Coordinate Geometry: Class 10 Maths Chapter 7 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
2 simulations
29 practice questions
19 MCQs

Is chapter mein aap kya seekhenge

Kisi point ka abscissa aur ordinate, aur axes par points ke forms yaad karo, aur chaaron quadrants mein points plot karo.
Pythagoras theorem se distance formula derive karo aur use kinhi bhi do points ke liye use karo, origin se distance bhi.
Distances se tay karo ki points collinear hain ya nahi, aur triangles aur quadrilaterals ko classify karo.
Distance conditions se unknown coordinates nikaalo, jaise do diye gaye points se equidistant point.
Similar triangles se internal division ke liye section formula derive karo, uske k : 1 aur midpoint forms ke saath.
Diye gaye ratio mein segment ko divide karne wala point, kisi point ya axis se segment kis ratio mein divide hota hai, aur points of trisection nikaalo.
Parallelograms aur circles ki properties ke saath midpoint formula use karke missing vertices aur endpoints nikaalo.

NCERT Mathematics Class 10, Chapter 7, pp. 99–112 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

7.1 Introduction

Koi map app kholo aur poochho ki sabse paas wala chemist kitni door hai. App tumhare ghar aur dukaan ko numbers ke pairs ki tarah store karta hai, aur ek chhota sa formula un numbers ko distance mein badal deta hai. Is chapter mein tum woh formula seekhoge, aur ek doosra formula bhi jo do jagahon ke beech raaste ke kisi hisse par point dhoondhta hai.

Pehle Class 9 ka ek quick recap. Do perpendicular number lines, x-axis aur y-axis, origin $O (0, 0)$ par ek doosre ko cross karti hain, aur plane ke har point ko ek ordered pair $(x, y)$ milta hai. Pehla number, yaani x-coordinate ya abscissa (भुज), batata hai ki point y-axis se kitni door hai: positive ho to right mein aur negative ho to left mein. Doosra number, yaani y-coordinate ya ordinate (कोटि), batata hai ki point x-axis se kitna upar ya neeche hai.

x-axis par point ki height zero hoti hai, isliye woh hamesha $(x, 0)$ jaisa dikhta hai. y-axis par point ka koi sideways shift nahi hota, isliye woh $(0, y)$ jaisa dikhta hai. Board questions aksar in forms ko "a point on the y-axis" jaise shabdon mein chhupa dete hain, isliye inpar nazar rakho.

Try karke dekhoNCERT 7.1 Introduction · p. 99

Ek kite plot karo

Tumhe chahiye

graph paper ya squared notebook ka ek page
ek nukeeli pencil
ek ruler

Pehle guess karo

Inmein se do points y-axis par hain. Kaun se do?

A (0, 7)

aur

C (0, - 5)

B (4, 3)

aur

D (- 4, 3)

P (- 1, - 6)

aur

Q (1, - 7)

1 unit ke liye 1 cm lekar dono axes banao: across $- 6$ se $6$ tak aur upar $- 9$ se $8$ tak.
$A (0, 7)$ , $B (4, 3)$ , $C (0, - 5)$ , $D (- 4, 3)$ , $P (- 1, - 6)$ , $Q (1, - 7)$ aur $R (- 0.5, - 8)$ plot karo.
A ko B se, B ko C se, C ko D se aur wapas A se jodo.
A ko C se aur B ko D se jodo. Phir C ko P se, P ko Q se aur Q ko R se jodo.
Tumhe kya picture milti hai? D, P aur Q kin quadrants mein hain?

Tumhe do sticks aur zigzag tail wali ek kite (patang) milti hai, Makar Sankranti ke liye taiyaar. D quadrant II mein hai, P quadrant III mein aur Q quadrant IV mein. R, $x = - 0.5$ par, y-axis se aadha square left mein, quadrant III mein hai.

3/3Sticks AC aur BD aur tail C, P, Q, R jodo.

Fig. 7.1Plot karo, jodo, aur kite dikh jaati hai.

Ek coordinate grid jo across minus 6 se 6 tak aur upar minus 9 se 8 tak hai; right mein points ki list hai: A(0, 7), B(4, 3), C(0, minus 5), D(minus 4, 3), P(minus 1, minus 6), Q(1, minus 7) aur R(minus 0.5, minus 8). Step 1 mein saaton points apne letters ke saath dots ki tarah plot hain; A aur C y-axis par hain. Step 2 mein A, B, C, D aur wapas A jude hain, jisse chaar sides wali kite ki outline banti hai. Step 3 mein do sticks AC aur BD jude hain, upar ka hissa yellow aur neeche ka hissa orange rang ka hai, aur tail C se P aur Q hote hue R tak zigzag bani hai.

Tum plane ko sirf akele points ke liye hi use nahi kar chuke ho. Ek linear equation $a x + b y + c = 0$ , jahan $a$ aur $b$ dono zero nahi hain, ek straight line banati hai, aur Chapter 2 mein $y = a x^{2} + b x + c$ ( $a \neq 0$ ) ka graph ek parabola nikla. Shapes ko equations mein badalna, aur equations ko wapas shapes mein, yahi coordinate geometry (निर्देशांक ज्यामिति) ka kaam hai. Kisi figure ke baare mein sawaal ek calculation ban jaata hai, aur equation ek picture ban jaati hai jise tum dekh sakte ho.

Kyunki yeh measuring ka kaam algebra se karwa deti hai, coordinate geometry tumhare use kiye kai tools ke andar hai: map aur GPS apps, computer graphics aur animation, engineering drawings, robot arms aur woh instruments jo earthquakes ki location dhoondhte hain.

Yeh chapter do tools jodta hai. Distance formula sirf coordinates se do points ke beech ka distance deta hai. Section formula woh point dhoondhta hai jo segment ko diye gaye ratio mein kaatta hai, jaise aadhe raaste par ya ek-tihaai raaste par.

7.2 Distance Formula

Pahadon mein ek district hospital drone se gaon tak medicines bhejta hai. Ek gaon, V, hospital H se 45 km east aur 24 km north mein hai. Drone straight line mein udta hai. Uski flight kitni lambi hai?

Hospital ko origin par rakho, east ko x-axis ke saath aur north ko y-axis ke saath, aur 1 unit ko 1 km maano. Tab H $(0, 0)$ hai aur V $(45, 24)$ hai. Safar ka east wala hissa aur north wala hissa ek right triangle ki do perpendicular sides hain, aur flight path uska hypotenuse hai (Fig. 7.2). Pythagoras theorem se, $H V^{2} = 45^{2} + 24^{2} = 2025 + 576 = 2601$ , isliye $H V = 51$ km.

3/3Seedha raasta hypotenuse hai: $\sqrt{45^{2} + 24^{2}} = 51$ km.

Fig. 7.2East aur north right triangle ke legs banate hain; flight hypotenuse hai.NCERT Fig. 7.1 · p. 100

Ek map jismein East right ki taraf aur North upar ki taraf hai. Step 1 mein hospital H origin par cross se marked hai, aur gaon V (45, 24) par hai. Step 2 mein dashed lines pehle 45 km east aur phir 24 km north jaati hain, aur V ke neeche right angle par milti hain. Step 3 mein H se V tak seedha flight path bana hai, jis par 51 km likha hai, saath mein working: 45 squared plus 24 squared ka square root 51 ke barabar hai.

Yahi idea kinhi bhi do points ke liye kaam karta hai. Chalo ise step by step banate hain, axes par points se shuru karke.

Axis par points

x-axis par $A (- 4, 0)$ aur $B (5, 0)$ lo (Fig. 7.3). A, O se 4 units left mein hai aur B 5 units right mein, isliye $A B = 4 + 5 = 9$ units. x-coordinates ko subtract karne se turant wahi answer milta hai: $5 - (- 4) = 9$ . y-axis par $C (0, 3)$ aur $D (0, - 2)$ ke beech $3 - (- 2) = 5$ units ka distance hai.

To x-axis par do points ke beech $∣ x_{2} - x_{1} ∣$ ka distance hota hai, aur y-axis par do points ke beech $∣ y_{2} - y_{1} ∣$ ka. Bars ka matlab hai "size lo aur sign hata do", kyunki distance kabhi negative nahi hota.

Ab x-axis wale A ko y-axis wale C se jodo. Axes perpendicular hain, isliye triangle AOC mein O par right angle hai, $O A = 4$ aur $O C = 3$ ke saath. Pythagoras theorem se $A C = \sqrt{4^{2} + 3^{2}} = 5$ units milta hai.

3/3Axes right angle par milte hain, isliye $A C = \sqrt{4^{2} + 3^{2}} = 5$ .

Fig. 7.3Axis par subtract karo; axes ke across, O par bane right angle ko use karo.NCERT Fig. 7.2 · p. 100

Ek grid jismein x-axis minus 5 se 7 tak aur y-axis minus 3 se 4 tak hai. Step 1 mein x-axis ka A(minus 4, 0) se B(5, 0) tak ka hissa highlight hai aur AB = 5 minus minus 4 = 9 dikhaya gaya hai. Step 2 mein y-axis ka C(0, 3) se neeche D(0, minus 2) tak ka hissa highlight hai aur CD = 3 minus minus 2 = 5 dikhaya gaya hai. Step 3 mein right triangle AOC shaded hai, uska right angle origin par hai, aur AC = 4 squared plus 3 squared ka square root = 5 dikhaya gaya hai.

Koi bhi do points

Axes se door ke points ke liye right triangle khud banao. Quadrant II mein $K (- 2, 5)$ aur quadrant IV mein $L (4, - 3)$ lo (Fig. 7.4). K se hokar ek horizontal line aur L se hokar ek vertical line $C (4, 5)$ par milti hain, jo right triangle KCL ka corner hai.

Horizontal side KC $x = - 2$ se $x = 4$ tak jaati hai: y-axis tak pahunchne ke liye 2 units aur uske aage 4 aur, isliye $K C = 4 - (- 2) = 6$ . Vertical side CL $y = 5$ se neeche $y = - 3$ tak jaati hai, isliye $C L = 5 - (- 3) = 8$ . Isliye $K L = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ units.

4/4Pythagoras se, $K L = \sqrt{6^{2} + 8^{2}} = 10$ .

Fig. 7.4Coordinates subtract karne se axis ke dono taraf ke units gin liye jaate hain.NCERT Fig. 7.4 · p. 101

Ek grid jo across minus 4 se 6 tak aur upar minus 5 se 7 tak hai. Step 1 mein K(minus 2, 5) quadrant II mein aur L(4, minus 3) quadrant IV mein hai. Step 2 mein corner C(4, 5) juda hai jahan K se hokar horizontal line aur L se hokar vertical line milti hain, saath mein right-angle mark hai. Step 3 mein legs rangeen hain aur likha hai KC = 2 + 4 = 6 aur CL = 5 + 3 = 8, kyunki har leg ek axis cross karta hai. Step 4 mein KL bana hai aur KL = 6 squared plus 8 squared ka square root = 10 dikhaya gaya hai.

Axis cross karne ke liye koi khaas dhyaan nahi dena pada: coordinates subtract karne se uske dono taraf ke units gin liye gaye. Same quadrant ke do points bhi isi tarah kaam karte hain. $(2, 3)$ aur $(7, 15)$ ke liye legs $7 - 2 = 5$ aur $15 - 3 = 12$ hain, isliye distance $\sqrt{25 + 144} = 13$ units hai.

DerivationDistance formula

Maan lo $P (x_{1}, y_{1})$ aur $Q (x_{2}, y_{2})$ koi bhi do points hain. P se hokar ek horizontal line aur Q se hokar ek vertical line banao. Woh corner $C (x_{2}, y_{1})$ par milti hain, jo P ke level par hai aur Q ki seedh mein hai (Fig. 7.5).

4/4C par right angle se $P Q^{2} = P C^{2} + C Q^{2}$ milta hai.

Fig. 7.5Corner

C

apna x

Q

se aur apna y

P

se leta hai.NCERT Fig. 7.5 · p. 102

Origin O wale axes. Step 1 mein P coordinates x1, y1 ke saath aur Q coordinates x2, y2 ke saath hai, aur dono axes tak dashed guides hain. Step 2 mein corner C(x2, y1) aur C par right angle juda hai. Step 3 mein horizontal leg PC par x2 minus x1 aur vertical leg CQ par y2 minus y1 likha hai. Step 4 mein PQ bana hai aur PQ squared = PC squared + CQ squared likha hai.

PC ke saath sirf x badalta hai, isliye $P C = ∣ x_{2} - x_{1} ∣$ . CQ ke saath sirf y badalta hai, isliye $C Q = ∣ y_{2} - y_{1} ∣$ . C par angle right angle hai, isliye Pythagoras theorem se $P Q^{2} = P C^{2} + C Q^{2}$ , yaani $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Square karne se koi bhi minus sign hat jaata hai, isliye bars ki zaroorat nahi rehti.

Length kabhi negative nahi hoti, isliye hum sirf positive square root rakhte hain:

Distance formula

P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

(7.1)

jahan

$(x_{1}, y_{1})$: P ke coordinates
$(x_{2}, y_{2})$: Q ke coordinates

Yeh distance formula (दूरी सूत्र) hai. Iske do nateeje yaad rakhne layak hain. Pehla, ek point ki jagah $(0, 0)$ rakhne se origin se $P (x, y)$ ka distance milta hai:

Origin se distance

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

(7.2)

Doosra, $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ , kyunki kisi number aur uske negative ka square same hota hai. Isliye isse koi fark nahi padta ki tum kis point ko P kehte ho aur kise Q.

SimulationDistance explorer

Dono axes par minus 8 se 8 tak ka coordinate grid, jismein teen draggable points hain: A (minus 2, 5) par, B (4, minus 3) par aur C (4, 5) par. Points whole numbers par snap hote hain. Sides AB, BC aur CA bani hain aur naapi gayi hain, har distance ke right-triangle legs ke saath, aur har length exact, jaise 2 root 5, aur do decimal places tak di gayi hai.

Fig. 7.6A, B aur C ko drag karo. Har distance apne horizontal aur vertical legs ke saath, exact surd aur decimal dono mein dikhta hai.

Try karo

C ko tab tak drag karo jab tak triangle ABC isosceles na ho jaaye. Ab kaun se do distances barabar hain?
C ko segment AB par drag karo, jaise $(1, 1)$ par. $A C + C B$ aur $A B$ ki tulna mein kya milta hai?
A ko $(- 3, - 1)$ par aur B ko $(3, 7)$ par le jao. Pehle AB khud nikaalo, phir check karo.

Quick check

3 mein se 1

(- 1, 3)

aur

(5, - 5)

ke beech ka distance nikaalo.

units

Answer dekho

Answer 10 units

Differences

5 - (- 1) = 6

aur

- 5 - 3 = - 8

hain, isliye distance

\sqrt{36 + 64} = 10

hai. Ek aam galti

- 5 - 3 = - 2

likhna hai, jisse

\sqrt{40}

aata hai.

Distance formula ka use

Distances se tum bina protractor ke shapes test kar sakte ho. Zyadatar kaam teen facts karte hain:

Triangle hai ya nahi? Teen points triangle tab banate hain jab teen distances mein se kinhi do ka sum teesre se bada ho. Agar do chhote distances ka sum bilkul sabse lambe ke barabar ho, to points collinear (संरेखी) hain: woh ek hi straight line par hain.
Right angle? Agar do sides ke squares ka sum teesri side ke square ke barabar ho, to us teesri side ke saamne ka angle $90^{\circ}$ hota hai. Yeh Pythagoras theorem ka converse hai.
Barabar sides? Barabar distances se isosceles aur equilateral triangles, aur quadrilaterals mein rhombus aur square pehchaane jaate hain.

Compare

Distances se quadrilaterals ko alag pehchaanna

Square

Chaaron sides barabar
Diagonals barabar

Rhombus

Chaaron sides barabar
Diagonals barabar nahi

Rectangle

Opposite sides barabar
Diagonals barabar

Parallelogram

Opposite sides barabar
Diagonals barabar nahi

Example 7.1NCERT Example 1 · p. 102

Triangle hai ya nahi? Kis tarah ka?

Kya points $P (2, - 1)$ , $Q (- 4, 2)$ aur $R (3, 1)$ ek triangle banate hain? Agar haan, to woh kis tarah ka triangle hai?

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

Example 7.2NCERT Example 2 · p. 103

Square saabit karna

Dikhao ki $A (- 2, 0)$ , $B (1, - 2)$ , $C (3, 1)$ aur $D (0, 3)$ ek square ke vertices hain.

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

3/3Dono diagonals $\sqrt{26}$ hain, isliye rhombus ek square hai.

Fig. 7.7Chaar barabar sides aur barabar diagonals: ek square, bhale hi woh tircha ho.

Dono axes par minus 3 se 4 tak ka grid, aur right mein coordinates A(minus 2, 0), B(1, minus 2), C(3, 1) aur D(0, 3) ki list. Step 1 mein chaaron points hain. Step 2 mein chaaron sides blue mein bani hain aur likha hai AB = BC = CD = DA = root 13. Step 3 mein dono diagonals dashed orange lines ki tarah jude hain, likha hai AC = BD = root 26, aur nateeja hai ki ABCD ek square hai.

Example 7.3NCERT Example 3 · p. 103

Kya ek cable seedhi ja sakti hai?

Ek colony ke map par ek electrician ne street lights ke liye teen poles $A (- 1, - 3)$ , $B (1, 1)$ aur $C (4, 7)$ par mark kiye hain. Woh chahti hai ki ek hi seedhi cable teenon se hokar jaaye. Kya yeh possible hai?

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

3/4 $A B + B C = 5 \sqrt{5} = A C$ : ek straight line.

Fig. 7.8Jab do chhote distances ka sum sabse lambe ke barabar ho, to points collinear hote hain.

Ek grid jo across minus 2 se 6 tak aur upar minus 4 se 8 tak hai, A(minus 1, minus 3), B(1, 1) aur C(4, 7) ke saath. Step 1 mein teenon points hain. Step 2 mein AB aur BC rangeen hain aur likha hai AB = root 20 = 2 root 5 aur BC = root 45 = 3 root 5. Step 3 mein AC seedhi B se hokar bani hai, likha hai AC = root 125 = 5 root 5, aur note hai ki AB + BC = AC, isliye points ek straight line par hain. Step 4 mein line se hatkar ek point E(5, 4) juda hai, B se hokar ek dotted muda hua raasta hai, aur note hai ki mude raaste ke liye AB + BE, AE se bada hai.

Dhyaan do ki simplify karna kyun zaroori tha: $2 \sqrt{5}$ aur $3 \sqrt{5}$ bilkul $2 x + 3 x$ ki tarah judte hain, lekin $\sqrt{20} + \sqrt{45}$ ko tab tak joda nahi ja sakta jab tak dono simplify na ho jaayein.

Example 7.4NCERT Example 4 · p. 104

A aur B se barabar door saare points

$P (x, y)$ ko $A (2, - 1)$ aur $B (6, 3)$ se equidistant (समदूरस्थ), yaani barabar door, hone ke liye $x$ aur $y$ ko kaun si condition satisfy karni chahiye?

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

$x + y = 5$ ko satisfy karne wale do points test karo, maan lo $(1, 4)$ aur $(5, 0)$ : har ek A se utna hi door hai jitna B se (Fig. 7.9). To answer points ki ek poori straight line hai. Yeh $(4, 1)$ se hokar jaati hai, jo A aur B ke beech aadhe raaste par hai, aur AB ko right angle par cross karti hai: yeh AB ka perpendicular bisector (लंब समद्विभाजक) hai jo tumne Class 9 ke constructions mein dekha tha.

4/4Aise saare points $x + y = 5$ par hain, jo AB ka perpendicular bisector hai.

Fig. 7.9A aur B se equidistant points AB ke perpendicular bisector par line mein aa jaate hain.NCERT Fig. 7.7 · p. 104

Ek grid jo across minus 1 se 7 tak aur upar minus 2 se 6 tak hai, jismein A(2, minus 1) aur B(6, 3) jude hain. Step 1 mein A aur B hain. Step 2 mein P(1, 4) juda hai, A aur B tak dashed blue lines ke saath, dono root 26. Step 3 mein Q(5, 0) juda hai, A aur B tak dotted orange lines ke saath, dono root 10, aur M(4, 1), yaani AB ka midpoint, marked hai. Step 4 mein green line x + y = 5 P, M aur Q se hokar bani hai, jo AB ko M par right angle par cross karti hai.

Example 7.5NCERT Example 5 · p. 104

y-axis par ek point

y-axis par woh point nikaalo jo $A (7, 3)$ aur $B (- 1, - 5)$ se equidistant hai.

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

Ise Example 7.4 se jodo: P woh jagah hai jahan AB ka perpendicular bisector y-axis se milta hai. Yahan bisector $x + y = 2$ hai, aur $x = 0$ rakhne se phir se $y = 2$ milta hai.

Speed practice ke liye taiyaar? Neeche wali drill ek baar mein do points deti hai. Distance ko whole number ya 2√5 jaise surd ki tarah type karo.

SimulationDistance drill

Minus 10 se 10 tak ke grid par ek challenge. Har round mein do points marked hote hain; tum unka distance whole number, decimal ya 2 root 5 jaise surd ki tarah type karte ho. Answer ke baad sim horizontal aur vertical legs banata hai aur exact distance dikhata hai, aur paanch rounds tak score rakhta hai.

Fig. 7.10Paanch rounds: do marked points ke beech ka distance nikaalo, phir right-triangle ke legs check karo.

Try karo

Type karne se pehle dono legs ki lengths zor se bolo.
Jab answer whole number na ho, to use simplified surd ki tarah chhodo.

Practice · 10 questions

Practice 7.1: answers ke saath

1
Short answerEasyNCERT Exercise 7.1, Q1
Har pair of points ke beech ka distance nikaalo.
1. $(3, - 2)$ aur $(- 5, 4)$
  Solution dikhao
  Answer10 units
  Differences: $- 5 - 3 = - 8$ aur $4 - (- 2) = 6$ .
  Distance $= \sqrt{64 + 36} = \sqrt{100} = 10$ .
2. $(- 6, - 1)$ aur $(- 2, 7)$
  Solution dikhao
  Answer $4 \sqrt{5}$ units
  Differences: $4$ aur $8$ .
  Distance $= \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5}$ .
3. $(a + b, a - b)$ aur $(a - b, a + b)$ , jahan $b > 0$
  Solution dikhao
  Answer $2 \sqrt{2} b$
  x-difference $(a - b) - (a + b) = - 2 b$ hai.
  y-difference $(a + b) - (a - b) = 2 b$ hai.
  Distance $= \sqrt{4 b^{2} + 4 b^{2}} = \sqrt{8 b^{2}} = 2 \sqrt{2} b$ .
Short answerEasyNCERT Exercise 7.1, Q2
$(0, 0)$ aur $(20, 21)$ ke beech ka distance nikaalo. Ek ferry jetty se nikalti hai aur usse 20 km east aur 21 km north mein jaakar rukti hai. Seedha crossing kitna lamba hai?
Solution dikhao
Answer29 units; 29 km
(7.2) se distance $\sqrt{20^{2} + 21^{2}} = \sqrt{400 + 441} = \sqrt{841} = 29$ hai.
Jetty $(0, 0)$ par aur dock $(20, 21)$ par ho, to seedha crossing 29 km hai.
3
Short answerMediumNCERT Exercise 7.1, Q3
Distances se tay karo ki har set ke points collinear hain ya nahi.
1. $A (- 2, 7)$ , $B (1, 1)$ , $C (4, - 5)$
  Solution dikhao
  AnswerCollinear
  $A B = \sqrt{9 + 36} = 3 \sqrt{5}$ aur $B C = \sqrt{9 + 36} = 3 \sqrt{5}$ .
  $A C = \sqrt{36 + 144} = \sqrt{180} = 6 \sqrt{5}$ .
  $A B + B C = 6 \sqrt{5} = A C$ , isliye points collinear hain (B beech wala point hai).
2. $A (- 1, 2)$ , $B (2, 6)$ , $C (6, 9)$
  Solution dikhao
  AnswerCollinear nahi
  $A B = \sqrt{9 + 16} = 5$ aur $B C = \sqrt{16 + 9} = 5$ .
  $A C = \sqrt{49 + 49} = 7 \sqrt{2} \approx 9.90$ .
  $A B + B C = 10$ , jo $A C$ se zyada hai, isliye points collinear nahi hain; woh ek triangle banate hain.
Short answerEasyNCERT Exercise 7.1, Q4
Kya $A (- 3, 1)$ , $B (1, 4)$ aur $C (4, 0)$ ek isosceles triangle banate hain? Kya tum iske baare mein kuch aur bhi bata sakte ho?
Solution dikhao
AnswerHaan; yeh B par right-angled bhi hai.
$A B^{2} = 16 + 9 = 25$ , $B C^{2} = 9 + 16 = 25$ aur $A C^{2} = 49 + 1 = 50$ .
To $A B = B C = 5$ : triangle isosceles hai.
Saath hi $A B^{2} + B C^{2} = 50 = A C^{2}$ , isliye $∠ B = 90^{\circ}$ . Yeh ek right isosceles triangle hai.
Short answerMediumNCERT Exercise 7.1, Q5
Diwali ke liye ek rangoli grid par chaar diye $A (2, 5)$ , $B (5, 7)$ , $C (8, 5)$ aur $D (5, 3)$ par rakhe gaye hain, aur chalk ki line se order mein jode gaye hain. Kya ABCD square hai? Agar nahi, to kya hai?
Solution dikhao
AnswerRhombus, square nahi
Har side ke differences 3 aur 2 hain, isliye $A B = B C = C D = D A = \sqrt{13}$ .
Diagonal $A C = 8 - 2 = 6$ (ek horizontal line) aur diagonal $B D = 7 - 3 = 4$ (ek vertical line).
Chaar barabar sides se rhombus banta hai, lekin diagonals barabar nahi hain, isliye yeh square nahi hai.
6
Long answerMediumNCERT Exercise 7.1, Q6
Chaar points ke har set ko diye gaye order mein jodne par kis tarah ka quadrilateral milta hai, agar milta hai to? Reasons do.
1. $(0, - 3)$ , $(4, 0)$ , $(1, 4)$ , $(- 3, 1)$
  Solution dikhao
  AnswerSquare
  Har side ke differences 4 aur 3 hain (kisi order aur sign mein), isliye chaaron sides 5 hain.
  Diagonals: $\sqrt{1 + 49} = \sqrt{50}$ aur $\sqrt{49 + 1} = \sqrt{50}$ .
  Barabar sides aur barabar diagonals: ek square.
2. $(- 2, - 1)$ , $(1, 1)$ , $(4, 3)$ , $(3, - 4)$
  Solution dikhao
  AnswerKoi quadrilateral nahi
  Inhe A, B, C, D kaho. $A B = \sqrt{13}$ , $B C = \sqrt{13}$ aur $A C = \sqrt{52} = 2 \sqrt{13}$ .
  $A B + B C = A C$ , isliye A, B aur C collinear hain.
  Teen points ek hi line par hain, isliye koi quadrilateral nahi banta.
3. $(- 2, 1)$ , $(3, 2)$ , $(5, - 1)$ , $(0, - 2)$
  Solution dikhao
  AnswerParallelogram
  Sides: $\sqrt{26}$ , $\sqrt{13}$ , $\sqrt{26}$ , $\sqrt{13}$ , isliye opposite sides barabar hain.
  Diagonals: $\sqrt{49 + 4} = \sqrt{53}$ aur $\sqrt{9 + 16} = 5$ , jo barabar nahi hain.
  Opposite sides barabar aur diagonals barabar nahi: ek parallelogram (rectangle nahi).
Short answerMediumNCERT Exercise 7.1, Q7
x-axis par woh point nikaalo jo $(7, - 2)$ aur $(- 1, 6)$ se equidistant hai.
Solution dikhao
Answer $(1, 0)$
Maan lo point $P (x, 0)$ hai. Tab $(x - 7)^{2} + 4 = (x + 1)^{2} + 36$ .
Expand karne par: $- 14 x + 53 = 2 x + 37$ , isliye $16 x = 16$ aur $x = 1$ .
Check: dono distances $\sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$ hain.
Short answerMediumNCERT Exercise 7.1, Q8
P $(1, - 2)$ hai aur Q $(6, y)$ hai. Agar $P Q = 13$ units hai, to $y$ kya ho sakta hai?
Solution dikhao
Answer $y = 10$ ya $y = - 14$
$(6 - 1)^{2} + (y + 2)^{2} = 13^{2}$ , isliye $(y + 2)^{2} = 169 - 25 = 144$ .
$y + 2 = 12$ ya $y + 2 = - 12$ .
To $y = 10$ ya $y = - 14$ . Dono points P se 13 units door hain, ek upar aur ek neeche.
Long answerMediumNCERT Exercise 7.1, Q9
Point $A (0, 2)$ , $B (3, - 2)$ aur $C (x, 5)$ se equidistant hai. $x$ ki possible values aur lengths AC aur BC nikaalo.
Solution dikhao
Answer $x = \pm 4$ ; $A C = 5$
$A B = \sqrt{9 + 16} = 5$ .
$A C^{2} = x^{2} + 9$ , aur yeh 25 hona chahiye, isliye $x^{2} = 16$ aur $x = 4$ ya $x = - 4$ .
Dono cases mein $A C = A B = 5$ .
$C (4, 5)$ ke liye: $B C = \sqrt{1 + 49} = \sqrt{50} = 5 \sqrt{2}$ .
$C (- 4, 5)$ ke liye: $B C = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2}$ .
Short answerMediumNCERT Exercise 7.1, Q10
Woh equation likho jo har us point $(x, y)$ ko satisfy karti hai jo $(4, 1)$ se utna hi door hai jitna $(- 2, 5)$ se.
Solution dikhao
Answer $3 x - 2 y + 3 = 0$
$(x - 4)^{2} + (y - 1)^{2} = (x + 2)^{2} + (y - 5)^{2}$ .
Expand karne par: $- 8 x + 17 - 2 y = 4 x + 29 - 10 y$ .
To $- 12 x + 8 y = 12$ , yaani $3 x - 2 y + 3 = 0$ .
Midpoint $(1, 3)$ se check karo: $3 - 6 + 3 = 0$ .

7.3 Section Formula

Section 7.2 wale medicine drone par wapas aao. Uski battery 51 km ki flight ke zyadatar hisse tak chalti hai, lekin poori nahi. District H se V tak ke seedhe route par ek battery-swap halt S banana chahta hai, aisi jagah ki drone jitna aur udna hai usse dugna pehle hi ud chuka ho: $H S : S V = 2 : 1$ . Halt kahan hona chahiye?

Route ko teen barabar hisson mein socho. S doosre hisse ke end par hai, isliye woh H se V tak ke raaste ka $\frac{2}{3}$ hai. Seedhe route par x aur y saath-saath ek jaise badhte hain, isliye S par drone east wale 45 km ka $\frac{2}{3}$ aur north wale 24 km ka $\frac{2}{3}$ tay kar chuka hai. Isse $x = \frac{2}{3} \times 45 = 30$ aur $y = \frac{2}{3} \times 24 = 16$ milta hai, isliye halt $S (30, 16)$ par hona chahiye.

Hum safar ko is tarah kyun baant sakte hain? x-axis par perpendiculars SD aur VE giraao (Fig. 7.11). Chhota right triangle HDS aur bada right triangle HEV, H par wala angle share karte hain, isliye woh similar hain (AA, Chapter 6). Isliye chhote triangle ki har side bade triangle ki matching side ka $\frac{H S}{H V} = \frac{2}{3}$ hai:

\begin{aligned} \frac{H D}{H E} & = \frac{x}{45} = \frac{2}{3} \\ \frac{D S}{E V} & = \frac{y}{24} = \frac{2}{3} \end{aligned}

Distance formula se check karo: $H S = \sqrt{30^{2} + 16^{2}} = 34$ km aur $S V = \sqrt{15^{2} + 8^{2}} = 17$ km, aur 34 sach mein 17 ka dugna hai.

4/4 $\frac{x}{45} = \frac{2}{3}$ aur $\frac{y}{24} = \frac{2}{3}$ se $S (30, 16)$ milta hai.

Fig. 7.11Chhota triangle bade triangle ko

\frac{2}{3}

se scale karke bana hai.NCERT Fig. 7.9 · p. 106

Origin wale H se V(45, 24) tak drone ka route, jo chhote dots se teen barabar hisson mein kata hai, aur halt S doosre dot par hai taaki HS aur SV ka ratio 2 aur 1 ho. Step 2 mein S aur V se x-axis par D aur E tak dashed perpendiculars gire hain. Step 3 mein bade triangle HEV ke andar chhota triangle HDS blue rang mein shaded hai, note hai ki dono similar hain, aur HD par x, DS par y aur EV par 24 likha hai. Step 4 mein likha hai x by 45 barabar 2 by 3 aur y by 24 barabar 2 by 3.

Hum kehte hain ki S, HV ko ratio $2 : 1$ mein internally divide karta hai: yeh H aur V ke beech hai, aur $H S : S V = 2 : 1$ . Ab yahi kaam kinhi bhi do points aur kisi bhi ratio ke liye karte hain.

General formula

DerivationSection formula

Maan lo $A (x_{1}, y_{1})$ aur $B (x_{2}, y_{2})$ do points hain, aur $P (x, y)$ segment AB par is tarah hai ki $A P : P B = m_{1} : m_{2}$ , jahan $m_{1}$ aur $m_{2}$ positive numbers hain. Tab $A B = A P + P B$ kul $m_{1} + m_{2}$ parts hai, isliye P, A se B tak ke raaste ka $\frac{m_{1}}{m_{1} + m_{2}}$ hai.

A se hokar ek horizontal line aur B se hokar ek vertical line banao; woh corner $C (x_{2}, y_{1})$ par milti hain. P se hokar vertical line AC ko $D (x, y_{1})$ par milti hai (Fig. 7.12).

4/4Har side coordinates ka difference hai, isliye $\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{m_{1}}{m_{1} + m_{2}}$ .

Fig. 7.12Bade right triangle ke andar ek chhota right triangle, jo A par wala angle share karta hai.NCERT Fig. 7.10 · p. 106

A(x1, y1) se upar B(x2, y2) tak ek segment jis par P(x, y) hai; hissa AP par m1 aur hissa PB par m2 marked hai. Step 2 mein A se B ke neeche wale corner C tak dashed horizontal line, dashed vertical line CB, aur P se neeche AC par D tak dashed vertical line judi hai. Step 3 mein bade triangle ACB ke andar chhota triangle ADP blue mein shaded hai aur likha hai ki dono similar hain. Step 4 mein AD = x minus x1, DP = y minus y1, AC = x2 minus x1 aur CB = y2 minus y1 likha hai; step caption ratio deta hai: x minus x1 by x2 minus x1 barabar m1 by m1 plus m2.

Triangles ADP aur ACB mein D aur C par right angles hain aur dono A par wala angle share karte hain, isliye woh similar hain. Matching sides same ratio mein hoti hain:

\begin{aligned} \frac{A D}{A C} & = \frac{D P}{C B} \\ = \frac{A P}{A B} \\ = \frac{m_{1}}{m_{1} + m_{2}} \end{aligned}

Coordinates se sides padhne par, $A D = x - x_{1}$ , $A C = x_{2} - x_{1}$ , $D P = y - y_{1}$ aur $C B = y_{2} - y_{1}$ . Pehla ratio deta hai

\begin{array}{c} x - x_{1} \\ = \frac{m_{1} (x_{2} - x_{1})}{m_{1} + m_{2}} \end{array}

$m_{1} + m_{2}$ se multiply karo aur saaf karo: $(m_{1} + m_{2}) x = m_{1} x_{2} - m_{1} x_{1} + (m_{1} + m_{2}) x_{1} = m_{1} x_{2} + m_{2} x_{1}$ . Ratio $\frac{D P}{C B}$ se x ki jagah y ke saath yahi working milti hai. To P yeh point hai

Section formula

\begin{aligned} x & = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ y & = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{aligned}

(7.3)

jahan

$m_{1} : m_{2}$: ratio $A P : P B$
$(x_{1}, y_{1}), (x_{2}, y_{2})$: A aur B ke coordinates

Figure mein B, A se upar aur right mein hai, lekin formula kisi bhi position ke liye sahi hai: agar B, A ke left mein ho, to $x - x_{1}$ aur $x_{2} - x_{1}$ dono negative hote hain, aur unka ratio phir bhi $\frac{m_{1}}{m_{1} + m_{2}}$ hi rehta hai.

Yeh section formula (विभाजन सूत्र) hai. Ise ek recipe ki tarah padho: P ka har coordinate endpoints ke coordinates ka weighted average hai, aur har endpoint ko ratio ka woh hissa milta hai jo doosre endpoint ke paas baitha hai. Isliye $m_{1}$ , jo A ke paas wala hissa hai, B ke coordinates se multiply hota hai. Yahi wajah hai ki jab $m_{1}$ bada hissa ho, to P, B ke zyada paas hota hai.

Derivation se ek quick mental picture bhi milti hai. Agar $A P : P B = m_{1} : m_{2}$ hai, to AB ko $m_{1} + m_{2}$ barabar steps mein kaato. Har step x ko $\frac{x_{2} - x_{1}}{m_{1} + m_{2}}$ aur y ko $\frac{y_{2} - y_{1}}{m_{1} + m_{2}}$ se badalta hai, aur A se $m_{1}$ steps ke baad P aata hai. Example 7.6 apna answer check karne ke liye yahi use karta hai.

Kabhi-kabhi ratio hi unknown hota hai. Tab use $k : 1$ ki tarah likhna zyada saaf hai, jahan $k = \frac{m_{1}}{m_{2}}$ . (7.3) ke upar aur neeche ko $m_{2}$ se divide karne par yeh point milta hai

Ratio k : 1

\begin{aligned} x & = \frac{k x_{2} + x_{1}}{k + 1} \\ y & = \frac{k y_{2} + y_{1}}{k + 1} \end{aligned}

(7.4)

Midpoint

AB ka midpoint (मध्य-बिंदु) use ratio $1 : 1$ mein divide karta hai. Section formula mein $m_{1} = m_{2} = 1$ rakhne par milta hai

Midpoint formula

\begin{aligned} x & = \frac{x_{1} + x_{2}}{2} \\ y & = \frac{y_{1} + y_{2}}{2} \end{aligned}

(7.5)

Midpoint ka har coordinate bas endpoints ke coordinates ka average hai. Jaise, $(- 3, 8)$ aur $(5, 2)$ ka midpoint $(\frac{2}{2}, \frac{10}{2}) = (1, 5)$ hai.

InteractiveSection lab

Dono axes par minus 8 se 8 tak ka grid, jismein do draggable points A aur B hain jo whole numbers par snap hote hain; keyboard se kisi point par focus karke arrow keys use karo. Plus aur minus buttons ratio m1 aur m2 ko 1 se 9 tak set karte hain. Segment AB chhote dots se m1 plus m2 barabar steps mein kata hai, aur P, A se m1 steps par hai. Ek panel numbers rakhkar section formula, P ke exact coordinates fractions mein, aur lengths AP aur PB unke ratio ke saath dikhata hai. Buttons midpoint (1 aur 1) set karte hain aur ratio ko swap karte hain, aur Hide P switch answer ko tab tak chhupata hai jab tak tum Reveal na dabao.

Fig. 7.13A aur B ko drag karo,

m_{1}

aur

m_{2}

set karo, aur P aur uske coordinates dekho. Pehle khud ko test karne ke liye Hide P on karo.

Quick check

3 mein se 1

Kaun sa point

A (1, 2)

B (7, 8)

tak ke segment ko ratio

1 : 2

mein divide karta hai?

(3, 4)

(5, 6)

(4, 5)

(3, 6)

Answer dekho

Answer (A) $(3, 4)$

x = \frac{1 (7) + 2 (1)}{3} = 3

aur

y = \frac{1 (8) + 2 (2)}{3} = 4

. Point

(5, 6)

har endpoint ko ratio ke uske apne hisse se multiply karne par aata hai; yeh AB ko

2 : 1

mein divide karta hai. Point

(4, 5)

midpoint hai.

Section formula se problems solve karna

Example 7.6NCERT Example 6 · p. 108

Diye gaye ratio mein divide karna

$A (- 2, 7)$ aur $B (6, - 5)$ ko jodne wale segment ko ratio $3 : 1$ mein internally divide karne wala point nikaalo.

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

3/3A se teen steps $P (4, - 2)$ tak pahunchte hain.

Fig. 7.14Ratio

3 : 1

ke liye A se 4 barabar steps mein se 3 chalo.

Ek grid jo across minus 3 se 7 tak aur upar minus 6 se 8 tak hai, jismein A(minus 2, 7), B(6, minus 5) se juda hai. Step 1 mein segment aur ratio 3 aur 1 hai. Step 2 mein ise 4 barabar steps mein kaata gaya hai, jo har ek 2 right aur 3 neeche wali dashed seedhi ki tarah bane hain, aur division points par dots hain. Step 3 mein pehle teen steps highlight hain aur jis point P tak woh pahunchte hain us par gol ghera hai, (minus 2 + 6, 7 minus 9) = (4, minus 2).

Example 7.7NCERT Example 7 · p. 108

Ratio nikaalna

Point $P (1, 4)$ , $A (- 3, 8)$ aur $B (9, - 4)$ ko jodne wale segment ko kis ratio mein divide karta hai?

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

Tum $m_{1} : m_{2}$ se bhi shuru kar sakte the: x-equation $1 = \frac{9 m_{1} - 3 m_{2}}{m_{1} + m_{2}}$ se $8 m_{1} = 4 m_{2}$ milta hai, isliye phir se $m_{1} : m_{2} = 1 : 2$ . Aur agar bataya gaya ho ki teenon points ek line par hain, to distances AP aur PB ko compare karke bhi ratio mil jaata hai.

Example 7.8NCERT Example 8 · p. 109

Points of trisection

$A (- 4, 5)$ aur $B (5, - 1)$ ko jodne wale segment ke points of trisection (समत्रिभाजन बिंदु) nikaalo, yaani woh do points jo use teen barabar hisson mein kaatte hain.

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

Example 7.9NCERT Example 9 · p. 110

Jahan axis segment ko kaatta hai

x-axis, $A (5, - 2)$ aur $B (- 1, 6)$ ko jodne wale segment ko kis ratio mein divide karta hai? Woh point nikaalo jahan yeh AB ko kaatta hai.

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

Fig. 7.15 ek kaam ka check dikhata hai. A aur B se x-axis tak giri verticals, jinki lengths 2 aur 6 hain, do similar right triangles banati hain, isliye x-axis AB ko ratio $2 : 6 = 1 : 3$ mein kaatta hai. General taur par, jab A aur B x-axis ke opposite sides par hon, to yeh AB ko ratio $∣ y_{1} ∣ : ∣ y_{2} ∣$ mein divide karta hai. Isse apna answer check karo, lekin exam mein section-formula wali working dikhao.

3/3Heights 2 aur 6 se $A P : P B = 2 : 6 = 1 : 3$ milta hai.

Fig. 7.15Axis ke dono taraf heights 2 aur 6 se ratio

1 : 3

milta hai.

Ek grid jo across minus 2 se 6 tak aur upar minus 3 se 7 tak hai, jismein A(5, minus 2) x-axis ke neeche aur B(minus 1, 6) uske upar hai, dono ek segment se jude hain. Step 2 mein us point P par gol ghera hai jahan segment x-axis ko cross karta hai, jahan y 0 hai. Step 3 mein A se 2 length ki blue vertical aur B se 6 length ki orange vertical x-axis tak giraayi gayi hai, do similar right triangles shaded hain, aur nateeja hai ki AP aur PB ka ratio 2 aur 6 hai, yaani 1 aur 3.

Example 7.10NCERT Example 10 · p. 110

Parallelogram ka missing vertex

Parallelogram ABCD ke teen vertices, order mein, $A (- 2, - 1)$ , $B (4, 0)$ aur $C (5, 4)$ hain. Chautha vertex D nikaalo.

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

4/4Sides jodo: ABCD ek parallelogram hai.

Fig. 7.16Diagonals ka midpoint same hai, isliye D, M ke aage utna hi door hai jitna B, M se pehle hai.

Ek grid jo across minus 3 se 6 tak aur upar minus 2 se 5 tak hai, jismein A(minus 2, minus 1), B(4, 0) aur C(5, 4) do sides se jude hain. Step 2 mein dashed diagonal AC bana hai aur uske midpoint M par (1.5, 1.5) par gol ghera hai. Step 3 mein B se M hote hue utni hi doori aage tak ek dotted line bani hai, jo D(minus 1, 3) par khatam hoti hai. Step 4 mein parallelogram ABCD poora aur shaded hai.

Quick check

3 mein se 1

x-axis,

A (1, - 4)

aur

B (3, 2)

ko jodne wale segment ko kis ratio mein divide karta hai?

2 : 1

1 : 2

1 : 1

4 : 3

Answer dekho

Answer (A) $2 : 1$

Ratio

k : 1

ke saath, y-coordinate

\frac{2 k - 4}{k + 1}

zero hai, isliye

k = 2

: ratio

2 : 1

hai. Heights wala check bhi yahi kehta hai:

∣ - 4 ∣ : ∣ 2 ∣ = 2 : 1

. Answer

1 : 2

ratio ko ulta kar deta hai.

Practice · 10 questions

Practice 7.2: answers ke saath

Short answerEasyNCERT Exercise 7.2, Q1
$(- 3, 8)$ aur $(7, - 2)$ ko jodne wale segment ko ratio $3 : 2$ mein internally divide karne wala point nikaalo.
Solution dikhao
Answer $(3, 2)$
$x = \frac{3 (7) + 2 (- 3)}{5} = \frac{15}{5} = 3$ .
$y = \frac{3 (- 2) + 2 (8)}{5} = \frac{10}{5} = 2$ .
Short answerEasyNCERT Exercise 7.2, Q2
$(- 5, 2)$ aur $(7, - 4)$ ko jodne wale segment ke points of trisection nikaalo.
Solution dikhao
Answer $(- 1, 0)$ aur $(3, - 2)$
Ratio $1 : 2$ : $x = \frac{7 + 2 (- 5)}{3} = - 1$ aur $y = \frac{- 4 + 2 (2)}{3} = 0$ .
Ratio $2 : 1$ : $x = \frac{2 (7) - 5}{3} = 3$ aur $y = \frac{2 (- 4) + 2}{3} = - 2$ .
Check: $(3, - 2)$ , $(- 1, 0)$ aur $(7, - 4)$ ka midpoint hai.
Long answerMediumNCERT Exercise 7.2, Q3
School hall ka farsh 1 m tiles ka grid hai. Aage wale left corner ko origin lo, x ko front wall ke across aur y ko 60 m lambi side wall ke saath neeche ki taraf naapo. Annual day ke liye microphone stand 3 m across aur hall ke ek-chauthaai neeche hai. Camera stand 9 m across aur 24 m neeche hai. Dono stands ek doosre se kitni door hain? Ek speaker box bilkul dono ke beech mein rakhna hai. Kahan?
Solution dikhao
Answer $3 \sqrt{13}$ m door; speaker $(6, 19.5)$ par
60 m ka ek-chauthaai 15 m hai, isliye microphone $(3, 15)$ par hai; camera $(9, 24)$ par hai.
Distance $= \sqrt{6^{2} + 9^{2}} = \sqrt{117} = 3 \sqrt{13} \approx 10.8$ m.
Beech ka point: $(\frac{3 + 9}{2}, \frac{15 + 24}{2}) = (6, 19.5)$ , yaani 6 m across aur 19.5 m neeche.
Short answerMediumNCERT Exercise 7.2, Q4
Point $(2, - 1)$ , $(- 4, 5)$ aur $(6, - 5)$ ko jodne wale segment ko kis ratio mein divide karta hai?
Solution dikhao
Answer $3 : 2$
$\frac{6 k - 4}{k + 1} = 2$ se $6 k - 4 = 2 k + 2$ milta hai, isliye $k = \frac{3}{2}$ .
Ratio $\frac{3}{2} : 1 = 3 : 2$ hai.
y se check karo: $\frac{- 5 k + 5}{k + 1} = \frac{- 2.5}{2.5} = - 1$ .
Short answerMediumNCERT Exercise 7.2, Q5
y-axis, $A (- 2, 3)$ aur $B (6, - 5)$ ko jodne wale segment ko kis ratio mein divide karta hai? Point of division nikaalo.
Solution dikhao
Answer $1 : 3$ ; $(0, 1)$
Ratio $k : 1$ ke saath: $\frac{6 k - 2}{k + 1} = 0$ , isliye $k = \frac{1}{3}$ aur ratio $1 : 3$ hai.
$y = \frac{1 (- 5) + 3 (3)}{4} = \frac{4}{4} = 1$ .
Shortcut se check karo: $∣ x_{1} ∣ : ∣ x_{2} ∣ = 2 : 6 = 1 : 3$ .
Short answerMediumNCERT Exercise 7.2, Q6
Is order mein jodne par $(- 1, 0)$ , $(3, y)$ , $(x, 5)$ aur $(1, 4)$ ek parallelogram banate hain. $x$ aur $y$ nikaalo.
Solution dikhao
Answer $x = 5$ , $y = 1$
Pehle diagonal ka midpoint: $(\frac{x - 1}{2}, \frac{5}{2})$ .
Doosre diagonal ka midpoint: $(\frac{4}{2}, \frac{y + 4}{2})$ .
To $x - 1 = 4$ , jisse $x = 5$ , aur $y + 4 = 5$ , jisse $y = 1$ .
Short answerEasyNCERT Exercise 7.2, Q7
AB ek circle ka diameter hai jiska centre $(- 1, 2)$ hai. Agar B $(3, - 4)$ hai, to A nikaalo.
Solution dikhao
Answer $A (- 5, 8)$
Maan lo $A = (p, q)$ . Tab $\frac{p + 3}{2} = - 1$ aur $\frac{q - 4}{2} = 2$ .
To $p = - 5$ aur $q = 8$ .
Short answerMediumNCERT Exercise 7.2, Q8
A $(- 3, 4)$ hai aur B $(7, - 1)$ hai. Segment AB par P aisa nikaalo ki $A P = \frac{2}{5} A B$ .
Solution dikhao
Answer $P (1, 2)$
$A P = \frac{2}{5} A B$ ka matlab hai $P B = \frac{3}{5} A B$ , isliye $A P : P B = 2 : 3$ .
$x = \frac{2 (7) + 3 (- 3)}{5} = 1$ aur $y = \frac{2 (- 1) + 3 (4)}{5} = 2$ .
Short answerMediumNCERT Exercise 7.2, Q9
Woh points nikaalo jo $A (- 4, 1)$ aur $B (4, 7)$ ko jodne wale segment ko chaar barabar hisson mein baantte hain.
Solution dikhao
Answer $(- 2, 2.5)$ , $(0, 4)$ , $(2, 5.5)$
Chaar barabar steps, har ek $\frac{8}{4} = 2$ across aur $\frac{6}{4} = 1.5$ upar.
A se ek step: $(- 2, 2.5)$ . Do steps: $(0, 4)$ , yaani midpoint. Teen steps: $(2, 5.5)$ .
Aakhri wale ko ratio $3 : 1$ se check karo: $x = \frac{12 - 4}{4} = 2$ aur $y = \frac{21 + 1}{4} = 5.5$ .
Long answerHardNCERT Exercise 7.2, Q10
Ek rhombus ke vertices, order mein, $(5, 3)$ , $(0, 3)$ , $(- 3, - 1)$ aur $(2, - 1)$ hain. Iska area nikaalo. (Rhombus ke liye dono diagonals ko multiply karo aur result ka aadha lo.)
Solution dikhao
Answer20 square units
$(5, 3)$ se $(- 3, - 1)$ tak ka diagonal: $\sqrt{64 + 16} = \sqrt{80} = 4 \sqrt{5}$ .
$(0, 3)$ se $(2, - 1)$ tak ka diagonal: $\sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$ .
Area $= \frac{1}{2} \times 4 \sqrt{5} \times 2 \sqrt{5} = \frac{1}{2} \times 40 = 20$ square units hai.

7.4 Summary

Ek segment poore chapter ko dikha sakta hai. Fig. 7.17 mein, wahi do points ek distance, ek dividing point aur ek midpoint dete hain.

3/3Chaar mein se do steps: midpoint $M (1, 2)$ .

Fig. 7.17Wahi do points se distance, section point aur midpoint.

Ek grid jismein A(minus 3, minus 1) aur B(5, 5) ek blue segment se jude hain, aur 8 across aur 6 upar ke dashed legs right angle par milte hain. Step 1 mein AB = 8 squared plus 6 squared ka square root = 10 diya hai. Step 2 mein woh teen points marked hain jo AB ko chaar barabar hisson mein kaatte hain, aur P(minus 1, 0.5) par gol ghera hai, jo AB ko ratio 1 aur 3 mein divide karta hai. Step 3 mein midpoint M(1, 2) par bhi gol ghera hai, jo AB ko ratio 1 aur 1 mein divide karta hai.

Recap

Summary aur key points

x-axis par point ka form $(x, 0)$ hota hai; y-axis par point ka form $(0, y)$ hota hai (Section 7.1).
Ek hi axis par points ke beech $∣ x_{2} - x_{1} ∣$ ya $∣ y_{2} - y_{1} ∣$ ka distance hota hai (Axis par points).
Distance formula: $P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ . Yeh ek horizontal aur ek vertical leg wale right triangle par Pythagoras theorem hi hai ((7.1)).
Origin se $(x, y)$ ka distance $\sqrt{x^{2} + y^{2}}$ hai ((7.2)).
Teen points collinear hote hain jab do chhote distances ka sum sabse lambe ke barabar ho; warna woh triangle banate hain (Distance formula ka use).
Sides aur diagonals se squares, rhombuses, rectangles aur parallelograms ko alag pehchaana jaata hai (comparison table).
A aur B se equidistant points $P A^{2} = P B^{2}$ ko satisfy karte hain, jo ek linear equation hai: AB ka perpendicular bisector (Example 7.4).
Section formula: AB ko ratio $m_{1} : m_{2}$ mein internally divide karne wale point ka $x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}$ aur $y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$ hota hai ((7.3)).
Unknown ratio ho to use $k : 1$ likho, ek coordinate se $k$ nikaalo aur doosre se check karo (Example 7.7).
Midpoint: yeh $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ hota hai ((7.5)).
Points of trisection AB ko $1 : 2$ aur $2 : 1$ mein divide karte hain; koi axis AB ko wahan kaatta hai jahan woh coordinate 0 ho jaata hai (Example 7.8, Example 7.9).
Parallelogram ke diagonals ka midpoint same hota hai, aur circle ka centre har diameter ka midpoint hota hai (Example 7.10).

Recap check

3 mein se 1

Segment AB ke baare mein har question ko us tool se match karo jo uska answer deta hai.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

AB kitna lamba hai? : distance formula
Diameter AB wale circle ka centre kahan hai? : midpoint formula
AB, y-axis ko kahan cross karta hai? : $k : 1$ ke saath section formula
Kaun se points A se utne hi door hain jitne B se? : $P A^{2} = P B^{2}$

Length ke liye distance formula chahiye; centre ek midpoint hai; crossing point ke liye unknown ratio chahiye; barabar distances se perpendicular bisector milta hai.

Flash cards · 7

Key terms

1 / 7

Board exam ki practice

Yeh questions CBSE Class 10 board paper ke formats follow karte hain: one-mark multiple-choice aur assertion–reason items, short aur long answers, aur ek case study. Har question ka solution kholne se pehle use khud try karo.

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 7.1, Q1	Practice 7.1 · Q1
NCERT Exercise 7.1, Q2	Practice 7.1 · Q2
NCERT Exercise 7.1, Q3	Practice 7.1 · Q3
NCERT Exercise 7.1, Q4	Practice 7.1 · Q4
NCERT Exercise 7.1, Q5	Practice 7.1 · Q5
NCERT Exercise 7.1, Q6	Practice 7.1 · Q6
NCERT Exercise 7.1, Q7	Practice 7.1 · Q7
NCERT Exercise 7.1, Q8	Practice 7.1 · Q8
NCERT Exercise 7.1, Q9	Practice 7.1 · Q9
NCERT Exercise 7.1, Q10	Practice 7.1 · Q10
NCERT Exercise 7.2, Q1	Practice 7.2 · Q1
NCERT Exercise 7.2, Q2	Practice 7.2 · Q2
NCERT Exercise 7.2, Q3	Practice 7.2 · Q3
NCERT Exercise 7.2, Q4	Practice 7.2 · Q4
NCERT Exercise 7.2, Q5	Practice 7.2 · Q5
NCERT Exercise 7.2, Q6	Practice 7.2 · Q6
NCERT Exercise 7.2, Q7	Practice 7.2 · Q7
NCERT Exercise 7.2, Q8	Practice 7.2 · Q8
NCERT Exercise 7.2, Q9	Practice 7.2 · Q9
NCERT Exercise 7.2, Q10	Practice 7.2 · Q10

Quick check: MCQs

7.1 Introduction3 questions
Koi bhi do points3 questions
Distance formula ka use4 questions
Midpoint3 questions
Section formula se problems solve karna3 questions
Recap check3 questions

Glossary · 11

Key terms (English + Hindi)

abscissaभुजbhuj: x-coordinate. Yeh batata hai ki point y-axis se kitna right (positive) ya left (negative) hai.Chapter mein pehli baar yahan aaya →

collinearसंरेखीsanrekhi: Ek hi straight line par hona. Teen points ke liye, do chhote distances ka sum sabse lambe ke barabar hota hai.Chapter mein pehli baar yahan aaya →
coordinate geometryनिर्देशांक ज्यामितिnirdeshank jyamiti: Coordinates ki madad se geometry ki padhai, taaki points, lines aur shapes ko algebra se handle kiya ja sake.Chapter mein pehli baar yahan aaya →

distance formulaदूरी सूत्रdoori sutra: Do points ke beech kitni doori hai, unke coordinates se nikaali hui: $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .Chapter mein pehli baar yahan aaya →

equidistantसमदूरस्थsamdoorasth: Do ya zyada diye gaye points se barabar door.Chapter mein pehli baar yahan aaya →

internal divisionअंतः विभाजनantah vibhajan: A aur B ke beech ka point P, jahan $A P : P B = m_{1} : m_{2}$ , AB ko us ratio mein internally kaatta hai.Chapter mein pehli baar yahan aaya →

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

ordinateकोटिkoti: y-coordinate. Yeh batata hai ki point x-axis se kitna upar (positive) ya neeche (negative) hai.Chapter mein pehli baar yahan aaya →

perpendicular bisectorलंब समद्विभाजकlamb samadvibhajak: Segment ke midpoint se hokar uske right angle par jaane wali line. Iske points dono ends se equidistant hote hain.Chapter mein pehli baar yahan aaya →
points of trisectionसमत्रिभाजन बिंदुsamtribhajan bindu: Woh do points jo segment ko teen barabar hisson mein kaatte hain, use $1 : 2$ aur $2 : 1$ mein divide karte hue.Chapter mein pehli baar yahan aaya →

section formulaविभाजन सूत्रvibhajan sutra: AB ko ratio $m_{1} : m_{2}$ mein internally divide karne wale point ke coordinates: $x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}$ , aur y ke liye bhi isi tarah.Chapter mein pehli baar yahan aaya →

Coordinate Geometry: Class 10 Maths Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Introduction

7.2 Distance Formula

Axis par points

Koi bhi do points

Distance formula ka use

Practice 7.1: answers ke saath

7.3 Section Formula

General formula

Midpoint

Section formula se problems solve karna

Practice 7.2: answers ke saath

7.4 Summary

Summary aur key points

Board exam ki practice

Practice questions answers ke saath (NCERT “Board exam ki practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye