Orienting Yourself: The Use of Coordinates (Apni Jagah Pehchano: Coordinates ka Use): Class 9 Maths Chapter 1 Hinglish mein

Ganita Manjari Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

75 min
3 simulations
28 practice questions
29 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki coordinate system ek fixed origin aur do reference directions ki madad se kisi bhi position ko kaise naam deta hai, aur yeh idea India mein aur duniya mein kaise aage badha.
Grid par ek scale floor plan banao aur padho, aur samjhao ki 2-D plan height kyun nahi dikha sakta.
Cartesian plane ke axes, origin aur quadrants pehchano, aur har coordinate ke sign se point ki jagah tay karo.
Coordinates se point plot karo aur plot kiye gaye point ke coordinates padho, yeh yaad rakhte hue ki (x, y) ek ordered pair hai.
Kisi axis ke parallel line par do points ke beech ki distance nikalo, aur Baudhāyana–Pythagoras theorem se kinhi bhi do points ke beech ki distance nikalo.
Distances se shapes ko test karo: barabar sides, right angles, collinear points aur circle par pade points.
Dikhao ki kisi axis mein reflection se positions badal jaati hain, lekin saari lengths wahi rehti hain.

NCERT Mathematics Class 9, Chapter 1, pp. 1–15 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

1.1 Introduction

Maan lo tum apni bicycle ek bade, khaali maidan mein kahin chhod aate ho aur ek dost ko phone karke use laane ko kehte ho. "Kahin beech mein hai" bolne se koi fayda nahi. Lekin "Gate se paani ki tanki ki taraf 40 kadam chalo, phir 15 kadam apne left mein" har baar kaam karta hai. Ek tay kiye gaye starting point se, tay ki gayi directions mein gine gaye do numbers kaafi hain, flat zameen par kisi bhi jagah ko pakka batane ke liye.

Yahi hai coordinate system (निर्देशांक पद्धति) ka poora idea: ek fixed starting point aur fixed reference lines, jinki madad se hum har location ko numbers se naam de sakte hain. Jab jagahon ke paas numbers aa jaate hain, to hum unse calculation kar sakte hain, jaise yeh pata karna ki do jagahein ek doosre se kitni door hain.

Hamare subcontinent mein yeh idea bahut purana hai. Sindhu-Sarasvatī Civilisation ke shehar, jaise Mohenjo-daro, plan karke banaye gaye the: chaudi main streets North–South aur East–West chalti thi aur right angles par ek doosre ko kaatti thi. Aise shehar mein koi jagah dhoondhna bas ginti ka kaam hai: itne blocks east, phir itne blocks north (Fig. 1.1).

4/4Stall ka address pair (3, 2) hai: pehle east ke blocks, phir north ke blocks.

Fig. 1.1Ek fixed crossing se blocks ginne par koi bhi location numbers ka ek pair ban jaati hai.

Upar se dekha gaya ek shehar ka map, jiski streets ek square grid banati hain; blocks mein kuch gol ped hain aur ek compass upar ki taraf north dikhata hai. Pehle step mein central crossing par 'centre' likha hai aur thoda upar aur right mein ek star wala stall mark kiya gaya hai. Doosre step mein ek aadmi main street par 3 blocks east chalta hai, jo ek dotted orange arrow se dikhaya gaya hai. Teesre step mein woh mudkar 2 blocks north chalta hai, jo ek dotted blue arrow se dikhaya gaya hai, aur stall tak pahunch jaata hai. Aakhri step mein ek panel stall ka address pair (3, 2) ke roop mein dikhata hai, matlab pehle east aur phir north.

Lagbhag 800 BCE mein, Baudhāyana ke Śulba-sūtra ne yagya ki vediyan (fire altars) banane ke exact rules diye. Banane wale sabse pehle ek rassi ko East–West line mein kheenchte the. Phir uske right angle par ek North–South line fix karte the. Uske baad vedi ka har corner in do lines se naapa jaata tha. Isi text mein woh rule bhi hai jise aaj hum Baudhāyana–Pythagoras theorem kehte hain; yahi theorem hamein Section 1.4 mein distances dega.

Gol Earth par kisi jagah ko mark karne ke liye hamein do angles chahiye, jinhe aaj latitude aur longitude kehte hain. Latitude equator se shuru hota hai, lekin longitude ka koi natural zero nahi hai, isliye kisi ko ek zero chunna padta hai. Indian Siddhānta texts ke astronomers ne Ujjayinī, yaani aaj ke Ujjain, se guzarne wali north–south line chuni. Lagbhag 150 CE mein, Alexandria ke Ptolemy ne ek gazetteer banaya jismein kai hazaar jagahon ke dono angles diye the, aur unmein Ujjain bhi tha. Āryabhaṭa (499 CE) ne apni astronomy half-chord par banayi; yahi idea aage chalkar aaj ka sine bana. Half-chords ki tables ne taaron aur jagahon ke mushkil angle wale sawaalon ko jaldi look-up karne wala kaam bana diya. Aasmaan ko bhi coordinates mile: kisi planet ki jagah ecliptic (Sun ka saal bhar ka raasta) ke saath aur usse door naape gaye angles se batayi jaati thi.

Jo plane tum ab dekhne wale ho, usmein dono axes jahan kaatte hain wahan 0 hai, aur har axis ke aadhe hisse par negative numbers hain. Is design ke liye zaroori hai ki zero aur negative numbers ko asli numbers maana jaaye. 628 CE mein Brahmagupta ne apne Brāhmasphuṭasiddhānta mein bilkul yahi kiya. Unhone positive numbers ko fortunes (dhan) aur negative numbers ko debts (rin) kaha. Unke rules mein “debt mein se zero ghatao to debt hi bachta hai” aur “do debts ka product ek fortune hai” shaamil hain.

Phir yeh gyaan west ki taraf gaya. Lagbhag 770 CE mein Baghdad mein, Brahmagupta ki parampara ka ek Sanskrit astronomy text Arabic mein Sindhind naam se translate hua. Iske saath Ujjain ka meridian bhi pahuncha, jise Arab map-makers ne Arin naam diya aur apne longitude ka zero banaya. Baad mein Al-Bīrūnī ne India mein kai saal bitaye aur Sanskrit seekhi. Unki Book of India (lagbhag 1030 CE) Indian astronomy ko detail mein samjhati hai, aur unki doosri kitaabon mein kai shehron ke latitudes aur longitudes ki list hai. Unhone astrolabe par ek manual bhi likha; astrolabe peetal ka ek instrument hai jisse Sun ya kisi taare ki oonchai naapi jaati hai. Kavi aur mathematician ʿUmar Khayyām (1048–1131) ne do curves banakar aur unke milne ki jagah dhoondhkar cubic equations solve kiye.

Europe mein sabse bada kadam 1630s mein aaya. Toulouse ke ek lawyer Pierre de Fermat, aur Netherlands mein rehne wale philosopher René Descartes, dono ko ek hi baat samajh aayi: kisi curve ko ek equation se describe kiya ja sakta hai jo do numbers ko jodti hai. Descartes ne apna method 1637 mein La Géométrie mein publish kiya. Tab se shapes ke kisi sawaal ko algebra mein badla ja sakta tha, aur kisi equation ko picture ki tarah draw kiya ja sakta tha.

Timeline

Coordinates ka idea kaise bada hua

lagbhag 2600–1900 BCE
Grid mein bane shehar
Mohenjo-daro jaise shehron ki main streets North–South aur East–West chalti hain.
lagbhag 800 BCE
Baudhāyana ka Śulba-sūtra
Vediyan ek East–West line aur ek North–South line se naapi gayin; rectangles ke diagonal ka rule.
lagbhag 150 CE
Ptolemy ki Geography
Hazaaron jagahon ke latitude aur longitude.
499 CE
Āryabhaṭa
Taaron aur shehron ki positions ke liye half-chord (sine) tables.
628 CE
Brahmagupta
Zero aur negative numbers ko rules ke saath numbers maana gaya.
lagbhag 1030 CE
Al-Bīrūnī
Kai shehron ke latitudes aur longitudes; astrolabe par ek manual.
lagbhag 1070 CE
ʿUmar Khayyām
Do curves ko milakar cubic equations solve kiye.
1637
Descartes ki La Géométrie
Points ko do numbers se naam diya gaya; geometry aur algebra mil gaye.

Is chapter mein hum do numbers wala yeh system step by step banayenge. Shuruaat ek room ke floor se karenge, phir ek aise anant flat plane par jayenge jismein negative numbers bhi hote hain. Aakhir mein hum isse kinhi bhi do points ke beech ki seedhi distance naapenge.

1.2 Settling In: Room ko set karna

Nashik ke ek school ke Grade 9 students ko ek khaali room aur ek challenge mila hai. Unki teacher chahti hain ki woh ise ek Ganit Lab bana dein: ek maths room, jismein ek work table ho, puzzles ki ek shelf ho aur ek noticeboard ho. Ek rule sabse zaroori hai. Unka classmate Farhan wheelchair use karta hai, isliye har darwaaza aur har raasta uske liye bhi kaam karna chahiye.

Koi bhi furniture hilane se pehle team room ko squared paper par draw karti hai. Woh scale chunte hain: paper par 1 cm matlab room mein 1 m, to 8 m lambi deewar paper par 8 cm lambi line ban jaati hai. Woh room ka ek corner apna starting point chunte hain aur sab kuch wahan milne wali do deewaron se naapte hain.

Aisi drawing ko floor plan kehte hain: room waisa dikhta hai jaisa chhat par baithi chhipkali ko seedha neeche dekhne par dikhega. Fig. 1.2 dikhata hai ki asli room se plan tak aate-aate kya bachta hai aur kya kho jaata hai.

4/4Plan dikhata hai ki khidki deewar par kahan hai, lekin yeh nahi ki woh kitni oonchi hai.

Fig. 1.2Floor plan length aur breadth rakhta hai, lekin height nahi.NCERT Fig. 1.1 · p. 3

Left mein ek room 3-D mein bana hai, jismein peeche ki deewar, left deewar aur floor hai; peeche ki deewar par ek khidki hai, left deewar par ek darwaaza hai aur floor par ek chhoti table hai. Doosre step mein 'top view' likha ek arrow right mein ek floor plan ki taraf ishaara karta hai: ek rectangle jiske upar wale edge par khidki ka mark hai aur left edge par darwaaze ka gap aur uska ghoomne ka raasta hai. Teesre step mein plan par grid squares aa jaate hain, saath mein note '1 square = 1 m', aur table ek green rectangle ki tarah dikhti hai. Aakhri step mein 3-D room par ek arrow khidki ke sill ki height dikhata hai, aur plan ke bagal mein 'height?' label batata hai ki plan ise record nahi kar sakta.

Dekho, plan hamein kya nahi bata sakta. Yeh khidki ko peeche ki deewar par ek mark ki tarah dikhata hai, isliye hum jaante hain ki woh deewar ke saath kitni door hai. Lekin yeh nahi dikhata ki khidki ka sill kitna ooncha hai, ya table kitni oonchi hai. Floor plan sirf do directions record karta hai: length aur breadth. Height ek teesri direction hai, aur uske liye ek teesra number chahiye.

Compare

Position fix karne ke liye kitne numbers chahiye?

Line par (1-D)

Ek number
Highway par kilometre ka patthar
Ruler par ek mark

Flat surface par (2-D)

Do numbers
Chessboard par ek square
Cinema ki seat: row aur seat number
Floor plan par ek corner

Space mein (3-D)

Teen numbers
Aasmaan mein ek patang
Room mein ceiling fan
Khidki ka sill: deewar ke saath aur floor se upar

Try karke dekho

Apne room ka plan khud banao

Tumhe chahiye

squared paper
ek measuring tape
ek pencil aur ek ruler

Pehle guess karo

Plan par apne bed ka ek corner fix karne ke liye tumhe kitne measurements chahiye?

ekdoteenchaar

Room ka ek corner apna starting point chuno aur use O kaho.
O par milne wali dono deewarein naapo, aur ek scale chuno, jaise 50 cm ke liye 1 cm.
Har furniture ke har corner ki dono deewaron se distance naapo.
Deewarein scale ke hisaab se draw karo, phir furniture ke corners mark karke unhe jodo.
Darwaaze aur khidkiyan deewaron par gap ya mark ki tarah dikhao.

Table 1.1Mere measurements

Corner	Deewar 1 se (m)	Deewar 2 se (m)
Bed, sabse paas wala corner
Table, sabse paas wala corner
Darwaaze ka hinge

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

Ek plan mein 50 cm ke liye 1 cm use hota hai. Ek table 3 cm lambi draw ki gayi hai. Asli table kitne metre lambi hai?

Answer dekho

Answer 1.5 m

Har centimetre 50 cm dikhata hai, to

3 \times 50 = 150

cm, yaani 1.5 m.

1.3 2-D Cartesian Coordinate System

Number line par ek number se ek point fix ho jaata hai: 3, 0 se teen units right mein hai, aur $- 2$ , do units left mein. Number line one-dimensional hoti hai. Flat surface ke points describe karne ke liye hamein ek doosra number chahiye, aur isliye ek doosri number line bhi.

Ab ek doosri number line lo, use seedha khada karo, aur tab tak khiskao jab tak uska 0 pehli line ke 0 par bilkul na baith jaaye (Fig. 1.3). Ab dono lines perpendicular hain aur unka zero ek hi hai. Horizontal line x-axis hai aur khadi line y-axis; dono ko milakar coordinate axes kehte hain. Unka common zero point origin (मूलबिंदु) hai, jise O likhte hain.

4/4Right aur upar positive hain; left aur neeche negative hain.

Fig. 1.3Right angle par rakhi do number lines coordinate axes banati hain.NCERT Fig. 1.2 · p. 4

Pehle step mein minus 5 se 5 tak ki ek horizontal number line hai, beech mein 0, aur us par x-axis likha hai. Doosre step mein minus 3 se 3 tak ki ek vertical number line use zero par kaatti hai, right angle ek chhote square se dikhaya gaya hai; us par y-axis likha hai. Teesre step mein kaatne wala point O se mark hai, label (0, 0) ke saath. Aakhri step mein x-axis ka right aadha hissa aur y-axis ka upar wala aadha hissa green mein highlight hai, labels '+ right' aur '+ upar' ke saath, aur left aur neeche wale hisse orange mein dashed hain, labels 'minus left' aur 'minus neeche' ke saath.

Har axis ek normal number line hai, aur dono ek hi unit length use karti hain. Isliye number line wala jaana-pehchana rule dono par chalta hai. x-axis par O ke right wale points positive hain aur O ke left wale negative. y-axis par upar positive hai aur neeche negative.

Definition

Origin

Woh point jahan x-axis aur y-axis ek doosre ko kaatte hain. Iske coordinates $(0, 0)$ hain.

ExampleGanit Lab ke plan par origin woh corner hai jahan saamne ki deewar left deewar se milti hai.

Axes par pade points

Shuruaat un points se karte hain jo kisi axis par pade hain. Fig. 1.4 mein point K x-axis par hai, O se saadhe teen units right mein. O se wahan pahunchne ke liye x-direction mein 3.5 units chalna padta hai aur y-direction mein bilkul nahi, isliye K ko pair $(3.5, 0)$ se describe karte hain. Hum likhte hain $K = (3.5, 0)$ , jise aksar chhota karke $K (3.5, 0)$ likhte hain. Point L, jo O se dhaai units left mein hai, $L (- 2.5, 0)$ hai.

y-axis ke points ulta kaam karte hain: unka pehla number 0 hota hai. Point M, jo O se 2.5 units upar hai, $M (0, 2.5)$ hai, aur N, jo O se 3 units neeche hai, $N (0, - 3)$ hai.

3/3x-axis ka har point $(x, 0)$ hai; y-axis ka har point $(0, y)$ hai.

Fig. 1.4Axis par do coordinates mein se ek hamesha 0 hota hai.

Unit marks wale coordinate axes. Pehle step mein x-axis par do blue points hain: O ke right mein K (3.5, 0) par aur left mein L (minus 2.5, 0) par. Doosre step mein y-axis par do orange points aate hain: O ke upar M (0, 2.5) par aur neeche N (0, minus 3) par. Aakhri step mein x-axis blue shade mein hai, label 'x-axis par: (x, 0)' ke saath, aur y-axis orange shade mein hai, label 'y-axis par: (0, y)' ke saath.

Wapas Ganit Lab mein

Ab students apne plan par axes lagate hain. Origin O woh corner hai jahan saamne ki deewar left deewar se milti hai. x-axis saamne ki deewar ke saath chalti hai aur y-axis left deewar ke saath (Fig. 1.5). Room 8 m chauda aur 6 m gehra hai, aur ek unit matlab 1 m.

3/3Ek shelf, ek noticeboard aur ek almaari se plan poora hota hai.

Fig. 1.5Ganit Lab, apne corners, darwaazon aur furniture ke saath. Ek unit 1 m hai.NCERT Fig. 1.3 · p. 5

Grid par bana ek rectangular room, jiski saamne ki deewar ke saath x-axis aur left deewar ke saath y-axis hai, 1 se 8 aur 1 se 6 tak marks ke saath. Pehle step mein corners ke labels hain: origin par O, A (8, 0), B (8, 6) aur C (0, 6). Doosre step mein do darwaaze aate hain: main darwaaza DE saamne ki deewar mein 5 aur 6 ke beech ek gap hai, dashed swing ke saath, aur store-room ka darwaaza ST left deewar mein lagbhag 3.5 aur 4.3 ke beech ek chhota gap hai. Aakhri step mein store ke darwaaze ke upar left deewar ke saath ek shelf, peeche ki deewar ke saath ek green noticeboard aur right deewar ke saath ek almaari aa jaati hai.

Ab room ke har corner ka do numbers wala naam hai: $O (0, 0)$ , $A (8, 0)$ , $B (8, 6)$ aur $C (0, 6)$ . Poora room left deewar ke right mein aur saamne ki deewar ke upar hai, isliye iske kisi bhi point ka koi coordinate negative nahi hai.

Example

Lab ka plan padhna

Lab ke plan par noticeboard peeche ki deewar ke saath $G (1.5, 6)$ se $H (4.5, 6)$ tak laga hai. (a) Yeh kitna lamba hai? (b) Iska right end right deewar AB se kitni door hai? (c) Iska midpoint kahan hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 4 questions

Practice 1.1: answers ke saath

Fig. 1.5 wala Ganit Lab ka plan use karo. Ek unit 1 m hai.

1
Short answerEasyNCERT Exercise Set 1.1 (i)–(ii)
Main darwaaza saamne ki deewar ke saath D se E tak hai.
1. D left deewar (y-axis) se kitne metre door hai?
  m
  Answer dekho
  Answer 5 m
  D saamne ki deewar par 5 m wale mark par hai.
2. D saamne ki deewar (x-axis) se kitne metre door hai?
  m
  Answer dekho
  Answer 0 m
  D saamne ki deewar par hi hai, isliye x-axis se uski distance 0 hai.
3. D aur E ke coordinates likho.
  Solution dikhao
  Answer $D (5, 0)$ aur $E (6, 0)$
  Dono points x-axis par hain, isliye unke y-coordinates 0 hain. Unke x-coordinates marks 5 aur 6 hain, to $D (5, 0)$ aur $E (6, 0)$ .
2
Short answerEasyNCERT Exercise Set 1.1 (iii)
Wheelchair ke liye darwaaze mein kam se kam 0.9 m ki khuli chaudai chahiye.
1. Main darwaaza DE kitne metre chauda hai?
  m
  Answer dekho
  Answer 1 m
  $6 - 5 = 1$ m.
2. Main darwaaza Farhan ki wheelchair ke liye kaafi chauda hai.
  SahiGalat
  Answer dekho
  Answer Sahi
  1 m, 0.9 m se zyada hai; 10 cm extra bhi bachte hain.
3
Short answerMediumNCERT Exercise Set 1.1 (iv)
Store-room ka darwaaza $S (0, 3.5)$ se $T (0, 4.3)$ tak hai.
1. Store-room ka darwaaza kitne metre chauda hai?
  m
  Answer dekho
  Answer 0.8 m
  $4.3 - 3.5 = 0.8$ m.
2. Kya yeh main darwaaze se patla hai ya chauda, aur kitna?
  Solution dikhao
  AnswerPatla: 0.2 m kam chauda.
  Main darwaaza 1 m chauda hai aur store-room ka darwaaza 0.8 m, isliye store-room ka darwaaza $1 - 0.8 = 0.2$ m patla hai.
3. Kya Farhan store-room ka darwaaza abhi jaisa hai waise use kar sakta hai? Tum kya suggest karoge?
  Solution dikhao
  AnswerNahi: 0.8 m, 0.9 m se kam hai.
  Yeh darwaaza wheelchair ke liye zaroori 0.9 m se 0.1 m patla hai. Team ise main darwaaze ki tarah lagbhag 1 m chauda kar sakti hai, ya Farhan ko jo cheezein chahiye unhe lab mein hi rakh sakti hai.
Quick answerEasy
Room ke kis corner ke coordinates $(0, 6)$ hain?
OABC
Answer dekho
Answer (D) C
$(0, 6)$ y-axis par hai, O se 6 m door: yaani peeche wala left corner C.

Socho zaraNCERT Think and Reflect · p. 5

Darwaaza kitna chauda hona chahiye?

Ghar ka main darwaaza aur bathroom ka darwaaza naapo, aur school mein ek classroom ka darwaaza bhi. Sabse patla kaunsa hai?
India ki accessibility guidelines kehti hain ki darwaaze kam se kam lagbhag 90 cm chaude hon, taaki wheelchair nikal sake. Tumhare kaunse darwaaze pass hote hain?
Apne school mein ghoom kar dekho. Farhan kahan atak jaayega: seedhiyon par, patle darwaazon par, bhaari darwaazon par, ya kahin aur?

Table 1.2Darwaazon ka survey

Darwaaza	Chaudai (cm)	Kaafi chauda?
Ghar ka main darwaaza
Ghar ke bathroom ka darwaaza
Classroom ka darwaaza
School washroom ka darwaaza

Tumhari entries apne aap save hoti hain.

Quick check

4 mein se 1

Kaunsa point x-axis par hai?

(0, - 2)

(2, 2)

(- 2, - 2)

(- 2, 0)

Answer dekho

Answer (D) $(- 2, 0)$

x-axis par y-coordinate 0 hota hai, isliye jawab

(- 2, 0)

hai.

(0, - 2)

ka x-coordinate 0 hai, isliye woh y-axis par hai.

Quadrants

Zyadatar points to kisi bhi axis par nahi hote. Jis poori flat surface par axes banaye jaate hain, uske teen common naam hain: Cartesian plane, coordinate plane aur xy-plane. Iske do axes khet ki do baadon ki tarah ek doosre ko kaatte hain aur ise chaar hisson mein baant dete hain, jinhe quadrants (चतुर्थांश) kehte hain. Hum inhe I, II, III aur IV number dete hain, top right se shuru karke anticlockwise (Fig. 1.6).

3/3P, Q, R aur S, har ek alag quadrant mein hai.

Fig. 1.6Chaar quadrants, top right se anticlockwise number kiye gaye.NCERT Fig. 1.4 · p. 6

Ek grid jiske beech mein x-axis aur y-axis kaatte hain. Pehle step mein chaaron hisse alag-alag rangon mein shaded hain aur bade Roman numerals se label kiye gaye hain: I top right, II top left, III bottom left aur IV bottom right; origin ke chaaron taraf ek curved arrow dikhata hai ki numbering anticlockwise chalti hai. Doosre step mein har quadrant apna sign pattern dikhata hai: I mein (+, +), II mein (minus, +), III mein (minus, minus) aur IV mein (+, minus). Aakhri step mein chaar points aate hain: Quadrant I mein P (5, 4), Quadrant II mein Q (minus 6, 1), Quadrant III mein R (minus 3, minus 4) aur Quadrant IV mein S (6, minus 3).

Table 1.3Har quadrant mein coordinates ke signs

Quadrant	x-coordinate	y-coordinate	Example
I (upar right)	positive	positive	$(5, 4)$
II (upar left)	negative	positive	$(- 6, 1)$
III (neeche left)	negative	negative	$(- 3, - 4)$
IV (neeche right)	positive	negative	$(6, - 3)$

Definition

Quadrant

Un chaar hisson mein se ek jinmein coordinate axes plane ko baant dete hain. Axes par pade points kisi bhi quadrant mein nahi hote.

Example $(- 3, 0)$ x-axis par hai, isliye yeh kisi quadrant mein nahi hai, bhale hi iska ek number negative hai.

Do numbers ka matlab kya hai

Fig. 1.7 mein point P ko dekho. O se P tak pahunchne ke liye 4 units right jao aur phir 3 units upar. 4, P ki y-axis se distance hai: horizontal dashed line ki length. 3, P ki x-axis se distance hai: vertical dashed line ki length. Signs side batate hain. Agar pehla number negative hai to point y-axis ke left mein hai, aur agar doosra number negative hai to point x-axis ke neeche hai.

Pair $(4, 3)$ ko P ke coordinates (निर्देशांक) kehte hain. Pehla number x-coordinate hai aur doosra y-coordinate. Kyunki order ka matlab hota hai, $(x, y)$ ek ordered pair hai: $(4, 3)$ aur $(3, 4)$ alag-alag points hain.

3/4P x-axis se 3 units door hai, isliye uska y-coordinate 3 hai. To P = (4, 3).

Fig. 1.7x-coordinate y-axis se distance hai; y-coordinate x-axis se distance hai.

First quadrant ka ek grid, x ke saath 1 se 6 aur y ke saath 1 se 5 tak numbered, aur right mein ek panel. Pehle step mein sirf point P mark hai aur panel poochta hai 'P kahan hai?'. Doosre step mein P se y-axis tak ek dashed blue line jaati hai, jis par 4 units likha hai, aur panel x = 4 dikhata hai. Teesre step mein P se neeche x-axis tak ek dashed orange line jaati hai, jis par 3 units likha hai, aur panel mein y = 3 aur P (4, 3) jud jaate hain. Aakhri step mein P ke upar-left mein ek purple point (3, 4) halki guide lines ke saath aata hai, aur panel batata hai ki (3, 4), (4, 3) ke barabar nahi hai.

Definition

Point ke coordinates

Ordered pair $(x, y)$ , jismein $x$ point ki y-axis se signed distance hai aur $y$ uski x-axis se signed distance hai.

Example $P (4, 3)$ y-axis ke 4 units right mein aur x-axis ke 3 units upar hai.

Step by step

Chaar steps mein point plot karna

Step 1 / 4

Origin se shuru karo

Apni pencil O par rakho. $B (- 3, 4)$ plot karne ke liye dono numbers order mein padho.

1 / 4

Example

Signs se quadrant tay hota hai

Har point kahan hai: (a) $(- 2.5, 7)$ (b) $(6, - 0.5)$ (c) $(0, - 4)$ (d) $(- 3, - 3)$ ? (e) Agar $a < 0$ aur $b > 0$ hai, to $(a, b)$ aur $(b, a)$ kaunse quadrants mein hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

SimulationChaaron quadrants explore karo

Ek coordinate plane jo left-right minus 10 se 10 tak aur upar-neeche minus 8 se 8 tak hai, chaaron quadrants shaded ke saath. Chaar labelled points A, B, C aur D alag-alag quadrants mein shuru hote hain; kisi bhi point ko grid ke crossing par drag karo, aur jaise-jaise tum use hilaate ho, uske coordinates aur quadrant update hote jaate hain.

Fig. 1.8Points ko drag karo aur dekho unke coordinates aur quadrants kaise badalte hain.

Try karo

A ko Quadrant II mein drag karo. Uske kaunse coordinate ka sign badla?
B ko bilkul y-axis par rakho. Ab uska x-coordinate kya hai?
Kya tum C ko aisi jagah rakh sakte ho jahan dono coordinates barabar hon? Yeh kaunse quadrants mein mumkin hai?

Lab ki deewaron ke bahar

Is floor par lab akeli nahi hai. Iske left mein, y-axis ke us paar, ek store room aur ek chhota wash area hai. Iske saamne, x-axis ke us paar, ek corridor hai (Fig. 1.9). Wahi origin aur wahi axes rakho, to in jagahon ke har point ka x-coordinate negative hoga, ya y-coordinate negative hoga, ya dono. Ab ek hi floor chaaron quadrants mein phaila hai.

4/4Ab ek hi floor chaaron quadrants mein phaila hai.

Fig. 1.9Ek hi origin ke saath, is floor ke rooms chaaron quadrants mein hain.NCERT Fig. 1.5 · p. 7

Grid par bana ek floor plan, jo neeche minus 3 se 8 tak aur left side par minus 2 se 6 tak numbered hai; dashed orange axes lab ke front-left corner par kaatte hain. Pehle step mein sirf Ganit Lab dikhti hai, y-axis ke right mein aur x-axis ke upar. Doosre step mein y-axis ke left mein y = 2 se 6 tak ek store room aur y = 0 se 2 tak ek wash area aate hain. Teesre step mein x-axis ke neeche 2 units gehra ek corridor aata hai, jo poori chaudai mein chalta hai. Aakhri step mein chaaron quadrants par unke sign patterns likhe hain: lab mein I (+, +), store mein II (minus, +), aur corridor mein III (minus, minus) aur IV (+, minus).

Example

Gayab corner

School ke garden ke ek map par, jahan 1 unit matlab 1 m, ek rectangular flower bed ke teen corners $K (- 3, 2)$ , $L (5, 2)$ aur $M (5, - 4)$ hain. Chautha corner N, bed ki length aur breadth, aur har corner ka quadrant pata karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 4 questions

Practice 1.2: answers ke saath

Fig. 1.9 wala poore floor ka plan use karo, jahan 1 unit matlab 1 m. Squared paper par x-axis $- 4$ se 9 tak aur y-axis $- 3$ se 7 tak banao (har unit ke liye 1 cm) aur plan copy karo.

1
Short answerMediumNCERT Exercise Set 1.2 Q1
Team ek rectangular work table rakhti hai jiske teen paaye $(1, 4)$ , $(3.5, 4)$ aur $(3.5, 5.2)$ par hain.
1. Chautha paaya kahan hona chahiye?
  Solution dikhao
  Answer $(1, 5.2)$
  Pehle do paayon ka $y = 4$ hai aur aakhri do ka $x = 3.5$ . Chautha paaya $(3.5, 5.2)$ ki level par aur $(1, 4)$ ke seedha upar hona chahiye, isliye woh $(1, 5.2)$ par hai.
2. Table kitni lambi aur kitni chaudi hai?
  Solution dikhao
  Answer2.5 m × 1.2 m
  Length: $3.5 - 1 = 2.5$ m. Chaudai: $5.2 - 4 = 1.2$ m.
3. Shelf left deewar ke saath $(0, 4.5)$ se $(0.4, 5.5)$ tak khadi hai. Farhan ko wahan pahunchne ke liye lagbhag 0.9 m ka khula gap chahiye. Kya table sahi jagah par hai?
  Solution dikhao
  AnswerNahi: gap sirf 0.6 m hai.
  Shelf ka saamne wala hissa $x = 0.4$ par hai aur table ka left edge $x = 1$ par, aur $y = 4.5$ se $y = 5.2$ ke beech dono aamne-saamne hain. Gap $1 - 0.4 = 0.6$ m hai, jo 0.3 m kam hai. Table ko kam se kam 0.3 m right mein khiskane se, jaise paaye $(1.3, 4)$ , $(3.8, 4)$ , $(3.8, 5.2)$ aur $(1.3, 5.2)$ par, problem theek ho jaati hai.
4. Kya tum plan se table ki height bata sakte ho?
  Solution dikhao
  AnswerNahi.
  Floor plan two-dimensional hota hai: yeh sirf length aur breadth record karta hai. Height ke liye ek teesra number chahiye hoga.
2
Short answerHardNCERT Exercise Set 1.2 Q2
Store-room ka darwaaza $S (0, 3.5)$ par hinge se laga hai aur lab ke andar khulta hai. Shelf left deewar ke saath $(0, 4.5)$ se $(0.4, 5.5)$ tak khadi hai.
1. Darwaaza 0.8 m chauda hai. Kya khulte waqt yeh shelf se takrayega?
  Solution dikhao
  AnswerNahi.
  Darwaaze ka har point hinge S se 0.8 m ke andar rehta hai. Shelf ka sabse paas wala point, $(0, 4.5)$ , S se $4.5 - 3.5 = 1$ m upar hai, isliye darwaaza shelf se 0.2 m bachakar nikal jaata hai.
2. Farhan ke liye darwaaze ko 1.2 m chauda kar diya jaata hai, hinge ab bhi S par hai. Kya badalna padega?
  Solution dikhao
  AnswerShelf ko kam se kam 0.2 m upar khiskana padega.
  Ab darwaaza $y = 3.5$ se $y = 3.5 + 1.2 = 4.7$ tak hoga, lekin shelf $y = 4.5$ se shuru hoti hai, yaani darwaaze ke andar. Shelf ko kam se kam 0.2 m upar khiskana hoga (asal mein thoda aur, taaki jagah bache), taaki woh $y = 4.7$ ya usse upar se shuru ho.
  Tab ghoomta hua darwaaza bhi shelf se nahi takrata: shelf ka neeche wala front corner $(0.4, 4.7)$ , S se lagbhag 1.26 m door hai, jo darwaaze ki 1.2 m ki pahunch se zyada hai. Aisi distances nikalna tum Section 1.4 mein seekhoge.
3
Short answerMediumNCERT Exercise Set 1.2 Q3
Store room aur wash area ko dekho.
1. Store room ke chaaron corners ke coordinates likho.
  Solution dikhao
  Answer $(- 3, 2)$ , $(0, 2)$ , $(0, 6)$ , $(- 3, 6)$
  Store room $x = - 3$ se $x = 0$ tak aur $y = 2$ se $y = 6$ tak hai. Iske corners $(- 3, 2)$ , $(0, 2)$ , $(0, 6)$ aur $(- 3, 6)$ hain.
2. Wash area ka shape kya hai? Uske corners, length aur breadth batao.
  Solution dikhao
  Answer3 m × 2 m ka ek rectangle.
  Iske corners $(- 3, 0)$ , $(0, 0)$ , $(0, 2)$ aur $(- 3, 2)$ hain. Yeh ek rectangle hai jo $0 - (- 3) = 3$ m lamba aur $2 - 0 = 2$ m chauda hai.
3. Wash area ke ek corner mein water cooler ke liye 1.5 m × 1 m ki jagah mark karo. Uske corners ke coordinates likho.
  Solution dikhao
  AnswerJaise $(- 3, 0)$ , $(- 1.5, 0)$ , $(- 1.5, 1)$ , $(- 3, 1)$ .
  Ek corner $(- 3, 0)$ par rakho. 1.5 m right aur 1 m upar jaane par $(- 3, 0)$ , $(- 1.5, 0)$ , $(- 1.5, 1)$ aur $(- 3, 1)$ milte hain. Wash area ke doosre corners se doosre sahi jawab bhi milte hain, jaise $(- 1.5, 1)$ , $(0, 1)$ , $(0, 2)$ , $(- 1.5, 2)$ .
4
Short answerMediumNCERT Exercise Set 1.2 Q4
Corridor 11 m lamba aur 2 m chauda hai. Yeh wash area aur lab ke saamne chalta hai.
1. Corridor ke chaaron corners ke coordinates likho.
  Solution dikhao
  Answer $(- 3, - 2)$ , $(8, - 2)$ , $(8, 0)$ , $(- 3, 0)$
  Yeh $x = - 3$ se $x = 8$ tak hai, yaani $8 - (- 3) = 11$ m lamba, aur $y = - 2$ se $y = 0$ tak. Isliye corners $(- 3, - 2)$ , $(8, - 2)$ , $(8, 0)$ aur $(- 3, 0)$ hain.
2. Ek 2 m lambi aur 1 m chaudi bench corridor ke bilkul centre mein, uski length ke parallel rakhi jaati hai. Bench ke corners ke coordinates nikalo.
  Solution dikhao
  Answer $(1.5, - 1.5)$ , $(3.5, - 1.5)$ , $(3.5, - 0.5)$ , $(1.5, - 0.5)$
  Left-right mein $- 3$ aur 8 ka beech $2.5$ hai, aur upar-neeche mein $- 2$ aur 0 ka beech $- 1$ hai: yaani centre point $(2.5, - 1)$ hai. Bench x ke saath har taraf 1 m aur y ke saath upar-neeche 0.5 m jaati hai, isliye corners $(1.5, - 1.5)$ , $(3.5, - 1.5)$ , $(3.5, - 0.5)$ aur $(1.5, - 0.5)$ hain.
3. Poori bench kis quadrant mein hai?
  Quadrant IQuadrant IIQuadrant IIIQuadrant IV
  Answer dekho
  Answer (D) Quadrant IV
  Har corner ka x-coordinate positive aur y-coordinate negative hai: $(+, -)$ , yaani Quadrant IV.

Chalo explore karein

Coordinates ki chhupan-chhupai

Tumhe chahiye

squared paper ki do sheets
har player ke liye ek pencil
ek partner

Pehle guess karo

Tumhara partner kehta hai ki uske saare ships aisi jagah hain jahan dono coordinates negative hain. Tumhe kahan dhoondhna chahiye?

Quadrant IQuadrant IIQuadrant IIIQuadrant IV

Har player chupke se dono directions mein $- 5$ se 5 tak axes banata hai.
Whole-number coordinates wale teen points mark karke teen ships chhupao.
Baari-baari se ek point bolo, jaise $(- 2, 3)$ . Agar wahan ship hai to partner "hit" bolta hai, agar koi ship grid line ke saath theek ek unit door hai to "near", aur warna "miss".
Har guess aur jawab apni sheet par likhte jao.
Jo player pehle teenon ships dhoondh le, woh jeetta hai.

SimulationPlot karo!

Ek coordinate plane par challenge, jo left-right minus 8 se 8 tak aur upar-neeche minus 6 se 6 tak hai. Har round mein ek point ke coordinates diye jaate hain; score karne ke liye grid ke us crossing par tap karo jahan woh point hona chahiye, aur galat tap par agli koshish se pehle ek hint dikhta hai.

Fig. 1.10Aath rounds: har point jahan hona chahiye wahan tap karo.

Try karo

Tap karne se pehle bolo ki point kis quadrant mein hona chahiye.
Naye points ke liye phir se khelo aur 8 mein se 8 score karne ki koshish karo.

1.4 2-D Plane mein Do Points ke Beech ki Distance

Farhan jaanna chahta hai ki uski wheelchair ko darwaaze se shelf tak kitna chalna padega. Team jaanna chahti hai ki lab ke ek corner se saamne wale corner tak seedhi jaane wali cable kitni lambi honi chahiye. Dono sawaal do points ke beech ki distance ke baare mein hain. Jab dono points ka ek coordinate same ho, to ek subtraction se jawab mil jaata hai. Jab aisa na ho, tab Baudhāyana–Pythagoras theorem kaam aata hai.

Axis ke parallel line ke saath

Points $A (- 4, 2)$ aur $B (3, 2)$ ka y-coordinate same hai, isliye segment AB x-axis ke parallel hai. A se B tak squares ginne par 7 aata hai (Fig. 1.11). x-coordinates ko subtract karne par bhi wahi jawab milta hai: $3 - (- 4) = 7$ . Hum absolute value lete hain, kyunki distance kabhi negative nahi hoti; ulta subtract karne par bhi $∣ (- 4) - 3 ∣ = ∣ - 7 ∣ = 7$ hi aata hai.

3/3C aur D ka x-coordinate same hai, isliye unke y-coordinates subtract karo: 5.

Fig. 1.11Jo coordinates badalte hain unhe subtract karne se wahi jawab milta hai jo squares ginne se.

Ek grid jiske dono axes numbered hain, aur uske neeche working ke liye ek patti hai. Pehle step mein ek blue segment A (minus 4, 2) aur B (3, 2) ko jodta hai, aur uske upar saat chhoti dashed chhalaangein 1 se 7 tak numbered hain, jo unit steps ginti hain. Doosre step mein grid ke neeche wali patti mein working AB = |3 minus (minus 4)| = 7 aati hai, jo dikhati hai ki subtraction se bhi wahi ginti milti hai. Aakhri step mein ek orange vertical segment C (5, minus 2) aur D (5, 3) ko jodta hai, bagal mein paanch chhalaangein hain, aur patti mein pehle result ke bagal mein working CD = |3 minus (minus 2)| = 5 aati hai.

Same y-coordinate

d = ∣ x_{2} - x_{1} ∣

jahan

$(x_{1}, y), (x_{2}, y)$: x-axis ke parallel line par do points

Same x-coordinate

d = ∣ y_{2} - y_{1} ∣

jahan

$(x, y_{1}), (x, y_{2})$: y-axis ke parallel line par do points

Isi tarah, $C (5, - 2)$ aur $D (5, 3)$ ka x-coordinate 5 same hai, isliye CD y-axis ke parallel hai aur $C D = ∣ 3 - (- 2) ∣ = 5$ . Jab ek point x-axis ke upar ho aur doosra neeche, to distance dono hisson ka jod hoti hai: 2 neeche aur 3 upar.

Quick check

3 mein se 1

(- 6, 3)

aur

(2, 3)

ek doosre se kitni door hain?

units

Answer dekho

Answer 8 units

Dono ka

y = 3

hai, isliye distance

∣ 2 - (- 6) ∣ = 8

hai.

Koi bhi do points

Grid wale raaste ke do hisse (legs) hain jo right angle par hain: 8 units right ki taraf aur 6 units upar. Seedha raasta PQ right-angled triangle PRQ ki teesri side hai, jiska corner $R (7, 2)$ hai (Fig. 1.12).

TheoremBaudhāyana–Pythagoras ka theorem

Right-angled triangle mein hypotenuse par bana square baaki do sides par bane squares ke sum ke barabar hota hai. Agar legs $a$ aur $b$ hain aur hypotenuse $c$ hai, to $a^{2} + b^{2} = c^{2}$ .

Triangle ki chaar copies ko $a + b$ side wale ek square ke andar rakho, har corner mein ek, taaki unke hypotenuses milkar $c$ side ka ek tircha square gherein. Yeh sach mein square hai: iske har corner par triangle ke do acute angles aur corner wala angle milkar ek straight angle banate hain. Dono acute angles ka jod $90^{\circ}$ hai, isliye har corner angle $90^{\circ}$ hai.

Bade square ka area do tarah se gino: $(a + b)^{2} = 4 \times \frac{1}{2} a b + c^{2}$ . To $a^{2} + 2 a b + b^{2} = 2 a b + c^{2}$ , jisse $a^{2} + b^{2} = c^{2}$ milta hai.

Baudhāyana ne ise rectangle ke liye kaha tha: diagonal par bana square, length aur breadth par bane squares ke jod ke barabar hota hai.

Iska converse bhi sach hai: agar kisi triangle ki sides $a^{2} + b^{2} = c^{2}$ satisfy karti hain, to side $c$ ke saamne wala angle right angle hai. Isse hum check karte hain ki koi triangle right-angled hai ya nahi.

4/4 $P Q^{2} = 8^{2} + 6^{2} = 100$ , isliye $P Q = 10$ units.

Fig. 1.12Legs coordinates ke differences hain; hypotenuse distance hai.NCERT Fig. 1.6–1.7 · p. 9

Axes wala ek grid, x-axis minus 2 se 8 tak aur y-axis 0 se 8 tak, aur right mein working ka ek panel. Pehle step mein points P (minus 1, 2) aur Q (7, 8) ek grey dashed line se jude hain jis par question mark hai, aur panel poochta hai PQ = ?. Doosre step mein P se right ki taraf ek blue horizontal leg jaata hai, jis par 8 likha hai, aur panel PR = 8 dikhata hai. Teesre step mein R (7, 2) se Q tak ek orange vertical leg upar jaata hai, jis par 6 likha hai, R par right-angle ka mark hai, aur panel mein RQ = 6 jud jaata hai. Aakhri step mein tirchi side PQ vermillion ho jaati hai aur us par 10 likha hota hai, aur panel PQ square = 64 + 36 aur PQ = 10 dikhata hai.

3/3Hypotenuse AB se distance formula milta hai.

Fig. 1.13Yahi right triangle kinhi bhi do points ke liye kaam karta hai.NCERT Fig. 1.8 · p. 10

Coordinate axes aur do points A aur B; har point se dono axes tak dashed lines girti hain, x-axis par x1, x2 aur y-axis par y1, y2 labels hain. Pehle step mein sirf A (x1, y1) aur B (x2, y2) mark hain, ek grey dashed line se jude hue. Doosre step mein A se ek blue horizontal leg aur B se neeche ek orange vertical leg C (x2, y1) par milte hain, jahan right-angle ka mark hai; legs par x2 minus x1 aur y2 minus y1 likha hai. Aakhri step mein side AB vermillion ho jaati hai aur upar formula aata hai: AB barabar hai (x2 minus x1) square plus (y2 minus y1) square ka square root.

DerivationDistance formula

Maan lo points $A (x_{1}, y_{1})$ aur $B (x_{2}, y_{2})$ hain, jaise Fig. 1.13 mein. A se hokar x-axis ke parallel ek line banao aur B se hokar y-axis ke parallel ek line. Dono $C (x_{2}, y_{1})$ par milti hain, aur C par angle right angle hai.

A aur C ka y-coordinate $y_{1}$ same hai, isliye $A C = ∣ x_{2} - x_{1} ∣$ . C aur B ka x-coordinate $x_{2}$ same hai, isliye $C B = ∣ y_{2} - y_{1} ∣$ .

Baudhāyana–Pythagoras theorem se, $A B^{2} = A C^{2} + C B^{2}$ . Square karne se absolute-value ki bars hat jaati hain, kyunki har number $a$ ke liye $∣ a ∣^{2} = a^{2}$ . Positive square root lene par distance formula (दूरी सूत्र) milta hai:

Distance formula

A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Is baat se fark nahi padta ki tum kis point ko A kehte ho aur kise B, kyunki $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ . Negative coordinates se bhi koi dikkat nahi hoti, kyunki square kabhi negative nahi hota. Khaas taur par, kisi point $P (x, y)$ ki origin se distance $(x_{1}, y_{1}) = (0, 0)$ rakhne se milti hai:

Origin se distance

O P = \sqrt{x^{2} + y^{2}}

Example

Negative coordinates ke saath distance

$A (- 2, 5)$ aur $B (4, - 3)$ ke beech ki distance nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Jab jawab whole number na ho

Ek food-delivery map par 1 unit matlab 1 km. Ek kitchen $K (- 1, - 2)$ par hai aur ek customer $C (3, 5)$ par. Seedhi distance KC ko nearest 0.1 km tak nikalo. Rider ka raasta shayad isse lamba kyun hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Kaisa triangle hai?

Dikhao ki $P (1, 2)$ , $Q (4, 6)$ aur $R (8, 3)$ ek right-angled isosceles triangle ke vertices hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Do points se barabar door

x-axis par woh point nikalo jo $A (2, 3)$ aur $B (6, 5)$ se barabar door hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Distances se yeh bhi pata chalta hai ki teen points collinear hain ya nahi, yaani woh ek hi seedhi line par hain ya nahi. Maan lo B segment AC par hai. Tab A se B aur phir C tak ka safar utna hi lamba hai jitna A se C tak ka seedha safar, isliye $A B + B C = A C$ . Agar B us line se hatkar hai, to safar ghoom kar jaata hai, isliye $A B + B C$ , AC se bada hota hai. Yeh test tum exercises mein use karoge.

Reflection se lengths nahi badalti

Ek graph sheet ko y-axis ke saath fold karo aur zor se dabao. Har point $(x, y)$ , point $(- x, y)$ par aa jaata hai: same height par aur fold se same distance par, lekin doosri taraf. Naya point pehle point ki y-axis mein mirror image, yaani reflection (परावर्तन), hai (Fig. 1.14).

3/3Image ki side lengths bhi wahi hain: 5, 5 aur $\sqrt{50}$ .

Fig. 1.14y-axis mein reflection har x-coordinate ka sign badal deta hai, lekin har length wahi rakhta hai.NCERT Fig. 1.9 · p. 11

x-axis ke upar minus 8 se 8 tak ek grid, beech mein y-axis. Pehle step mein y-axis ke right mein ek blue triangle TUV bana hai, T (1, 1), U (4, 5) aur V (8, 2) ke saath, aur uski sides par 5, 5 aur root 50 likha hai. Doosre step mein y-axis ko mirror ki tarah highlight kiya gaya hai, har vertex se seedhi dashed lines doosri taraf jaati hain, aur (minus 1, 1), (minus 4, 5) aur (minus 8, 2) par ek purple dashed triangle T′U′V′ aata hai. Aakhri step mein image ki sides par bhi 5, 5 aur root 50 likha hai, aur ek note kehta hai 'lengths same hain'.

Distance formula se ek side check karo: $T U = \sqrt{(4 - 1)^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = 5$ aur $T^{'} U^{'} = \sqrt{(- 4 - (- 1))^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = 5$ . x-difference ka sign badla, 3 se $- 3$ , lekin uska square nahi badla. Points ke har pair ke saath yahi hota hai, isliye reflection saari lengths wahi rakhta hai.

To badalta kya hai? Positions badalti hain, aur quadrant bhi: TUV Quadrant I mein hai aur uski image Quadrant II mein. Ghoomne ki direction bhi palat jaati hai. $T \to U \to V$ jaana original par clockwise hai, lekin image par anticlockwise. Shape aur size bilkul wahi rehte hain. x-axis mein reflection bhi isi tarah kaam karta hai: $(x, y)$ , $(x, - y)$ ban jaata hai.

SimulationReflect karo aur naapo

Dono directions mein minus 10 se 10 tak ek coordinate plane, jis par right-angled triangle A (2, 1), B (8, 9), C (8, 1) hai. Iski sides AB = 10, BC = 8 aur AC = 6 distance formula se naapi jaati hain, aur y-axis mein har vertex ki mirror image ek khokhle point ki tarah dikhti hai. Triangle ka shape badalne ke liye vertices ko drag karo, aur mirror ko x-axis ya origin par switch karke dekho ki images kahan aati hain.

Fig. 1.15Triangle ABC ki sides live naapi jaati hain; har drag ke saath mirror images bhi chalti hain.

Try karo

Images ke beech squares gino: kya image ki do chhoti sides ab bhi 6 aur 8 units lambi hain?
Mirror ko x-axis par switch karo. Ab kaunse coordinate ka sign badalta hai, aur images kis quadrant mein aati hain?
B ko kisi nayi jagah drag karo aur phir se gino. Kya image hamesha triangle ki side lengths wahi rakhti hai?

Recap

Summary aur key points

Coordinate system positions ko numbers se fix karta hai, jo ek chune hue origin se tay ki gayi directions mein naape jaate hain. Yeh idea grid wale shehron, vediyon ki geometry aur astronomy se bada hua (Section 1.1).
Floor plan ek 2-D scale drawing hai: yeh length aur breadth rakhta hai, lekin height kho deta hai (Section 1.2).
Horizontal x-axis aur vertical y-axis origin $O (0, 0)$ par right angle mein kaatte hain (Section 1.3).
O ke right aur O ke upar positive hai; O ke left aur O ke neeche negative hai (Fig. 1.3).
x-axis ke points $(x, 0)$ form ke hote hain aur y-axis ke points $(0, y)$ form ke (Axes par pade points).
Axes plane ko chaar quadrants mein baant dete hain, jinke sign patterns I se IV tak $(+, +)$ , $(-, +)$ , $(-, -)$ aur $(+, -)$ hain. Axes par pade points kisi quadrant mein nahi hote (Quadrants).
$(x, y)$ mein $x$ , y-axis se signed distance hai aur $y$ , x-axis se signed distance. $(x, y) = (y, x)$ sirf tab jab $x = y$ (Do numbers ka matlab kya hai).
Axis ke parallel line ke saath distance $∣ x_{2} - x_{1} ∣$ ya $∣ y_{2} - y_{1} ∣$ hoti hai (Axis ke parallel line ke saath distance).
Kinhi bhi do points ke beech ki distance $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ hai, jo Baudhāyana–Pythagoras theorem se aati hai (distance formula).
Distances se shapes test hote hain: barabar sides, right angles (theorem ke converse se) aur collinear points (Koi bhi do points).
y-axis mein reflection $(x, y)$ ko $(- x, y)$ bana deta hai, aur x-axis mein $(x, - y)$ . Reflection har length wahi rakhta hai (Reflection se lengths nahi badalti).

Flash cards · 8

Key terms

1 / 8

Practice · 20 questions

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

Questions 1 se 16 textbook ki end-of-chapter exercises ke hisaab se hain; 17 se 20 jaldi ki extra practice hain. Jab bhi ho sake, ek sketch zaroor banao.

Quick answerEasyNCERT End-of-chapter Q1
Dono axes jis point par kaatte hain use kehte hain. Uska x-coordinate hai aur y-coordinate hai.
Answer dekho
Answer (1) origin, (2) 0, (3) 0
Axes origin O par kaatte hain, jiske coordinates $(0, 0)$ hain.
2
Short answerMediumNCERT End-of-chapter Q2
Point K ka y-coordinate 3 hai. Line $ℓ$ K se hokar jaati hai aur x-axis ke parallel hai.
1. $ℓ$ par kisi general point H ke coordinates likho.
  Solution dikhao
  Answer $H (h, 3)$ , jahan $h$ koi bhi number ho sakta hai.
  x-axis ke parallel line par chalne se sirf x-coordinate badalta hai. Isliye $ℓ$ ka har point x-axis se 3 units upar hai, aur kisi number $h$ ke liye $H = (h, 3)$ .
2. (b)
  H kaunse quadrants mein ho sakta hai?
  Quadrant IQuadrant IIQuadrant IIIQuadrant IV
  Saare sahi options chuno
  Answer dekho
  Answer
  (A) Quadrant I
  (B) Quadrant II
  y-coordinate hamesha $+ 3$ hai. Agar $h > 0$ hai to H Quadrant I mein hai; agar $h < 0$ hai to Quadrant II mein. Yeh kabhi x-axis ke neeche nahi ho sakta.
3. Kya H kisi axis par ho sakta hai? Kahan?
  Solution dikhao
  AnswerSirf y-axis par, $(0, 3)$ par.
  Jab $h = 0$ hai, to H $(0, 3)$ hai, jo y-axis par hai. H kabhi x-axis par nahi ho sakta, kyunki uska y-coordinate 3 hai, 0 nahi.
3
Short answerMediumNCERT End-of-chapter Q3
Points $S (4, 2)$ , $H (- 4, 2)$ , $I (- 4, - 3)$ aur $P (2, - 3)$ ko isi order mein jodkar shape SHIP banti hai. Plot karne se pehle neeche ke jawab predict karo. Phir points plot karke check karo.
1. SHIP ki do aisi sides ke naam batao jo ek doosre ke perpendicular hain.
  Solution dikhao
  AnswerSH aur HI (HI aur IP bhi).
  S aur H dono ka $y = 2$ hai, isliye SH horizontal hai. H aur I dono ka $x = - 4$ hai, isliye HI vertical hai. Horizontal aur vertical sides perpendicular hoti hain, isliye $S H ⊥ H I$ . Isi tarah $H I ⊥ I P$ .
2. x-axis ya y-axis ke parallel ek side ka naam batao.
  Solution dikhao
  AnswerSH (ya IP) x-axis ke parallel hai; HI y-axis ke parallel hai.
  SH aur IP dono aise points ko jodti hain jinke y-coordinates barabar hain, isliye yeh x-axis ke parallel hain. HI aise points ko jodti hai jinke x-coordinates barabar hain, isliye yeh y-axis ke parallel hai. Sirf PS tirchi hai.
3. Kaunse do points kisi axis mein ek doosre ki mirror image hain, aur kis axis mein?
  Solution dikhao
  AnswerS aur H, y-axis mein.
  $S (4, 2)$ aur $H (- 4, 2)$ mein sirf x-coordinate ke sign ka fark hai, isliye dono ek doosre ka y-axis mein reflection hain. Kyunki SH aur IP dono horizontal hain, SHIP ek trapezium hai.
4
NumericalMediumNCERT End-of-chapter Q4
$R (- 6, 8)$ plot karo. x-axis par ek point S aisa chuno ki triangle ORS mein S par right angle ho, jahan O origin hai.
1. S kahan hona chahiye?
  Solution dikhao
  Answer $S (- 6, 0)$
  S par OS (x-axis ke saath) aur SR ke beech right angle ka matlab hai ki SR vertical hai. Isliye S ka x-coordinate R wala hi hai aur S x-axis par hai: $S (- 6, 0)$ .
2. Lengths OS, SR aur OR nikalo.
  Solution dikhao
  Answer6, 8 aur 10 units
  $O S = ∣ - 6 - 0 ∣ = 6$ aur $S R = ∣ 8 - 0 ∣ = 8$ . Baudhāyana–Pythagoras theorem se, $O R = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ . Distance formula se bhi yahi milta hai: $\sqrt{(- 6)^{2} + 8^{2}} = 10$ .
5
Long answerMediumNCERT End-of-chapter Q5
Ek aise map ki kalpana karo jiske coordinates sirf zero aur positive numbers use karte hain, ek fixed origin se naape hue.
1. Yeh map plane ke kaunse hisse ko describe kar sakta hai?
  Solution dikhao
  AnswerSirf Quadrant I, axes ke positive halves aur origin ke saath.
  Negative numbers ke bina koi point y-axis ke left ya x-axis ke neeche nahi ho sakta. Map sirf Quadrant I wala chauthai plane aur uske kinaare cover karta hai.
2. Kya yeh ek anant plane ke har point ki jagah bata sakta hai? Samjhao.
  Solution dikhao
  AnswerNahi.
  Origin ke left ya neeche wale points ka koi naam nahi hoga. Origin ko aur door khiskane se kuch der madad milti hai, lekin anant plane mein kisi bhi chune hue corner ke us paar hamesha points rahenge.
3. Screens sirf non-negative pixel coordinates use karti hain. Unke liye yeh kaam kyun karta hai?
  Solution dikhao
  Screen ek limited size ka rectangle hai. Origin ek corner par ho, to har pixel dono axes ki ek hi taraf hota hai, isliye uske coordinates kabhi negative nahi hote. Negative numbers ki zaroorat tabhi padti hai jab points origin ki dono taraf ho sakte hon.
Short answerHardNCERT End-of-chapter Q6
Kya points $P (- 2, 3)$ , $Q (2, 0)$ aur $R (10, - 6)$ ek seedhi line par hain? Aisa method batao jismein plot karne ki zaroorat na ho.
Solution dikhao
AnswerHaan: $P Q + Q R = 5 + 10 = 15 = P R$ .
$P Q = \sqrt{4^{2} + (- 3)^{2}} = 5, Q R = \sqrt{8^{2} + (- 6)^{2}} = 10, P R = \sqrt{12^{2} + (- 9)^{2}} = 15$
Kyunki $P Q + Q R = P R$ hai, point Q segment PR par hai, isliye teenon points collinear hain.
Ek aur check: P se Q jaane mein move $(+ 4, - 3)$ hai, aur Q se R jaane mein move $(+ 8, - 6)$ , jo same direction mein theek do guna hai.
Short answerHardNCERT End-of-chapter Q7
Question 6 wale apne method se check karo ki $D (- 1, 2)$ , $E (3, 5)$ aur $F (11, 10)$ ek seedhi line par hain ya nahi. Phir unhe plot karo. Drawing tumhe dhoka kyun de sakti hai?
Solution dikhao
AnswerNahi: $D E + E F \approx 14.434$ lekin $D F \approx 14.422$ .
$D E = \sqrt{4^{2} + 3^{2}} = 5, E F = \sqrt{8^{2} + 5^{2}} = \sqrt{89} \approx 9.434, D F = \sqrt{12^{2} + 8^{2}} = \sqrt{208} \approx 14.422$
$D E + E F \approx 14.434$ , jo $D F \approx 14.422$ se zyada hai. Isliye E segment DF par nahi hai, aur points collinear nahi hain.
Steps bhi yahi batate hain: D se E ka move $(+ 4, + 3)$ hai, jabki E se F ka move $(+ 8, + 6)$ ki jagah $(+ 8, + 5)$ hai. Drawing par E line DF se itna kam hatta hai ki sab seedha dikhta hai; isiliye sketch se zyada calculation par bharosa kiya ja sakta hai.
8
DiagramMediumNCERT End-of-chapter Q8
Origin O ko ek vertex maankar, har triangle ke baaki do vertices batao aur use plot karo.
1. Ek right-angled isosceles triangle, jiska right angle O par ho aur ek vertex negative x-axis par ho.
  Solution dikhao
  AnswerJaise $(- 5, 0)$ aur $(0, 5)$ .
  Legs ko axes ke saath rakho: $A (- 5, 0)$ aur $B (0, 5)$ . Tab $O A = O B = 5$ aur O par axes ke beech ka angle right angle hai. $k > 0$ ke liye koi bhi $(- k, 0)$ , $(0, k)$ ya $(0, - k)$ ke saath, bhi kaam karta hai.
2. Ek isosceles triangle jismein barabar sides ke beech ka vertex O ho, ek vertex Quadrant I mein ho aur teesra Quadrant II mein.
  Solution dikhao
  AnswerJaise $(4, 3)$ aur $(- 4, 3)$ .
  y-axis mein mirror images lo: Quadrant I mein $A (4, 3)$ aur Quadrant II mein $B (- 4, 3)$ . Tab $O A = O B = \sqrt{16 + 9} = 5$ . General roop mein $a, b > 0$ ke saath $(a, b)$ aur $(- a, b)$ kaam karte hain.
9
Short answerHardNCERT End-of-chapter Q9
Har row ke liye tay karo ki M segment ST ka midpoint hai ya nahi. Phir ek rule dhoondho jo midpoint ke coordinates ko dono ends ke coordinates se jodta ho.
Table 1.4Kya M, ST ka midpoint hai?
Row S M T
a $(- 6, 1)$ $(0, 1)$ $(6, 1)$
b $(2, - 3)$ $(4, 0)$ $(6, 4)$
c $(0, 3)$ $(0, 0)$ $(0, - 6)$
d $(- 5, 4)$ $(- 1, 1)$ $(3, - 2)$
1. Row a: M, ST ka midpoint hai.
  SahiGalat
  Answer dekho
  Answer Sahi
  $S M = 6$ aur $M T = 6$ , aur teenon points line $y = 1$ par hain.
2. Row b: M, ST ka midpoint hai.
  SahiGalat
  Answer dekho
  Answer Galat
  $S M = \sqrt{4 + 9} = \sqrt{13}$ lekin $M T = \sqrt{4 + 16} = \sqrt{20}$ . Midpoint $(4, 0.5)$ hota.
3. Row c: M, ST ka midpoint hai.
  SahiGalat
  Answer dekho
  Answer Galat
  Teenon y-axis par hain, lekin $S M = 3$ aur $M T = 6$ . Midpoint $(0, - 1.5)$ hota.
4. Row d: M, ST ka midpoint hai.
  SahiGalat
  Answer dekho
  Answer Sahi
  $S M = \sqrt{16 + 9} = 5$ aur $M T = \sqrt{16 + 9} = 5$ , aur $S \to M$ aur $M \to T$ ek hi move $(+ 4, - 3)$ hain.
5. Kaunsa rule midpoint ke coordinates ko S aur T ke coordinates se jodta hai?
  Solution dikhao
  AnswerMidpoint ka har coordinate dono ends ke coordinates ka average hai.
  Rows a aur d mein M ka har coordinate S aur T ke coordinates ke bilkul beech mein hai. $S (x_{1}, y_{1})$ aur $T (x_{2}, y_{2})$ ke liye midpoint hai
  $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
  Midpoint check karne ke liye dono halves barabar hone chahiye aur teenon points ek line mein hone chahiye; average wala test dono ek saath check kar leta hai.
10
NumericalMediumNCERT End-of-chapter Q10
$M (2, - 3)$ , $A (- 4, 5)$ aur $B (x, y)$ ka midpoint hai. B nikalo.
1. $x$ nikalo.
  Answer dekho
  Answer 8
  $\frac{- 4 + x}{2} = 2$ , isliye $- 4 + x = 4$ aur $x = 8$ .
2. $y$ nikalo.
  Answer dekho
  Answer -11
  $\frac{5 + y}{2} = - 3$ , isliye $5 + y = - 6$ aur $y = - 11$ .
NumericalHardNCERT End-of-chapter Q11
P aur Q, $A (- 5, 9)$ se $B (10, - 3)$ tak ke segment ko teen barabar hisson mein baantte hain, aur P, A ke zyada paas hai. Midpoints ki madad se P aur Q nikalo.
Solution dikhao
Answer $P (0, 5)$ aur $Q (5, 1)$
A se B tak ka poora move $(10 - (- 5), - 3 - 9) = (15, - 12)$ hai. Iska ek-tihaai $(5, - 4)$ hai.
Isliye $P = (- 5 + 5, 9 - 4) = (0, 5)$ , jo sanyog se y-axis par hai, aur $Q = (0 + 5, 5 - 4) = (5, 1)$ .
Midpoints se check karo: A aur Q ka midpoint $(\frac{- 5 + 5}{2}, \frac{9 + 1}{2}) = (0, 5) = P$ hai, aur P aur B ka midpoint $(\frac{0 + 10}{2}, \frac{5 - 3}{2}) = (5, 1) = Q$ hai.
12
NumericalHardNCERT End-of-chapter Q12
Circle K ka centre origin par hai.
1. $A (2, 9)$ , $B (- 6, 7)$ aur $C (- 7, - 6)$ teenon K par hain. K ka radius 2 decimal places tak nikalo.
  Answer dekho
  Answer 9.22
  $O A^{2} = 4 + 81 = 85$ , $O B^{2} = 36 + 49 = 85$ , $O C^{2} = 49 + 36 = 85$ . Teenon O se barabar door hain, isliye woh $\sqrt{85} \approx 9.22$ radius wale ek hi circle par hain.
2. (b)
  Har point ko uski jagah se match karo.
  Left mein koi item tap karo, phir right mein uska match.
  Answer dekho
  Answer
  $D (- 8, 5)$ : K ke bahar
  $E (0, - 9)$ : K ke andar
  $F (9, 2)$ : K par
  Squared distances ko 85 se compare karo: $64 + 25 = 89 > 85$ (bahar), $0 + 81 = 81 < 85$ (andar), $81 + 4 = 85$ (circle par).
NumericalHardNCERT End-of-chapter Q13
Triangle ABC ki sides BC, CA aur AB ke midpoints $D (3, 2)$ , $E (- 1, 4)$ aur $F (1, - 2)$ hain. A, B aur C nikalo.
Solution dikhao
Answer $A (- 3, 0)$ , $B (5, - 4)$ , $C (1, 8)$
Midpoint ka do guna dono ends ke jod ke barabar hota hai, ek-ek coordinate karke. $B + C$ ka matlab likho: “B aur C ke x-coordinates jodo, aur unke y-coordinates jodo”. Har midpoint ko double karne par $B + C = (6, 4)$ , $C + A = (- 2, 8)$ aur $A + B = (2, - 4)$ milte hain.
Teenon ko jodne par: $2 (A + B + C) = (6, 8)$ , isliye $A + B + C = (3, 4)$ .
Har pair ko subtract karo: $A = (3, 4) - (6, 4) = (- 3, 0)$ , $B = (3, 4) - (- 2, 8) = (5, - 4)$ aur $C = (3, 4) - (2, - 4) = (1, 8)$ .
Check: B aur C ka midpoint $(\frac{5 + 1}{2}, \frac{- 4 + 8}{2}) = (3, 2) = D$ hai.
14
Short answerMediumNCERT End-of-chapter Q14
Ek mela ground mein 8 lanes North–South aur 6 lanes East–West chalti hain, har lane agli lane se 50 m door hai, aur crossings par stalls hain. Crossing $(a, b)$ woh jagah hai jahan $a$ th North–South lane (west se gini hui) $b$ th East–West lane (south se gini hui) se milti hai.
1. 50 m ke liye 1 cm lekar, lane grid ka model, har direction mein pehli lane se aakhri lane tak, kitna lamba aur kitna chauda hoga?
  Solution dikhao
  Answer7 cm × 5 cm
  8 lanes ke beech 50 m ke 7 gap hain, isliye grid 350 m chauda hai: model par 7 cm. 6 lanes ke beech 5 gap hain: 250 m, yaani 5 cm.
2. Kul kitne crossings hain?
  Answer dekho
  Answer 48
  8 North–South lanes mein se har ek, 6 East–West lanes mein se har ek se milti hai: $8 \times 6 = 48$ .
3. Kitne crossings ka naam $(5, 2)$ hai, aur kitnon ka $(2, 5)$ ? Kya yeh ek hi crossing hai?
  Solution dikhao
  AnswerDono ek-ek hain, aur yeh alag-alag crossings hain.
  $(5, 2)$ matlab 5th North–South lane aur 2nd East–West lane; $(2, 5)$ matlab 2nd North–South lane aur 5th East–West lane. Har naam theek ek crossing par fit hota hai, aur dono crossings alag jagah par hain.
4. Inmein se kaunsa naam is ground ki asli crossing ka hai?
  $(7, 6)$ $(6, 7)$ donokoi nahi
  Answer dekho
  Answer (A) $(7, 6)$
  $(7, 6)$ mein 8 mein se 7th North–South lane aur 6 mein se 6th East–West lane hai: yeh exist karti hai. $(6, 7)$ ke liye 7th East–West lane chahiye, jo hai hi nahi. Order matter karta hai.
15
NumericalHardNCERT End-of-chapter Q15
Ek game screen 720 pixels chaudi aur 480 pixels oonchi hai, aur origin uske bottom-left corner par hai. Icon 1, 100 pixels radius ka circle hai jiska centre $(90, 120)$ par hai. Icon 2, 120 pixels radius ka circle hai jiska centre $(330, 190)$ par hai.
1. Kya kisi bhi icon ka koi hissa screen se bahar jaata hai?
  Solution dikhao
  AnswerIcon 1 left mein kat jaata hai; icon 2 fit ho jaata hai.
  Icon 1, $x = 90 - 100 = - 10$ se $x = 190$ tak aur $y = 20$ se $y = 220$ tak jaata hai. Iske sabse left wale 10 pixels screen se bahar hain.
  Icon 2, $x = 210$ se $x = 450$ tak aur $y = 70$ se $y = 310$ tak jaata hai, jo $0 \leq x \leq 720$ aur $0 \leq y \leq 480$ ke kaafi andar hai.
2. Kya dono icons overlap karte hain?
  Solution dikhao
  AnswerNahi: centres 250 pixels door hain, jo $100 + 120 = 220$ se zyada hai.
  $\sqrt{(330 - 90)^{2} + (190 - 120)^{2}} = \sqrt{57 600 + 4 900} = \sqrt{62 500} = 250$
  Circles ek doosre ko tabhi touch karte jab centres $100 + 120 = 220$ pixels ya usse kam door hote. 250 pixels par 30 pixels ka gap hai.
NumericalMediumNCERT End-of-chapter Q16
$A (1, - 2)$ , $B (5, 1)$ , $C (2, 5)$ aur $D (- 2, 2)$ plot karo. Kya ABCD ek square hai? Samjhao, aur uska area nikalo.
Solution dikhao
AnswerHaan; area 25 square units.
$A B^{2} = 4^{2} + 3^{2} = 25, B C^{2} = (- 3)^{2} + 4^{2} = 25, C D^{2} = (- 4)^{2} + (- 3)^{2} = 25, D A^{2} = 3^{2} + (- 4)^{2} = 25$
$A C^{2} = 1^{2} + 7^{2} = 50, B D^{2} = (- 7)^{2} + 1^{2} = 50$
Chaaron sides 5 units ki hain aur dono diagonals barabar hain, isliye ABCD ek square hai. Saath hi $A B^{2} + B C^{2} = 50 = A C^{2}$ , isliye B par angle right angle hai.
Area $= 5^{2} = 25$ square units. Iski koi bhi side kisi axis ke parallel nahi hai, phir bhi distance formula har side ko naap leta hai.
Quick answerEasy
Kaunsa point origin se sabse door hai?
$(5, 0)$ $(- 3, 4)$ $(- 4, - 4)$ $(0, - 6)$
Answer dekho
Answer (D) $(0, - 6)$
Squared distances: 25, 25, 32 aur 36. $(0, - 6)$ 6 units door hai; $(- 4, - 4)$ sirf $\sqrt{32} \approx 5.66$ door hai.
Quick answerMedium
Vertices $(0, 0)$ , $(6, 0)$ aur $(0, 8)$ wale triangle ka perimeter nikalo.
units
Answer dekho
Answer 24 units
Sides 6, 8 aur $\sqrt{36 + 64} = 10$ hain, isliye perimeter 24 hai.
19
Quick answerMedium
Point $(2, - 5)$ ke har reflection ko uski image se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
x-axis mein : $(2, 5)$
y-axis mein : $(- 2, - 5)$
origin mein : $(- 2, 5)$
x-axis y ko palatti hai, y-axis x ko palatti hai, aur origin dono ko.
20
Quick answerMedium
In points ko origin se distance ke hisaab se lagao, sabse paas wala pehle.
1. $(1, 1)$
2. $(- 2, 0)$
3. $(3, 3)$
4. $(0, - 3)$
Answer dekho
Answer
$(1, 1)$
$(- 2, 0)$
$(0, - 3)$
$(3, 3)$
Distances $\sqrt{2} \approx 1.41$ , 2, 3 aur $\sqrt{18} \approx 4.24$ hain.

Row	S	M	T
a	$(- 6, 1)$	$(0, 1)$	$(6, 1)$
b	$(2, - 3)$	$(4, 0)$	$(6, 4)$
c	$(0, 3)$	$(0, 0)$	$(0, - 6)$
d	$(- 5, 4)$	$(- 1, 1)$	$(3, - 2)$

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise Set 1.1 (i)–(ii)	Practice 1.1 · Q1
NCERT Exercise Set 1.1 (iii)	Practice 1.1 · Q2
NCERT Exercise Set 1.1 (iv)	Practice 1.1 · Q3
NCERT Exercise Set 1.2 Q1	Practice 1.2 · Q1
NCERT Exercise Set 1.2 Q2	Practice 1.2 · Q2
NCERT Exercise Set 1.2 Q3	Practice 1.2 · Q3
NCERT Exercise Set 1.2 Q4	Practice 1.2 · Q4
NCERT End-of-chapter Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Q4	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Q5	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Q6	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Q7	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Q8	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Q9	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Q10	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Q11	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Q12	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Q13	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Q14	Chapter ke end ki exercises · Q14
NCERT End-of-chapter Q15	Chapter ke end ki exercises · Q15
NCERT End-of-chapter Q16	Chapter ke end ki exercises · Q16

Quick check: MCQs

1.1 Introduction4 questions
1.2 Settling In: Room ko set karna3 questions
Wapas Ganit Lab mein4 questions
Do numbers ka matlab kya hai4 questions
Lab ki deewaron ke bahar3 questions
Axis ke parallel line ke saath3 questions
Koi bhi do points4 questions
Reflection se lengths nahi badalti4 questions

Glossary · 14

Key terms (English + Hindi)

Cartesian planeकार्तीय तलkaarteey tal: Woh flat surface jis par coordinate axes banaye jaate hain; is par har point ke coordinates hote hain.Chapter mein pehli baar yahan aaya →
collinearसंरेखsanrekh: Ek hi seedhi line par hona. Agar $A B + B C = A C$ hai, to A, B, C collinear hain aur B beech mein hai.Chapter mein pehli baar yahan aaya →
coordinate systemनिर्देशांक पद्धतिnirdeshank paddhati: Ek fixed starting point aur reference lines, jinki madad se hum har location ko numbers se naam de sakte hain.Chapter mein pehli baar yahan aaya →
coordinatesनिर्देशांकnirdeshank: Ordered pair $(x, y)$ jo point ki jagah fix karta hai: y-axis se aur x-axis se uski signed distances.Chapter mein pehli baar yahan aaya →

distance formulaदूरी सूत्रdoori sootra: $(x_{1}, y_{1})$ aur $(x_{2}, y_{2})$ ke beech ki distance $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ hoti hai.Chapter mein pehli baar yahan aaya →

midpointमध्य-बिंदुmadhya-bindu: Segment ke bilkul beech ka point; iske coordinates dono ends ke coordinates ka average hote hain.Chapter mein pehli baar yahan aaya →

ordered pairक्रमित युग्मkramit yugm: Numbers ka aisa pair jismein order matter karta hai: $(a, b) = (b, a)$ sirf tab jab $a = b$ .Chapter mein pehli baar yahan aaya →
originमूलबिंदुmoolbindu: Woh point $O (0, 0)$ jahan dono axes ek doosre ko kaatte hain.Chapter mein pehli baar yahan aaya →

quadrantचतुर्थांशchaturthansh: Axes se bane chaar hisson mein se ek; top right se anticlockwise I se IV tak number diye jaate hain.Chapter mein pehli baar yahan aaya →

reflectionपरावर्तनparaavartan: Kisi line mein point ya shape ki mirror image. y-axis mein $(x, y)$ , $(- x, y)$ ban jaata hai.Chapter mein pehli baar yahan aaya →

x-axisx-अक्षx-aksh: Coordinate plane ki horizontal number line. Iske points $(x, 0)$ form ke hote hain.Chapter mein pehli baar yahan aaya →
x-coordinateभुजbhuj: $(x, y)$ ka pehla number: point y-axis ke right (+) ya left (−) mein kitni door hai.Chapter mein pehli baar yahan aaya →

y-axisy-अक्षy-aksh: Coordinate plane ki vertical number line. Iske points $(0, y)$ form ke hote hain.Chapter mein pehli baar yahan aaya →
y-coordinateकोटिkoti: $(x, y)$ ka doosra number: point x-axis ke upar (+) ya neeche (−) kitni door hai.Chapter mein pehli baar yahan aaya →

1.1 Introduction

1.2 Settling In: Room ko set karna

1.3 2-D Cartesian Coordinate System

Axes par pade points

Wapas Ganit Lab mein

Quadrants

Do numbers ka matlab kya hai

Lab ki deewaron ke bahar

1.4 2-D Plane mein Do Points ke Beech ki Distance

Axis ke parallel line ke saath

Koi bhi do points

Reflection se lengths nahi badalti

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye