Class 9 Maths · Chapter 1
Short answer:
Class 9 Maths Chapter 1 "Orienting Yourself: The Use of Coordinates" naye Ganita Manjari textbook (session 2026-27) ka opening chapter hai. Ye chapter Cartesian coordinate system — x-axis, y-axis, origin, quadrants, abscissa, ordinate — sikhata hai, aur phir Baudhāyana-Pythagoras Theorem se Distance Formula derive karta hai, jisse midpoint aur collinearity jaise real problems solve hote hain. Neeche is chapter ke saare in-text aur exercise questions ke step-by-step "ganita manjari class 9 maths ncert solutions" mil jayenge.
Note: agar aap purani NCERT "Mathematics" book dhoondh rahe the, wo session 2026-27 se retire ho chuki hai — "class 9 maths chapter 1 orienting yourself use of coordinates solutions" ab isi naye Ganita Manjari Part 1 se milenge.
Chapter 1 poore Class 9 coordinate geometry ki neev hai. Ganita Manjari Part 1 ka pehla chapter — "Orienting Yourself: The Use of Coordinates" — sirf formula ratt-a nahi karwata, balki sikhata hai ki kisi bhi point ki exact position ko numbers se kaise describe karte hain. Chapter ek story se shuru hota hai (Reiaan ke room ka layout — furniture, door, window kahan hain), jisse students naturally samajh jaate hain ki position batane ke liye do reference lines (axes) chahiye hoti hain.
Uske baad chapter formally Cartesian coordinate system introduce karta hai: x-axis, y-axis, origin, char quadrants, aur har point ka abscissa (x-coordinate) + ordinate (y-coordinate). Ek historical angle bhi hai — kaise Sindhu-Sarasvati Civilisation ke grid-based city planning se lekar Baudhāyana, Āryabhaṭa aur Brahmagupta tak coordinate-jaisi soch Bharat me evolve hui. Chapter ka sabse important formal result hai Distance Formula, jo Baudhāyana-Pythagoras Theorem se derive hota hai — isi se aage midpoint, collinearity (teen points ek line pe hain ya nahi), triangle-type check, aur city-grid design jaisi contextual problems solve hoti hain.
Ye chapter isliye critical hai kyunki iske concepts — plotting points, distance formula, midpoint — aage geometry ke har chapter me (jaise Chapter 5 "I'm Up and Down, and Round and Round" — Circles, aur Chapter 6 "Measuring Space" — Perimeter and Area) baar-baar use hote hain. Isliye "class 9 maths chapter 1 orienting yourself use of coordinates solutions" sirf is chapter ke marks ke liye nahi, poore session ki foundation ke liye zaroori hai.
Chapter 1 Summary — 5 Minute Revision
Is chapter ka core summary:
- Position describe karna: Kisi plane pe point ki position batane ke liye do perpendicular reference lines (x-axis aur y-axis) chahiye — inka intersection point origin O(0, 0) kehlata hai.
- Coordinates: Har point P ko ordered pair (x, y) se likhte hain, jahan x = abscissa (x-axis se distance) aur y = ordinate (y-axis se distance).
- Char quadrants: Plane 4 quadrants (I, II, III, IV) me divide hota hai, har quadrant me x aur y ke signs (+, −) ka apna fixed pattern hota hai.
- Distance Formula: Do points A(x₁, y₁) aur B(x₂, y₂) ke beech distance = Baudhāyana-Pythagoras Theorem use karke derive hoti hai — ek right triangle banakar jiske legs (x₂ − x₁) aur (y₂ − y₁) hote hain.
- Applications: Distance formula se midpoint check, collinearity (teen points ek seedhi line pe hain ya nahi), triangle ka type (equilateral/isosceles/right-angled) identify karna, aur city-grid / room-layout jaise real-life design problems solve hote hain.
- Historical thread: Sindhu-Sarasvati grid cities → Baudhāyana Śulba-sūtra → Āryabhaṭa/Brahmagupta tak coordinate-jaisi soch ka evolution — ye NCF-2023 ka signature "story + history + formal math" approach hai.
In-Text Questions — Solutions
Reiaan apne room ke ek corner ko origin maan kar apni study table ki position 'dayi taraf 3 units, upar 2 units' bata raha hai. Isko coordinate form me likho.
Origin se dayi taraf (positive x-direction) 3 units aur upar (positive y-direction) 2 units — dono positive directions hain, isliye ye point Quadrant I me hoga.
Coordinates = (3, 2)
Point P ka x-coordinate (abscissa) 0 hai aur y-coordinate (ordinate) −5 hai. P kahan sthit hoga?
Jab x-coordinate = 0 ho, point hamesha y-axis par hota hai (kisi quadrant me nahi).
P = (0, −5) → y-axis par, origin se 5 units neeche
Point Q(−4, 0) kis axis par hoga?
Jab y-coordinate = 0 ho, point hamesha x-axis par hota hai.
Q(−4, 0) → x-axis par, origin se 4 units baayi taraf
Point (−2, 5) kis quadrant me hoga? Sign-pattern se justify karo.
x = −2 (negative), y = 5 (positive).
Quadrant sign-pattern: Quadrant II me x negative aur y positive hota hai.
(−2, 5) → Quadrant II
Do points A(3, 4) aur B(3, −4) diye hain. Ye dono kis axis ke respect me symmetric (mirror image) hain?
A aur B ka x-coordinate same (3) hai, sirf y-coordinate ka sign badla hai (+4 se −4).
Jab sirf y-coordinate ka sign flip ho, points x-axis ke respect me mirror image hote hain.
A(3, 4) aur B(3, −4) → x-axis ke sapeksh symmetric
Ek right triangle banao jiske legs coordinate axes ke parallel hon, points A(2, 2) aur B(6, 5) diye hain — third vertex C ka coordinate batao jisse angle C = 90° ho aur AC, x-axis ke parallel ho.
AC ko x-axis ke parallel banane ke liye C ka y-coordinate A jaisa (= 2) hona chahiye, aur BC ko y-axis ke parallel banane ke liye C ka x-coordinate B jaisa (= 6) hona chahiye.
C = (6, 2)
Check: AC = horizontal segment (length 4 units), BC = vertical segment (length 3 units) — right angle C par bin diagonal measure kiye hi confirm ho jata hai.
Think and Reflect: Kya do alag-alag points ke coordinates same ho sakte hain?
Nahi. Cartesian plane me har point ki ek hi unique (x, y) coordinate pair hoti hai — ye ek one-to-one correspondence hai. Agar do 'points' ke coordinates same hain, to wo asal me ek hi point hain.
Ek naksha (map) me school origin par hai. Bus-stop (5, −3) par hai. Bus-stop school se kis direction me hai — sirf sign dekh kar batao (calculation ke bina).
x = 5 (positive) → school se dayi taraf (East)
y = −3 (negative) → school se neeche (South)
Isliye bus-stop school se South-East direction me hai (exact distance ke liye Distance Formula chahiye hogi, jo aage section me aata hai).
Exercise Questions — Solutions (Q1–Q12)
Exercise 1.1, Q1: Plot points A(4, 2), B(−3, 2), C(−3, −2), D(4, −2) ek graph par. In char points ko join karne par kaunsa figure banta hai?
Points plot karne par:
- A(4, 2) — Quadrant I
- B(−3, 2) — Quadrant II
- C(−3, −2) — Quadrant III
- D(4, −2) — Quadrant IV
AB aur CD dono horizontal lines hain (length = 4 − (−3) = 7 units), BC aur AD dono vertical lines hain (length = 2 − (−2) = 4 units). Opposite sides equal aur parallel hain, saare angles 90° hain.
Figure = Rectangle (length 7 units × breadth 4 units)
Exercise 1.1, Q2: Bina plot kiye batao ki points (−7, 0), (0, 9), (3, 3), (0, −6) me se kaun-kaunse points kisi axis par sthit hain?
Rule: y = 0 → x-axis par; x = 0 → y-axis par.
- (−7, 0): y = 0 → x-axis par
- (0, 9): x = 0 → y-axis par
- (3, 3): dono non-zero → kisi axis par nahi, Quadrant I me
- (0, −6): x = 0 → y-axis par
Exercise 1.1, Q3: In-text signs ke aadhar par batao ki (−8, −5), (6, −2), (−1, 7), (4, 9) kaunse quadrant me honge.
Sign-pattern table use karke:
- (−8, −5): (−, −) → Quadrant III
- (6, −2): (+, −) → Quadrant IV
- (−1, 7): (−, +) → Quadrant II
- (4, 9): (+, +) → Quadrant I
Exercise 1.2, Q1: Baudhāyana-Pythagoras Theorem use karke Distance Formula derive karo, do points A(x₁, y₁) aur B(x₂, y₂) ke liye.
A(x₁, y₁) aur B(x₂, y₂) se ek third point C(x₂, y₁) lo, jisse ABC ek right triangle ban jaaye (angle C = 90°).
AC = |x₂ − x₁| (horizontal leg), BC = |y₂ − y₁| (vertical leg)
Baudhāyana-Pythagoras Theorem (hypotenuse² = base² + height²) lagane par:
AB² = AC² + BC² = (x₂ − x₁)² + (y₂ − y₁)²
∴ AB = √[(x₂ − x₁)² + (y₂ − y₁)²]
Exercise 1.2, Q2: Points A(2, 3) aur B(5, 7) ke beech distance nikalo.
Distance formula lagane par:
AB = √[(5 − 2)² + (7 − 3)²] = √[(3)² + (4)²] = √(9 + 16) = √25
AB = 5 units
Exercise 1.2, Q3: Point P(6, 8) ki origin se distance nikalo.
Origin O(0, 0) se distance formula:
OP = √[(6 − 0)² + (8 − 0)²] = √(36 + 64) = √100
OP = 10 units
Exercise 1.2, Q4: Check karo ki points A(1, 2), B(3, 6), C(5, 10) collinear hain ya nahi.
Teeno pairs ke beech distance nikalte hain:
AB = √[(3−1)² + (6−2)²] = √(4 + 16) = √20
BC = √[(5−3)² + (10−6)²] = √(4 + 16) = √20
AC = √[(5−1)² + (10−2)²] = √(16 + 64) = √80 = 2√20
Check: AB + BC = √20 + √20 = 2√20 = AC
AB + BC = AC, isliye A, B, C ek hi seedhi line par hain — points collinear hain.
Exercise 1.2, Q5: Points A(0, 0), B(4, 0), C(2, 2√3) se bana triangle kaunsa type ka hai — pata karo distance formula se.
Teeno sides nikalte hain:
AB = √[(4−0)² + (0−0)²] = √16 = 4
BC = √[(2−4)² + (2√3−0)²] = √(4 + 12) = √16 = 4
AC = √[(2−0)² + (2√3−0)²] = √(4 + 12) = √16 = 4
Teeno sides equal (AB = BC = AC = 4 units) hain.
Triangle = Equilateral Triangle
Exercise 1.2, Q6: Midpoint formula use karke A(2, 3) aur B(6, 7) ka midpoint M nikalo, aur verify karo ki AM = MB.
Midpoint formula:
M = ((x₁+x₂)/2, (y₁+y₂)/2) = ((2+6)/2, (3+7)/2) = (4, 5)
Verification (distance formula se):
AM = √[(4−2)² + (5−3)²] = √(4+4) = √8
MB = √[(6−4)² + (7−5)²] = √(4+4) = √8
AM = MB = √8 units, isliye M(4, 5) sach me midpoint hai.
★ Starred Question: Ek city-planner ek naye grid-based colony design kar raha hai jisme har plot ek coordinate point se represent hota hai. Agar Community Hall H(0, 0) par hai aur do proposed School locations S₁(6, 8) aur S₂(9, 0) hain, to Community Hall se kaunsa school zyada paas hai?
Distance formula se dono distances nikalte hain:
HS₁ = √[(6−0)² + (8−0)²] = √(36+64) = √100 = 10 units
HS₂ = √[(9−0)² + (0−0)²] = √81 = 9 units
Kyunki 9 < 10, isliye S₂ zyada paas hai Community Hall se — city-grid design me ye distance formula direct real-world decision (kaunsi location choose karein) me help karta hai.
★ Starred Question: Circle ke center O(0, 0) se ek point P(3, 4) diya hai, jo circle par sthit hai. Circle ka radius aur equation-jaisa form batao (is chapter ke level tak).
Radius = center se point ki distance:
r = OP = √[(3−0)² + (4−0)²] = √(9+16) = √25 = 5 units
Is chapter ke level par: circle par sthit koi bhi point (x, y) ke liye distance formula se OP hamesha 5 units barabar rahegi. (Poora circle-equation formal treatment Chapter 5 "I'm Up and Down, and Round and Round" me aata hai.)
End of Chapter Exercise: Points A(−2, −1), B(1, 0), C(4, 3), D(1, 2) diye hain. Distance formula se check karo ki ABCD ek parallelogram hai.
Char sides nikalte hain:
AB = √[(1−(−2))² + (0−(−1))²] = √(9+1) = √10
BC = √[(4−1)² + (3−0)²] = √(9+9) = √18
CD = √[(1−4)² + (2−3)²] = √(9+1) = √10
DA = √[(−2−1)² + (−1−2)²] = √(9+9) = √18
AB = CD = √10 aur BC = DA = √18 — opposite sides equal hain.
ABCD = Parallelogram (opposite sides equal)
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Coordinates of a point | P(x, y), jahan x = abscissa, y = ordinate |
| Origin | O(0, 0) — dono axes ka intersection point |
| Point on x-axis | y-coordinate = 0, form: (x, 0) |
| Point on y-axis | x-coordinate = 0, form: (0, y) |
| Quadrant I sign | (x, y) = (+, +) |
| Quadrant II sign | (x, y) = (−, +) |
| Quadrant III sign | (x, y) = (−, −) |
| Quadrant IV sign | (x, y) = (+, −) |
| Baudhāyana-Pythagoras Theorem | hypotenuse² = base² + height² |
| Distance Formula (two points) | d = √[(x₂ − x₁)² + (y₂ − y₁)²] |
| Distance from origin | d = √(x² + y²) |
| Midpoint of a segment | M = ((x₁+x₂)/2, (y₁+y₂)/2) |
| Collinearity check (3 points A, B, C) | Points collinear ⟺ AB + BC = AC (ya koi ek sum baaki barabar ho) |
| Equilateral triangle check | Saare teen sides distance formula se equal aane chahiye |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Coordinate likhte waqt x aur y ka order ulta kar dena — (x, y) hamesha x pehle, y baad me likhte hain; (3, 5) aur (5, 3) alag-alag points hain.
- Quadrant sign-pattern yaad rakhte waqt confusion — Quadrant II aur IV ka sign pattern mix ho jaata hai. Hamesha table se cross-check karo: II = (−,+), IV = (+,−).
- Distance Formula lagate waqt (x₂ − x₁) aur (y₂ − y₁) ko galat point se subtract karna — sign chahe jo bhi aaye, square karne ke baad negative khatam ho jaata hai, isliye order matter nahi karta, par students confuse ho jaate hain aur beech me hi galat calculation kar dete hain.
- Square root simplify karte waqt galti — √20 ko galat tareeke se simplify karna (sahi: √20 = √(4×5) = 2√5), ya √18 ko 3√2 ki jagah kuch aur likh dena.
- Collinearity check me sirf ek pair ki distance nikal kar conclusion nikal lena — teeno pairs (AB, BC, AC) nikalna zaroori hai, tabhi AB + BC = AC verify ho sakta hai.
- Origin se distance (d = √(x²+y²)) aur do general points ke beech distance formula ko mix kar dena — origin wala case general formula ka hi special case hai (x₁=0, y₁=0), isko alag ratt-a maarne ki zaroorat nahi.
Board-Style Important Questions
- Direct Concept: Point (0, −7) kis axis par sthit hoga?
- Direct Concept: Bina plot kiye batao ki point (−5, 3) kis quadrant me hoga, sign-rule se justify karo.
- Formula Application: Points A(0, 0) aur B(8, 6) ke beech distance nikalo.
- Multi-step: Distance formula use karke check karo ki points (1, 5), (2, 3), (−2, −1) collinear hain ya nahi.
- Multi-step: A(4, 3), B(4, −1) ka midpoint nikalo aur verify karo ki midpoint se dono points equidistant hain.
- Long Answer / HOTS: Points A(0, 0), B(6, 0), C(3, 3√3) se bana triangle equilateral hai — distance formula se poora step-by-step prove karo.
Aksar Poochhe Jaane Wale Sawaal
Ganita Manjari kya hai aur ye purani NCERT Class 9 Maths book se kaise alag hai?
Ganita Manjari, NCF-2023 ke aadhar par bana bilkul naya Class 9 Maths textbook hai, jo session 2026-27 se purani "Mathematics" book ki jagah le chuka hai. Abhi sirf Ganita Manjari Part 1 (8 chapters) published aur taught ho raha hai. Ye story-driven approach use karta hai — jaise Chapter 1 me Reiaan ke room se coordinate system explain hota hai — aur Class 9 Maths ke ncert solutions dhoondhne wale students ko ab yehi naya book follow karna hai.
Class 9 Maths new syllabus 2026-27 me total kitne chapters hain?
Ganita Manjari Part 1 me abhi 8 chapters hain: (1) Orienting Yourself: The Use of Coordinates, (2) Introduction to Linear Polynomials, (3) The World of Numbers, (4) Exploring Algebraic Identities, (5) I'm Up and Down, and Round and Round (Circles), (6) Measuring Space (Perimeter and Area), (7) Probability, (8) Sequences and Progressions. Class 9 maths new syllabus 2026-27 chapter list pdf official NCERT website (ncert.nic.in) par available hai.
Ganita Manjari Part 2 kab aayega?
Abhi (4 Aug 2026 tak) Ganita Manjari Part 2 release nahi hua hai aur NCERT ne koi official release date announce nahi ki hai. Multiple secondary sources ke hisaab se Part 2 me purane-syllabus-jaisi topics (Linear Equations in Two Variables, Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area and Volume, Statistics) expected hain, par ye sirf expected list hai, official confirmation nahi — isliye in par abhi chapters banana start nahi karna chahiye.
Chapter 1 ke coordinate aur distance formula concepts aage kaunse chapters me use hote hain?
Coordinate geometry ke basics is chapter me build hote hain aur aage geometry-heavy chapters me kaam aate hain — jaise Chapter 5 "I'm Up and Down, and Round and Round" (Circles) me center-radius distance concept, aur Chapter 6 "Measuring Space" (Perimeter and Area) me shape ke vertices se side-lengths nikalna. Isliye Chapter 1 ko strong banana zaroori hai.
Class 9 Maths ganita manjari all chapters pdf kahan se download karein?
Official PDFs NCERT ki website (ncert.nic.in) ke textbook section me "GANITA MANJARI Textbook of Mathematics for GRADE 9 Part I" ke naam se available hain — wahi sabse reliable source hai chapter list aur content verify karne ke liye.
Class 9 me koi board exam hota hai, to ye "important questions" kis basis par hain?
Class 9 CBSE board exam wali class nahi hai (board sirf Class 10 aur 12 ke liye hota hai), isliye upar diye important questions kisi specific past board paper se nahi hain — ye typical school-level exam-pattern practice questions hain, chapter ke actual concepts (coordinates, quadrants, distance formula, collinearity) ko cover karne ke liye design kiye gaye hain.
Class 9 Maths — Saare Chapters
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