives

110+10\sqrt{85}=202

, so

\sqrt{85}=\frac{46}{5}

. Squaring again would give

2125=2116

, a contradiction. Therefore the points are not collinear. Rounded distances are not a reliable proof when values are this close. On the line through R and B,

x=4

would give

y=-13

; C has

y=-12

, one unit above that line.

Q6–7: M, A, G are collinear; R, B, C are not. The second set is nearly straight, so the exact calculation is essential. Coordinates are given in the solutions.

*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

\mathrm{OA}=\mathrm{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:

\begin{aligned}\mathrm{OP}&=\mathrm{OQ}\\&=\sqrt{3^{2}+4^{2}}\\&=5\text{,}\end{aligned}

so the triangle is isosceles. (Other correct answers are possible.)

Q8: sample constructions. The blue triangle has two perpendicular sides of length 4; the orange triangle has two sides of length 5.

*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

S	M	T	Midpoint of S, T	Is 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=\left(\frac{x_{S}+x_{T}}{2},\frac{y_{S}+y_{T}}{2}\right)\text{.}

*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

\begin{aligned}-7&=\frac{3+x}{2}\\\Rightarrow 3+x&=-14\\\Rightarrow x&=-17\text{.}\end{aligned}

\begin{aligned}1&=\frac{-4+y}{2}\\\Rightarrow -4+y&=2\\\Rightarrow y&=6\text{.}\end{aligned}

\therefore B=\left(-17,6\right)

*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 Using midpoints as requested: P is the midpoint of AQ and Q is the midpoint of PB. For either coordinate write

2p=a+q

and

2q=p+b

. Substitute

q=2p-a

to obtain

3p=2a+b

; hence

p=\frac{2a+b}{3}

and

q=\frac{a+2b}{3}

. Thus P divides AB in the ratio 1 : 2, while Q divides it in the ratio 2 : 1.

\begin{aligned}P&=\left(4+\frac{16-4}{3},7+\frac{-2-7}{3}\right)\\&=\left(8,4\right)\text{.}\end{aligned}

\begin{aligned}Q&=\left(4+\frac{2\left(16-4\right)}{3},7+\frac{2\left(-2-7\right)}{3}\right)\\&=\left(12,1\right)\text{.}\end{aligned}

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)

\mathrm{OA}=\sqrt{1^{2}+8^{2}}=\sqrt{65}

;

\mathrm{OB}=\sqrt{4^{2}+7^{2}}=\sqrt{65}

;

\mathrm{OC}=\sqrt{7^{2}+4^{2}}=\sqrt{65}

. All equal, so A, B, C lie on a circle centred at O with radius $\sqrt{65}\approx 8.06$ units. (ii)

\begin{aligned}\mathrm{OD}&=\sqrt{5^{2}+6^{2}}\\&=\sqrt{61}\\&\approx 7.81<\sqrt{65}\text{,}\end{aligned}

so D is inside K.

\mathrm{OE}=\sqrt{0+9^{2}}=9>\sqrt{65}

, so E is outside K.

Q12: circle K has radius

\sqrt{65}

. A, B and C lie on it; D lies inside and E outside. Equal scales on both axes.

*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 Labelling ambiguity: the question does not specify which side corresponds to each midpoint. Assume D, E, F are midpoints of BC, CA, AB respectively. Then

2D=B+C

2E=C+A

and

2F=A+B

(coordinate-wise). Adding the two adjacent equations and subtracting the opposite one gives each vertex. Different side assignments permute the same three vertices.

\begin{aligned}A&=F+E-D\\&=\left(0+6-5,3+5-1\right)\\&=\left(1,7\right)\text{.}\end{aligned}

\begin{aligned}B&=F+D-E\\&=\left(0+5-6,3+1-5\right)\\&=\left(-1,-1\right)\text{.}\end{aligned}

\begin{aligned}C&=D+E-F\\&=\left(5+6-0,1+5-3\right)\\&=\left(11,3\right)\text{.}\end{aligned}

Check: midpoint of

\mathrm{BC}=\left(5,1\right)=D

, of

\mathrm{CA}=\left(6,5\right)=E

, of

\mathrm{AB}=\left(0,3\right)=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). (i) Draw a model using $1\,\mathrm{cm}=200m$ and single lines for roads. (ii) Using the stated intersection convention, find: (a) (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 (i) Drawing: mark the main N–S and E–W roads as perpendicular axes. Draw the parallel streets at 1 cm intervals, since 1 cm represents 200 m. Label the street families consistently, state your numbering convention and mark north. The wording does not specify how the ten streets are distributed on opposite sides of a main road; use a clearly stated arrangement, rather than claiming that one distribution is compulsory.(ii) 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]\times [0,600]

, so no part of either circle is outside the screen. (ii)

\begin{aligned}\mathrm{AB}&=\sqrt{\left(250-100\right)^{2}+\left(230-150\right)^{2}}\\&=\sqrt{150^{2}+80^{2}}\\&=\sqrt{28900}\\&=170\text{.}\end{aligned}

Sum of radii =

80+100=180

. Also

|100-80|=20

. Since $20<170<180$ , the circumferences intersect at two points. Both the difference and sum of radii must be checked: one circle entirely inside another need not intersect it.

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

\begin{aligned}\mathrm{AB}&=\mathrm{BC}\\&=\mathrm{CD}\\&=\mathrm{DA}\\&=\sqrt{3^{2}+1^{2}}\\&=\sqrt{10}\text{,}\end{aligned}

so all four sides are equal. Diagonals

\begin{aligned}\mathrm{AC}&=\mathrm{BD}\\&=\sqrt{4^{2}+2^{2}}\\&=\sqrt{20}\end{aligned}

are also equal. Equal sides and equal diagonals ⇒ ABCD is a square. Area = side² =

\left(\sqrt{10}\right)^{2}=10

square units.

Q16: equal scales on both axes show the square accurately. Each side is

\sqrt{10}

units and its area is 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

Additional ClearStudy practice; these questions are not part of the NCERT exercise sets.

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) $\sqrt{x^{2}+y^{2}}$

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

(a) 10 (b) 14 (c) $\sqrt{14}$ (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 $\sqrt{x^{2}+y^{2}}$ .

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

Reason: $\left(x,y\right)=\left(y,x\right)$ 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): the distances from the y-axis and x-axis are |x| and |y| respectively. Coordinates carry signs; distances do not.
Quadrant signs: Q I (+, +), Q II (−, +), Q III (−, −), Q IV (+, −).
x-axis points are (x, 0); y-axis points are (0, y).
$\left(x,y\right)=\left(y,x\right)$ only when $x=y$ .
Distance = $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ (Baudhāyana–Pythagoras theorem).
Midpoint = averages of the coordinates; points are collinear if the part-distances add up (or slopes are equal).

How to present clear mathematical working

Always write coordinates as (x, y), show each step of the distance formula, and keep answers in surd form (e.g. $\sqrt{65}$ ) 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 $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\text{.}$

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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