Chapter ke text par jao
MathematicsCh 9 Straight Lines

Straight Lines: Class 11 Maths Chapter 9 Hinglish mein

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

Published · Updated

  • 75 min
  • 1 simulation
  • 75 practice questions
  • 20 MCQs

Is chapter mein aap kya seekhenge

  • Distance, section, midpoint aur triangle-area formulas yaad karna aur lagaana, collinear points ke liye zero-area test samet.
  • Kisi angle se ya line ke do points se uska inclination aur slope nikaalna, aur horizontal aur vertical lines ko pehchaanna.
  • Slopes se decide karna ki do lines parallel hain ya perpendicular, aur teen points collinear hain ya nahi.
  • Do lines ke slopes se unke beech ka acute angle nikaalna, aur diye gaye angle se missing slope nikaalna.
  • Line ki equation point-slope, two-point, slope-intercept, intercept aur general form mein likhna, aur ek form se doosre mein badalna.
  • Kisi point ki line se distance aur do parallel lines ke beech ki distance nikaalna.
  • Concurrent lines, line mein image, kisi direction mein distance aur straight line wale loci par sawaal solve karna.

NCERT Mathematics Class 11, Chapter 9, pp. 151–175 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

9.1 Introduction

Graph paper par ek straight line kheencho to tumhare paas ek picture hai. y=2x+1 likho to wahi cheez algebra ke ek sentence ke roop mein hai. Coordinate geometry in dono ke beech aane-jaane ki kala hai: har point numbers ka ek pair ban jaata hai, har curve ek equation ban jaata hai, aur phir kai geometric sawaal calculation se suljhaaye ja sakte hain.

René Descartes ne yeh programme La Géométrie (1637) mein rakha, aur Pierre de Fermat unhi saalon mein alag se milte-julte ideas tak pahunche. Curves ko unki equations ke zariye padhna ab analytical geometry kehlaata hai. Pichhli classes mein tumne points plot kiye aur distances, section points aur areas nikaale. Yeh chapter un tools ko sabse simple curve, yaani straight line, par lagaata hai, aur sabse zyada kaam karne wala number hai uska slope.

Pichhli classes ke tools

Point (x,y) y-axis se ∣x∣ units aur x-axis se ∣y∣ units door hota hai, aur signs batate hain ki kis taraf. To (−5,2) y-axis se 5 units left aur x-axis se 2 units upar hai, jabki (0,−3) khud y-axis par baitha hai. Fig. 9.1 un chaar results ko dikhata hai jo hum baar-baar use karte hain.

2/4Legs 8 aur 6 wala right triangle AB=10 deta hai.

Fig. 9.1Points ka ek pair, chaar purane formulas.NCERT Fig. 9.1 · p. 151

Ek coordinate grid jo across −4 se 6 aur upar −3 se 5 tak hai. Step 1 mein A(−3, −2) aur B(5, 4) plot hote hain aur ek blue segment se jude hain. Step 2 mein length 8 ka horizontal leg aur length 6 ka vertical leg judta hai jo (5, −2) par right angle banaate hain, to AB = 10. Step 3 mein AB par do barabar tick marks ke saath midpoint M(1, 1) mark hota hai. Step 4 mein C(5, −2) ke saath triangle ABC shade hota hai aur uska area 6 guna 8 ka aadha, yaani 24, diya gaya hai.

Distance. P(x1,y1) aur Q(x2,y2) ko jodo aur ek right triangle poora karo jiske legs axes ke parallel hon. Legs ∣x2−x1∣ aur ∣y2−y1∣ hain, to Pythagoras' theorem se milta hai

distance formula
PQ2=(x2−x1)2+(y2−y1)2
(9.1)

(1,−2) aur (4,2) ke liye PQ2=9+16=25 milta hai, to PQ=5.

Section formula. Jo point R, PQ ko P se gin kar ratio m:n mein internally divide karta hai, uske liye

section formula
x=mx2+nx1m+ny=my2+ny1m+n
(9.2)

Jaise, (2,1) aur (8,10) ke join ko ratio 1:2 mein divide karne wala point (8+43,10+23)=(4,4) hai. m=n lene par formula midpoint (x1+x22,y1+y22) deta hai.

Triangle ka area. Vertices (x1,y1), (x2,y2) aur (x3,y3) ke liye area Δ hai

triangle ka area
Δ=12 ∣ x1(y2−y3)+ x2(y3−y1)+ x3(y1−y2) ∣
(9.3)

Modulus area ko positive rakhta hai, chahe tum vertices kisi bhi order mein likho. (1,1), (4,3) aur (2,6) ke liye bracket −3+20−4=13 hai, to area 6.5 square units hai. Agar bracket 0 hai, to triangle pichak gaya hai: teenon points collinear hain.

SimulationDistances aur collinear points

Dono axes par −6 se 6 tak ka coordinate grid jismein teen draggable points A(−4, −2), B(2, 1) aur C(4, 2) hain; har point apne coordinates dikhata hai aur whole numbers par snap hota hai. Distances AB, BC aur AC kheenchi aur naapi jaati hain. Shuru mein teenon points ek line par hain, to AB + BC, AC ke barabar hai; kisi bhi point ko us line se hataane par AB + BC, AC se bada ho jaata hai.

Fig. 9.2Points ko drag karo aur teenon distances dekho. Jab B segment AC par hota hai, to do chhoti distances jud kar lambi distance ke barabar ho jaati hain.
Try karo
  • Shuru mein check karo ki AB+BC=AC hai. Ab C ko ek square upar drag karo. Sum ka kya hota hai?
  • A ko (−4,−2) par rakho aur B ko tab tak hilao jab tak AB=5 na ho jaaye. Tumne kaun sa right triangle use kiya?
  • B ko AC ke midpoint par rakho. Kya uske coordinates midpoint formula se match karte hain?
Quick check
3 mein se 1
(−2,3) aur (4,−5) ke beech ki distance nikaalo.
Answer dekho

Answer 10

Legs ∣4−(−2)∣=6 aur ∣−5−3∣=8 hain, to (9.1) se distance 36+64=10 hai.

9.2 Line ka Slope

Do seedhi sadakein bilkul alag mehsoos ho sakti hain: ek dheere chadhti hai, doosri tez chadhti hai, teesri neeche utarti hai. Ek number is jhukaav ko pakad leta hai. Hum ise line aur x-axis ke beech ke angle se banaate hain.

Jhukaav ko ek angle se naapo. Wahan khade ho jahan line x-axis ko kaatti hai, positive x-direction ki taraf mooh karo, aur anticlockwise tab tak ghoomo jab tak line ke saath mooh na ho jaaye. Jitna angle tum ghoome, θ, woh line ka inclination hai. Poora straight angle ghoomne se tum wapas usi line par aa jaoge, isliye 0∘≤θ<180∘. Horizontal line ke liye ghoomna nahi padta (θ=0∘) aur vertical line ke liye quarter turn chahiye (θ=90∘).

Definition
Slope

Agar line ka inclination θ hai aur θ≠90∘, to uska slope (ya gradient) m=tan⁡θ hai. Vertical line ka koi slope nahi hota.

ExampleInclination 45∘ wali line ka slope tan⁡45∘=1 hai.

slope
m=tan⁡θ,θ≠90∘
(9.4)

Slope (ढाल) ka sign batata hai ki left se right jaate hue line kis taraf jaati hai. Positive slope ka matlab line chadhti hai (θ acute), negative slope ka matlab girti hai (θ obtuse), aur slope 0 ka matlab woh level rehti hai. To x-axis ka slope 0 hai, jabki y-axis ka koi slope hi nahi hai kyunki tan⁡90∘ undefined hai. Fig. 9.3 ek line ko chaaron cases se ghumata hai.

2/4θ=125∘: line girti hai aur m<0.

Fig. 9.3Inclination positive x-axis se anticlockwise naapa jaata hai.NCERT Fig. 9.2 · p. 153

Axes ke saath point K(1, 0) se guzarti ek blue line. Step 1: line positive x-axis ke saath 35 degrees ka acute angle banati hai, jo ek arc se mark hai, aur uska slope tan 35° lagbhag 0.70 hai. Step 2: line 125 degrees tak ghoom jaati hai, arc ab obtuse angle dikhata hai, aur slope lagbhag −1.43 hai. Step 3: line x-axis ke saath padi hai, inclination 0 aur slope 0. Step 4: line right-angle mark ke saath vertical khadi hai; tan 90° undefined hai, isliye line ka koi slope nahi.

9.2.1 Do points ke coordinates se slope

Do points ek line ko fix kar dete hain, to unhe uska slope bhi fix karna chahiye. Ek non-vertical line par P(x1,y1) aur Q(x2,y2) lo, taaki x1≠x2. P se Q jaate hue horizontal change x2−x1 ko run aur vertical change y2−y1 ko rise kaho (Fig. 9.4).

DerivationSlope yaani rise upon run

Line chadhti hai (θ acute). Q ko P ke right mein aur uske upar rakho, aur M(x2,y1) se right triangle PMQ poora karo. Kyunki PM x-axis ke parallel hai, triangle ka P par angle θ ke barabar hai (corresponding angles). To

tan⁡θ=MQPM=y2−y1x2−x1

Line girti hai (θ obtuse). Ab Q, P ke right mein aur uske neeche hai. Triangle PMQ ka P par angle 180∘−θ hai aur vertical side MQ=y1−y2 hai. tan⁡(180∘−θ)=−tan⁡θ use karke,

tan⁡θ=−y1−y2x2−x1=y2−y1x2−x1

Dono cases ek hi rule dete hain. P ko Q aur Q ko P naam dene se dono differences ka sign palat jaata hai, to ratio wahi rehta hai.

do points se slope
m=y2−y1x2−x1
(9.5)
jahan
x1≠x2
line vertical nahi hai

1/3Rise 3 upon run 5: m=35.

Fig. 9.4Slope rise divided by run hai, chahe tum kisi bhi point se shuru karo.NCERT Fig. 9.3 · p. 153

Step 1: ek blue line P(−2, −1) aur Q(3, 2) se hokar chadhti hai; uske neeche ek right triangle hai jiska horizontal run x2 − x1 aur vertical rise y2 − y1 likha hai, jisse slope 3/5 milta hai. Step 2: ek dashed line P(−2, 2) se Q(3, −2) tak girti hai; run 5 hai aur vertical side par arrow neeche ki taraf hai, yaani rise −4, to slope −4/5 hai. Step 3: P aur Q ke naam badal diye jaate hain aur arrow upar ki taraf ho jaata hai; ab run −5 aur rise 4 hai, aur slope phir bhi −4/5 hai.

9.2.2 Parallel aur perpendicular lines ki conditions

Do lines l1 aur l2 lo, jinmein se koi vertical nahi hai. Unke slopes ko m1, m2 aur inclinations ko α, β kaho.

PropertyParallel lines

Do non-vertical lines tabhi aur sirf tabhi parallel hoti hain jab unke slopes barabar hon: m1=m2.

PropertyPerpendicular lines

Do non-vertical lines tabhi aur sirf tabhi perpendicular hoti hain jab m1m2=−1, yaani jab m2=−1m1.

Fig. 9.5 dikhata hai ki −1 kahan se aata hai. Run-aur-rise wale triangle ko 90∘ ghumao: run rise ban jaata hai, rise ulti direction mein run ban jaata hai, aur slope 23 badal kar −32 ho jaata hai. Perpendicular lines ke slopes negative reciprocals hote hain.

3/323×(−32)=−1.

Fig. 9.5Barabar slopes ka matlab parallel; jin slopes ka product −1 ho unka matlab perpendicular.NCERT Fig. 9.5 · p. 155

Step 1: do parallel lines, ek solid aur ek dashed; har ek par run 3 aur rise 2 wala slope triangle hai, to dono slopes 2/3 hain. Step 2: origin se guzarti ek dotted line solid line ke right angle par dikhti hai, aur solid line wala triangle origin ke charon taraf 90 degrees anticlockwise ghoom kar us par aa jaata hai; uska run 3 rise 3 ban jaata hai aur rise 2 left ki taraf run 2 ban jaata hai, jisse slope −3/2 milta hai. Step 3: origin par right-angle mark aur product (2/3)(−3/2) = −1.

Collinear points. Points A, B aur C tabhi aur sirf tabhi ek line par hote hain jab AB ka slope BC ke slope ke barabar ho (ya dono segments vertical hon). Barabar slopes segments ko parallel banaate hain, aur kyunki dono mein point B common hai, woh ek hi line par hote hain. Yeh test aksar (9.3) wale area test se jaldi ho jaata hai.

Example 9.1NCERT Example 1 · p. 155

Chaar slopes

Har line ka slope nikaalo, jahan woh exist karta ho: (a) (1,−3) aur (5,7) se guzarti line; (b) (1,−3) aur (−6,−3) se guzarti line; (c) (1,−3) aur (1,8) se guzarti line; (d) inclination 150∘ wali line.

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

Quick check
4 mein se 1
Ek line ka inclination 120∘ hai. Uska slope hai
Answer dekho

Answer (B) −3

tan⁡120∘=−tan⁡60∘=−3. Slope negative hai kyunki inclination obtuse hai. −13 to tan⁡150∘ hoga, aur 12 mein tan⁡ aur cos⁡ mix ho gaye hain.

9.2.3 Do lines ke beech ka angle

Jahan do lines kaatti hain, wahan chaar angles bante hain lekin size sirf do: opposite angles barabar hote hain aur paas wale supplementary. To ek hi angle θ sab kuch tay kar deta hai, aur uska padosi ϕ=180∘−θ hai.

DerivationAngle formula

Maan lo L1 aur L2 ke slopes m1=tan⁡α1 aur m2=tan⁡α2 hain, α1<α2, aur koi bhi 90∘ nahi hai (Fig. 9.6). Lines x-axis ke saath jo triangle banati hain, usmein α2 ek exterior angle hai. To α2=α1+θ, jahan θ crossing point par us triangle ka angle hai, aur tan⁡θ=tan⁡(α2−α1). tan⁡(A−B) ke formula se expand karo aur tan⁡α1=m1, tan⁡α2=m2 rakho:

tan⁡θ=m2−m11+m1m2

Iske liye 1+m1m2≠0 chahiye; agar yeh 0 hai, to lines perpendicular hain. Padosi angle ke liye tan⁡ϕ=tan⁡(180∘−θ)=−tan⁡θ. tan⁡θ aur tan⁡ϕ mein se ek positive hota hai, aur positive tangent wala angle acute hota hai. Modulus lene se wahi chun liya jaata hai.

do lines ke beech ka acute angle
tan⁡θ=∣m2−m11+m1m2∣
(9.6)
jahan
θ
acute angle; obtuse angle 180∘−θ hai
1+m1m2
non-zero (warna lines perpendicular hain)

2/3θ=α2−α1; yahan tan⁡θ=3−1/21+3/2=1.

Fig. 9.6Exterior-angle wali baat se θ=α2−α1 milta hai.NCERT Fig. 9.6 · p. 156

Line L1 (solid, slope 1/2) x-axis ko (−4, 0) par kaatti hai aur line L2 (dashed, slope 3) use (1, 0) par kaatti hai; dono (2, 3) par milti hain. Step 1: x-axis ke do crossing points par arcs inclinations alpha 1 aur alpha 2 mark karte hain. Step 2: milne wale point par ek arc theta mark karta hai, jo wahan triangle ka angle hai, aur theta = alpha 2 − alpha 1 aur tan theta = 1. Step 3: ek doosra arc padosi angle phi mark karta hai; theta 45 degrees hai aur phi 135 degrees.

Example 9.2NCERT Example 2 · p. 157

Angle se slope nikaalna

Do lines 45∘ par milti hain, aur unmein se ek ka slope 2 hai. Doosri line ke kaun se slopes ho sakte hain?

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

Do answers kyun? Kisi bhi point se do lines aisi guzarti hain jo di gayi line ke saath 45∘ banati hain, uske dono taraf ek-ek (Fig. 9.7). Check: ∣−3−21−6∣=1 aur ∣1/3−21+2/3∣=1.

3/345∘ aage ghumane se slope −3 milta hai.

Fig. 9.7Ek line, ek angle, do possible partners.NCERT Fig. 9.7 · p. 158

Axes ke saath origin se guzarti slope 2 wali ek solid blue line. Step 1 mein sirf yahi line hai aur sawaal hai ki kaun sa slope iske saath 45 degree ka angle banata hai. Step 2: ek dashed copy 45 degrees clockwise ghoom kar slope 1/3 par aati hai, aur ek arc lines ke beech 45 degrees mark karta hai. Step 3: ek dotted copy 45 degrees anticlockwise ghoom kar slope −3 par aati hai, ek doosre 45 degree arc ke saath, to dono answers ek saath dikhte hain.

Example 9.3NCERT Example 3 · p. 158

Perpendicular slopes ka use

Points (1,2) aur (5,4) ek line tay karte hain, aur (3,−1) aur (k,7) doosri. Kis k ke liye lines right angle par milti hain?

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

Quick check
3 mein se 1
y=3x+1 aur y=−2x+5 ke beech ka acute angle hai
Answer dekho

Answer (C) 45∘

tan⁡θ=∣−2−31+(3)(−2)∣=∣−5−5∣=1, to θ=45∘. 135∘ lines ke beech ka doosra angle hai, lekin woh obtuse hai.
Practice · 11 questions

Practice 9.1: answers ke saath

  1. Short answerMediumNCERT Exercise 9.1, Q1

    Grid par A(−3,4), B(2,6), C(6,−2) aur D(−2,−4) mark karo, unhe order mein jodo, aur ABCD ka area nikaalo.

    Solution dikhao

    Answer57 square units

    Order mein plot karne par ABCD ek convex quadrilateral hai, to diagonal AC ise triangles ABC aur ACD mein baant deta hai.

    (9.3) se triangle ABC: teen terms −3(8)=−24, 2(−6)=−12 aur 6(−2)=−12 hain. Inka sum −48 hai, to area 12⋅48=24 hai.

    Triangle ACD: terms −3(2)=−6, 6(−8)=−48 aur −2(6)=−12 hain. Inka sum −66 hai, to area 12⋅66=33 hai.

    ABCD ka area =24+33=57 square units.

  2. Short answerMediumNCERT Exercise 9.1, Q2

    Sides 2b wala ek equilateral triangle aise rakha hai ki uski ek side x-axis par hai aur centre origin par. Uske corners kahan hain?

    Solution dikhao

    Base (−b,0) se (b,0) tak hai. Teesra vertex dono siron se barabar door hai, isliye woh y-axis par (0,h) par hai.

    (b,0) se uski distance 2b hai: b2+h2=4b2, to h=±3 b.

    Vertices: (−b,0), (b,0) aur (0,3 b), ya axis ke neeche wale triangle ke liye (0,−3 b).

  3. Short answerEasyNCERT Exercise 9.1, Q3

    Aise do points ke beech ki distance nikaalo jinka ek coordinate same hai.

    1. A(5,−3) aur B(5,9), aur general mein P(p,y1) aur Q(p,y2) (yaani PQ y-axis ke parallel hai).

      Solution dikhao

      x-difference 0 hai, to AB=∣9−(−3)∣=12. General mein PQ=∣y2−y1∣.

    2. C(−4,7) aur D(6,7), aur general mein P(x1,q) aur Q(x2,q) (yaani PQ x-axis ke parallel hai).

      Solution dikhao

      y-difference 0 hai, to CD=∣6−(−4)∣=10. General mein PQ=∣x2−x1∣.

  4. Short answerMediumNCERT Exercise 9.1, Q4

    y-axis ka kaun sa point (5,2) se utna hi door hai jitna (−1,6) se?

    Solution dikhao

    Answer(0,1)

    Maan lo point (0,k) hai. To 25+(k−2)2=1+(k−6)2.

    Expand karne par: 29−4k+k2=37−12k+k2, to 8k=8 aur k=1.

    Point (0,1) hai. Check: dono squared distances 26 hain.

  5. Quick answerEasyNCERT Exercise 9.1, Q5

    M, A(−2,3) aur B(6,−7) ka midpoint hai. (1,5) ko M se jodne wali line ka slope kya hai?

    Answer dekho

    Answer -7

    Midpoint (−2+62,3−72)=(2,−2) hai. Slope =−2−52−1=−7.
  6. ProofMediumNCERT Exercise 9.1, Q6

    Pythagoras' theorem nahi, slopes use karke dikhao ki A(1,2), B(4,3), C(2,9) wale triangle ABC mein ek right angle hai, aur batao woh kahan hai.

    Solution dikhao

    Slopes: AB=3−24−1=13, BC=9−32−4=−3 aur AC=9−22−1=7.

    Kyunki 13×(−3)=−1, AB⊥BC. Triangle mein right angle B par hai.

  7. Short answerMediumNCERT Exercise 9.1, Q7

    Positive y-axis ki taraf mooh karke shuru karo aur 60∘ anticlockwise ghoomo. Us direction mein point karne wali line ka slope kya hai?

    Solution dikhao

    Answer−13

    Positive y-axis positive x-axis se 90∘ par point karta hai. Aur 60∘ anticlockwise ghoomne par inclination 150∘ ho jaata hai.

    To m=tan⁡150∘=−tan⁡30∘=−13.

  8. ProofMediumNCERT Exercise 9.1, Q8

    Sirf slopes use karke dikhao ki P(−1,−2), Q(5,1), R(6,5) aur S(0,2) wala PQRS ek parallelogram hai.

    Solution dikhao

    Slopes: PQ=36=12, QR=41=4, RS=−3−6=12, SP=−4−1=4.

    To PQ∥RS aur QR∥SP. Slopes 12 aur 4 alag hain, isliye chaaron points ek line par nahi hain, aur PQRS ek parallelogram hai.

  9. Short answerEasyNCERT Exercise 9.1, Q9

    (2,−3) aur (3,0) se guzarti line ka inclination kya hai?

    Solution dikhao

    Answer60∘

    m=0−(−3)3−2=3. To 0∘≤θ<180∘ ke saath tan⁡θ=3, jisse θ=60∘ milta hai.

  10. Long answerHardNCERT Exercise 9.1, Q10

    Do lines ke slopes m aur 3m hain, aur unke beech ke acute angle θ ke liye tan⁡θ=12. Slopes ke saare possible pairs nikaalo.

    Solution dikhao

    Maan lo slopes m aur 3m hain. (9.6) se, 3m−m1+3m2=±12, yaani 4m=±(1+3m2).

    + ke saath: 3m2−4m+1=0, to m=1 ya m=13.

    − ke saath: 3m2+4m+1=0, to m=−1 ya m=−13.

    Pairs: 1 aur 3; 13 aur 1; −1 aur −3; −13 aur −1.

  11. ProofEasyNCERT Exercise 9.1, Q11

    Points (x1,y1) aur (x2,y2) slope m wali line par hain. Dikhao ki y2=y1+m(x2−x1). Iska use karke (1,3) se guzarti slope 23 wali line par x=7 wale point ka y-coordinate nikaalo.

    Solution dikhao

    Answer7

    (9.5) se, m=y2−y1x2−x1, to y2−y1=m(x2−x1), jise rearrange karne par result milta hai. (Agar x2=x1 ho to points ek hi hain, kyunki line vertical nahi hai.)

    Yahan y=3+23(7−1)=3+4=7.

9.3 Line ki Equation ke Alag-Alag Forms

Tum kaise check karoge ki (2,7) us line par hai ya nahi jo (0,1) se guzarti hai aur jiska slope 3 hai? Tumhe ek aisa rule chahiye jise line ke points ke coordinates maanein aur baaki har point tod de. Woh rule, x aur y mein likha hua, line ki equation hai: koi point line par tabhi hota hai jab uske coordinates use satisfy karein. Yahan rule y=3x+1 nikalta hai, aur 7=3⋅2+1, to answer hai haan.

Line ke baare mein humein kya bataya gaya hai, isse tay hota hai ki uski equation ka kaun sa form sabse jaldi likha jaayega: ek point aur slope, do points, slope aur ek intercept, ya dono intercepts.

9.3.1 Horizontal aur vertical lines

Points (−3,3), (0,3) aur (2,3) ko dekho. Inmein bas y=3 common hai, aur yahi unse guzarti horizontal line ki equation hai (Fig. 9.8). General mein, height k par horizontal line y=k hai, x-axis ke upar k>0 aur neeche k<0; to axis se a units neeche wali line y=−a hai. Isi tarah x-axis ko h par milne wali vertical line x=h hai, aur y-axis se b units left wali line x=−b hai. Khud axes y=0 aur x=0 hain.

3/3Vertical lines x ko fix karti hain: x=4 aur x=−3.

Fig. 9.8Horizontal line y ko fix karti hai; vertical line x ko.NCERT Fig. 9.8 · p. 159

Ek grid jo across −5 se 5 aur upar −4 se 4 tak hai. Step 1: x-axis se 3 units upar ek horizontal line, jis par x = −3, 0 aur 2 par teen marked points hain, sab height 3 par, aur ek double arrow 3 ka gap dikhata hai; iski equation y = 3 hai. Step 2: x-axis se 2 units neeche ek dashed horizontal line, jis par y = −2 likha hai. Step 3: do vertical lines judti hain, y-axis ke right mein x = 4 aur uske left mein x = −3.

Example 9.4NCERT Example 4 · p. 160

Axes ke parallel lines

(4,−1) se ek horizontal aur ek vertical line kheencho. Unki equations likho.

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

9.3.2 Point-slope form

Maan lo humein line L ka ek point P0(x0,y0) aur uska slope m pata hai. Koi aur point P(x,y) L par tabhi aur sirf tabhi hai jab segment P0P ka slope m ho, yaani y−y0x−x0=m. Cross-multiply karne se fraction hat jaata hai, aur result ko P0 bhi satisfy karta hai:

point-slope form
y−y0=m(x−x0)
(9.7)

L ka har point (9.7) ko satisfy karta hai. L ke bahar ka point nahi karta, kyunki P0 se uska slope m se alag hai (ya, P0 ke theek upar ya neeche wale point ke liye, left side non-zero hai jabki right side zero hai). Fig. 9.9 dikhata hai ki slope kyun chahiye: ek point se infinitely many lines guzarti hain, aur slope unmein se ek ko chunta hai.

2/3Line par kisi bhi P(x,y) ke liye y−y0x−x0=m.

Fig. 9.9Ek point aur ek slope, ya do points, line ko fix karte hain.NCERT Fig. 9.10 · p. 160

Step 1: alag-alag slopes wali paanch patli lines point P0(1, 2) se guzarti hain, jo dikhata hai ki ek point line ko fix nahi karta. Step 2: sirf slope 1/2 wali blue line bachti hai; us par ek general point P(x, y) = (4, 3.5) ek dashed run x − x0 aur rise y − y0 se P0 se juda hai, jinka ratio slope m hai. Step 3: fixed point ka naam P1 ho jaata hai aur line par ek doosra point P2(−3, 0) mark hota hai; do points se slope (2 − 0)/(1 + 3) = 1/2 hai.

Example 9.5NCERT Example 5 · p. 161

Ek point aur ek slope

Ek line ka slope 5 hai aur woh (3,−2) se guzarti hai. Uski equation general form mein likho.

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

9.3.3 Two-point form

Do points se humein slope mil jaata hai, isliye two-point form asal mein point-slope form hi hai. Agar L P1(x1,y1) aur P2(x2,y2) se guzarti hai aur x1≠x2, to (9.5) se uska slope y2−y1x2−x1 hai. (Doosre shabdon mein, P(x,y) L par tabhi hai jab P1, P2 aur P collinear hon.) Is slope ko P1 par (9.7) mein lagaane se milta hai

two-point form
y−y1=m(x−x1),m=y2−y1x2−x1
(9.8)
Example 9.6NCERT Example 6 · p. 161

Do points se guzarti line

Kaun si line (−1,4) aur (3,−4) ko jodti hai?

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

9.3.4 Slope-intercept form

Jahan line y-axis ko kaatti hai, woh jagah khaas kaam ki hai: yeh x=0 hone par y ki value hai. Agar crossing point (0,c) hai, to number c line ka y-intercept hai; agar line x-axis ko (d,0) par milti hai, to d uska x-intercept hai. Intercepts coordinates hain, lengths nahi, isliye origin ke neeche ya left mein crossing ke liye woh negative hote hain.

Slope m, y-intercept c. Point (0,c) par (9.7) use karo: y−c=m(x−0), jo simplify hokar banta hai

slope-intercept form
y=mx+c
(9.9)

Slope m, x-intercept d. Yahi step (d,0) par karne se milta hai

y=m(x−d)
(9.10)

y=mx+c mein slope aur y-intercept seedhe equation se padh sakte hain, isliye line ka sketch banane ke liye yeh sabse handy form hai.

Example 9.7NCERT Example 7 · p. 162

Intercept ke saath slope

Ek line ka slope −23 hai. Uski equation nikaalo agar (a) uska y-intercept 4 hai; (b) uska x-intercept −3 hai.

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

9.3.5 Intercept form

Maan lo ek line x-axis ko (a,0) par aur y-axis ko (0,b) par kaatti hai, aur a aur b dono non-zero hain. Uska slope b−00−a=−ba hai aur y-intercept b hai, to y=−bax+b. b se divide karke x-term ko doosri taraf le jaane par milta hai

intercept form
xa+yb=1
(9.11)

Yeh form origin se guzarti line (dono intercepts 0 hain) ya kisi axis ke parallel line (ek intercept hota hi nahi) ko describe nahi kar sakta. Fig. 9.10 ek hi line 3x+4y=12 par intercept forms ko compare karta hai.

3/4Intercepts a=4, b=3: x4+y3=1.

Fig. 9.10Intercepts coordinates hain, isliye negative bhi ho sakte hain.NCERT Fig. 9.13 · p. 163

Step 1: blue line 3x + 4y = 12, jiska y-axis crossing (0, c) = (0, 3) mark hai; slope −3/4 ke saath equation y = mx + c hai. Step 2: x-axis crossing (d, 0) = (4, 0) mark hota hai aur equation y = m(x − d) ban jaati hai. Step 3: dono crossings double arrows se mark hote hain, x-axis ke saath a = 4 aur y-axis ke saath b = 3, jisse x/4 + y/3 = 1 milta hai. Step 4: blue line ki jagah (−2, 0) aur (0, 3) se guzarti ek dashed line aati hai, jiske intercepts a = −2 aur b = 3 se x/(−2) + y/3 = 1 milta hai.

Example 9.8NCERT Example 8 · p. 163

Do intercepts

Ek line x-axis ko 4 par aur y-axis ko −6 par kaatti hai. Uski equation nikaalo.

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

Line ki general equation

Upar ka har form rearrange hokar Ax+By+C=0 ban jaata hai, jahan numbers A, B dono zero nahi hain, aur aisi har equation ek line describe karti hai. Yeh line ki general linear equation hai.

general equation
Ax+By+C=0
(9.12)
jahan
A,B
real numbers, dono zero nahi

Agar B≠0, to yeh rearrange hokar y=−ABx−CB banti hai, to slope −AB aur y-intercept −CB hai. Agar B=0, to yeh vertical line x=−CA hai. Jab A≠0, to x-intercept −CA hai. 3x−2y−12=0 ke liye isse slope 32, y-intercept −6 aur x-intercept 4 milta hai, jo Example 9.8 se match karta hai.

InteractiveLine lab

Dono axes par −6 se 6 tak ka grid jismein do draggable points P aur Q hain jo half units par snap hote hain. Equation mode mein panel slope ko fraction ke roop mein (ya vertical line ke liye "not defined"), inclination degrees mein, aur equation point-slope, slope-intercept, intercept aur general form mein dikhata hai, aur batata hai ki current line ke liye kaun se forms exist nahi karte. Distance mode mein ek teesra point R aata hai, line par uske perpendicular aur perpendicular ke foot ke saath, aur distance formula se nikaali jaati hai. Two lines mode mein R se guzarti ek doosri line aati hai jiska slope badla ja sakta hai, aur panel lines ke beech ka acute angle deta hai aur batata hai ki woh kab parallel ya perpendicular hain.

Fig. 9.11Do points drag karke ek line banao aur uska slope, inclination aur equation ka har form padho. Phir ek teesra point jodo aur line se uski distance naapo, ya ek doosri line jodo aur dono ke beech ka angle naapo.
Quick check
4 mein se 1
Kaun si line ka slope −2 aur y-intercept 3 hai?
Answer dekho

Answer (D) 2x+y−3=0

2x+y−3=0 yaani y=−2x+3. Option 2x−y+3=0 yaani y=2x+3, jiska slope +2 hai, aur y=3x−2 mein dono numbers ulat gaye hain.
Practice · 19 questions

Practice 9.2: answers ke saath

  1. Quick answerEasyNCERT Exercise 9.2, Q1

    Har line ko uski equation se match karo.

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

    Answer dekho
    Answer
    • x-axis : y=0
    • y-axis : x=0
    • (0,4) se guzarti, x-axis ke parallel : y=4
    • (−2,0) se guzarti, y-axis ke parallel : x=−2
    x-axis ke points par y=0 hota hai aur y-axis ke points par x=0. Horizontal line apne points ka y-coordinate same rakhti hai; vertical line x-coordinate.
  2. Short answerEasyNCERT Exercise 9.2, Q2

    Slope −34 wali ek line point (5,−2) se guzarti hai. Uski equation general form mein likho.

    Solution dikhao

    Answer3x+4y−7=0

    y+2=−34(x−5). 4 se multiply karo: 4y+8=−3x+15.

    To 3x+4y−7=0.

  3. Short answerEasyNCERT Exercise 9.2, Q3

    Origin se guzarti inclination θ wali line likho. θ=90∘ hone par kya hota hai?

    Solution dikhao

    θ≠90∘ ke liye: y−0=tan⁡θ (x−0), to y=xtan⁡θ.

    θ=90∘ ke liye line y-axis hai, x=0.

  4. Short answerMediumNCERT Exercise 9.2, Q4

    (−1,3) se guzarti ek line ka inclination 105∘ hai. Uski equation nikaalo.

    Solution dikhao

    tan⁡105∘=tan⁡60∘+tan⁡45∘1−tan⁡60∘tan⁡45∘=3+11−3=−(2+3).

    To y−3=−(2+3)(x+1), jisse y=−(2+3)x−2 milta hai.

    Equation: (2+3)x+y+2=0.

  5. Short answerEasyNCERT Exercise 9.2, Q5

    Slope 32 wali ek line x-axis ko O ke 4 units right mein kaatti hai. Use nikaalo.

    Solution dikhao

    (9.10) se, y=32(x−4), to 2y=3x−12.

    Equation: 3x−2y−12=0.

  6. Short answerEasyNCERT Exercise 9.2, Q6

    Inclination 60∘ wali ek line y-axis ko (0,−3) par milti hai. Uski equation likho.

    Solution dikhao

    m=tan⁡60∘=3 aur c=−3, to y=3x−3, yaani 3x−y−3=0.

  7. Short answerEasyNCERT Exercise 9.2, Q7

    (−3,−2) ko (1,6) se jodne wali line ki equation likho.

    Solution dikhao

    m=6+21+3=2, to y+2=2(x+3).

    Equation: 2x−y+4=0. Check: 2(1)−6+4=0.

  8. Short answerMediumNCERT Exercise 9.2, Q8

    Triangle ABC mein A=(1,4), B=(−3,2) aur C=(5,−2). B se median AC ke midpoint tak jaata hai. Uski equation nikaalo.

    Solution dikhao

    AC ka midpoint (3,1) hai. Median B(−3,2) ko (3,1) se jodta hai, slope 1−23+3=−16 ke saath.

    y−2=−16(x+3), to 6y−12=−x−3.

    Equation: x+6y−9=0.

  9. Short answerMediumNCERT Exercise 9.2, Q9

    Ek line (2,−4) se guzarti hai aur (−1,3) se (4,5) tak ke segment ke right angle par hai. Uski equation likho.

    Solution dikhao

    Diya gaya slope 5−34+1=25 hai, to chahiye wala slope −52 hai.

    y+4=−52(x−2), to 2y+8=−5x+10.

    Equation: 5x+2y−2=0.

  10. Long answerHardNCERT Exercise 9.2, Q10

    A(−1,2) se B(3,4) tak ke segment ko ek line ratio m:n mein (A se gin kar) kaatti hai, aur woh use right angle par cross karti hai. Woh line nikaalo.

    Solution dikhao

    Dividing point (3m−nm+n,4m+2nm+n) hai. AB ka slope 24=12 hai, to chahiye wala slope −2 hai.

    y−4m+2nm+n=−2(x−3m−nm+n). m+n se multiply karo aur simplify karo:

    2(m+n)x+(m+n)y=10m. (m=n se check karo: line midpoint (1,3) se guzarti hai.)

  11. Short answerMediumNCERT Exercise 9.2, Q11

    (−4,7) se guzarti ek line dono axes par same non-zero intercept banati hai. Use nikaalo.

    Solution dikhao

    Line x+y=a hai. Yeh (−4,7) se guzarti hai, to a=3.

    Equation: x+y=3. (Agar dono intercepts 0 hone ki chhoot hoti, to origin se guzarti line 7x+4y=0 bhi chalti.)

  12. Long answerMediumNCERT Exercise 9.2, Q12

    (1,4) se guzarti ek line ke dono intercepts ka sum 10 hai. Aisi saari lines nikaalo.

    Solution dikhao

    Line xa+y10−a=1 hai. (1,4) substitute karne par: 1a+410−a=1.

    To 10−a+4a=a(10−a), jisse a2−7a+10=0 aur a=2 ya a=5 milta hai.

    a=2, b=8: 4x+y=8. a=5, b=5: x+y=5.

  13. Short answerMediumNCERT Exercise 9.2, Q13

    Line ℓ ka inclination π6 aur y-intercept −1 hai. Uski equation likho, aur ℓ ke parallel y-intercept 3 wali line ki equation bhi.

    Solution dikhao

    m=13 aur c=−1: y=x3−1, yaani x−3y−3=0.

    c=3 wali parallel line: y=x3+3, yaani x−3y+33=0.

  14. Short answerMediumNCERT Exercise 9.2, Q14

    Ek line ka origin ke sabse paas wala point (3,−4) hai. Line ki equation likho.

    Solution dikhao

    Sabse paas wala point O se daale gaye perpendicular ka foot hai. O se (3,−4) ka slope −43 hai, to line ka slope 34 hai aur woh (3,−4) se guzarti hai.

    y+4=34(x−3), to 4y+16=3x−9.

    Equation: 3x−4y−25=0.

  15. NumericalMediumNCERT Exercise 9.2, Q15

    Ek steel rail 10 °C par 12.000 m lambi hai aur 35 °C par 12.0036 m lambi. Maan lo uski length L (metres mein) temperature T (°C mein) ka linear function hai; L ko T ke terms mein likho aur 45 °C par length nikaalo.

    Solution dikhao

    Slope: 12.0036−12.00035−10=0.000144 metre prati degree.

    L−12=0.000144(T−10), to L=0.000144 T+11.99856.

    45 °C par: L=12+0.000144×35=12.00504 m. Isi wajah se rails ke beech chhote gaps chhode jaate hain.

  16. Quick answerMediumNCERT Exercise 9.2, Q16

    Ek juice stall ₹30 prati glass par hafte mein 450 glasses bechta hai aur ₹35 par hafte mein 390 glasses. Maan lo price badhne par sales linearly girti hai, to ₹40 par woh kitne glasses bechega?

    glasses
    Answer dekho

    Answer 330 glasses

    (price, sales) ko points maano: (30,450) aur (35,390). Slope =390−4505=−12 glasses per rupee. ₹40 par: 390−12×5=330.
  17. ProofMediumNCERT Exercise 9.2, Q17

    Ek line ka segment dono axes ke beech hai.

    1. Uska midpoint (3,−2) hai. Line ki equation nikaalo.

      Solution dikhao

      Sire (a,0) aur (0,b) hain aur midpoint (a2,b2)=(3,−2) hai, to a=6 aur b=−4.

      x6+y−4=1, yaani 2x−3y=12.

    2. Dikhao ki agar midpoint (p,q) hai, p,q≠0 ke saath, to line xp+yq=2 hai.

      Solution dikhao

      (a) ki tarah, intercepts 2p aur 2q hain, to x2p+y2q=1. 2 se multiply karo.

  18. Long answerHardNCERT Exercise 9.2, Q18

    Ek line x-axis ko X par aur y-axis ko Y par milti hai. XY par point R(h,k) aisa hai ki XR:RY=2:3. Line nikaalo.

    Solution dikhao

    (9.2) se, dividing point (2⋅0+3p5,2q+3⋅05)=(3p5,2q5) hai.

    To p=5h3 aur q=5k2, aur line 3x5h+2y5k=1 hai.

    Equation: 3kx+2hy=5hk.

  19. ProofEasyNCERT Exercise 9.2, Q19

    (−1,−4) aur (2,2) se guzarti line nikaalo, aur uski madad se decide karo ki (5,8) usi line par hai ya nahi.

    Solution dikhao

    (−1,−4) aur (2,2) se: slope 2, to y+4=2(x+1), yaani y=2x−2.

    (5,8) ke liye: 2(5)−2=8. Teesra point equation ko satisfy karta hai, isliye teenon ek line par hain.

9.4 Point ki Line se Distance

Ek gaon seedhe highway se kitna door hai? Gaon se sadak tak kai raaste hain, lekin sabse chhota raasta sadak ko right angle par milta hai; koi bhi tircha raasta ek right triangle ka hypotenuse hai, isliye lamba hai. Isi wajah se point ki line se distance ka matlab hai point se line par daale gaye perpendicular ki length (Fig. 9.12).

3/4PQR ka area 5 hai aur QR=25.

Fig. 9.12Perpendicular base QR par triangle PQR ki height hai.NCERT Fig. 9.14 · p. 164

Line x − 2y + 4 = 0 x-axis ko Q(−4, 0) par aur y-axis ko R(0, 2) par kaatti hai. Step 1: P(3, 1) se teen dashed tirche raaste line tak jaate hain, aur solid perpendicular PM sabse chhota hai. Step 2: foot M(2, 3) par right angle ke saath PM mark hai, aur d = PM. Step 3: triangle PQR shaded hai; uska area 5 hai aur base QR 2√5 hai, aur us base par height PM hai. Step 4: d = 2 × 5 ÷ 2√5 = √5.

DerivationAx+By+C=0 se (x1,y1) ki distance

Maan lo point P(x1,y1) hai aur line L:Ax+By+C=0, jahan A, B aur C sab non-zero hain, taaki L axes ko do alag points Q(−CA,0) aur R(0,−CB) par mile. Maan lo M, P se daale gaye perpendicular ka foot hai, to d=PM.

QR ko triangle PQR ka base lo. Tab uski height PM hai, to

d=2⋅area(PQR)QR

S=Ax1+By1+C aur k=CAB likho, taaki Q=(−kB,0) aur R=(0,−kA). Teeno vertices ko (9.3) mein daalne par bracket x1(kA)−kB(−kA−y1)=k(Ax1+By1+kAB) hai, jo kS hai kyunki kAB=C. To

2⋅area=∣k∣ ∣S∣

Distance formula se, QR2=k2B2+k2A2, to QR=∣k∣A2+B2.

Divide karne par factor ∣k∣ cancel ho jaata hai aur humein (9.13) milta hai. Yeh formula tab bhi sahi hai jab L origin se guzarti ho ya kisi axis ke parallel ho; jaise (x1,y1) ki x−k=0 se distance ∣x1−k∣ aati hai, jaisa hona chahiye.

point ki line se distance
d=∣Ax1+By1+C∣A2+B2
(9.13)

Figure ke liye, P(3,1) aur x−2y+4=0 se d=∣3−2+4∣1+4=55=5 milta hai. Khaas taur par, origin ki Ax+By+C=0 se distance ∣C∣A2+B2 hai.

9.4.1 Do parallel lines ke beech ki distance

Parallel lines ka slope same hota hai, isliye hum unhe y=mx+c1 aur y=mx+c2 likh sakte hain. Unke beech ka gap har jagah same hai, isliye hum ise pehli line ke kisi bhi suvidha wale point se naap sakte hain, jaise uska y-axis wala point A(0,c1) (Fig. 9.13). Doosri line ko mx−y+c2=0 likho aur A par (9.13) lagao:

d=∣m⋅0−c1+c2∣m2+1
slope form mein parallel lines
d=∣c1−c2∣1+m2
(9.14)

3/4d=∣3−(−2)∣1+1/4=25.

Fig. 9.13Gap ko ek line ke kisi bhi point se naapo.NCERT Fig. 9.15 · p. 166

Step 1: parallel lines y = x/2 + 3 (solid) aur y = x/2 − 2 (dashed). Step 2: point A(0, 3) mark hota hai jahan upar wali line y-axis se milti hai. Step 3: A se perpendicular neeche wali line ko (2, −1) par milta hai, right-angle mark ke saath; uski length 2√5 hai. Step 4: upar wali line ke point (4, 5) se ek doosra perpendicular neeche wali line ko (6, 1) par milta hai aur uski length bhi same hai, lagbhag 4.47.

Agar dono lines general form mein same A aur B ke saath likhi hon, taaki woh sirf constants C1 aur C2 mein alag hon, to wahi idea (try karo) deta hai

general form mein parallel lines
d=∣C1−C2∣A2+B2
(9.15)
Example 9.9NCERT Example 9 · p. 167

Point ki line se distance

Point (−2,3) line 5x−12y+7=0 se kitna door hai?

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

Example 9.10NCERT Example 10 · p. 167

Parallel lines ke beech ka gap

Parallel lines 6x+8y−9=0 aur 6x+8y+11=0 ek doosre se kitni door hain?

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

Quick check
3 mein se 1
Line 3x+4y−10=0 se (1,−2) ki distance nikaalo.
Answer dekho

Answer 3

d=∣3(1)+4(−2)−10∣9+16=∣−15∣5=3. Modulus ke bina −3 milta, lekin distance kabhi negative nahi hoti.
Practice · 17 questions

Practice 9.3: answers ke saath

  1. Short answerEasyNCERT Exercise 9.3, Q1

    Har equation ko jahan possible ho y=mx+c ke roop mein likho, aur m aur c batao.

    1. 2x−5y=0

      Solution dikhao

      y=25x: slope 25 hai, y-intercept 0 hai.

    2. 4x+2y−7=0

      Solution dikhao

      y=−2x+72: slope −2, y-intercept 72.

    3. y−3=0

      Solution dikhao

      y=0⋅x+3: slope 0 hai, y-intercept 3 hai.

    4. x=4

      Solution dikhao

      Possible nahi: line vertical hai, isliye uska koi slope nahi hai aur woh y-axis se milti hi nahi.

  2. Short answerEasyNCERT Exercise 9.3, Q2

    Har equation ko jahan possible ho xa+yb=1 ke roop mein likho, aur intercepts batao.

    1. 5x+3y−15=0

      Solution dikhao

      5x+3y=15, to x3+y5=1: intercepts 3 aur 5.

    2. 2x−7y=14

      Solution dikhao

      x7+y−2=1: intercepts 7 aur −2.

    3. 2x+5=0

      Solution dikhao

      x=−52, ek vertical line: x-intercept −52 aur koi y-intercept nahi, isliye koi intercept form nahi hai.

  3. Quick answerEasyNCERT Exercise 9.3, Q3

    (2,−3) line 3(x−1)=4(y+2) se kitna door hai?

    Answer dekho

    Answer 1.4

    Line 3x−4y−11=0 hai. To d=∣6+12−11∣5=75=1.4.
  4. Short answerMediumNCERT Exercise 9.3, Q4

    x-axis ke kaun se points line x6+y8=1 se 4 units door hain?

    Solution dikhao

    Line 4x+3y−24=0 hai. (h,0) ke liye: ∣4h−24∣5=4, to 4h−24=±20.

    h=11 ya h=1. Points (11,0) aur (1,0) hain.

  5. Short answerMediumNCERT Exercise 9.3, Q5

    Parallel lines ke har pair ke beech ki distance nikaalo.

    1. 12x−5y+20=0 aur 12x−5y−6=0

      Solution dikhao

      d=∣20−(−6)∣144+25=2613=2.

    2. a(x−y)+p=0 aur a(x−y)−q=0, jahan a≠0

      Solution dikhao

      Yahan A=a, B=−a, C1=p, C2=−q, to d=∣p+q∣∣a∣2.

  6. Short answerEasyNCERT Exercise 9.3, Q6

    (1,−4) se guzarti kaun si line 5x+2y−7=0 se kabhi nahi milti?

    Solution dikhao

    Ek parallel line 5x+2y+k=0 hai. (1,−4) se guzarne ke liye: 5−8+k=0, to k=3.

    Equation: 5x+2y+3=0.

  7. Short answerMediumNCERT Exercise 9.3, Q7

    Ek line 2x+3y−6=0 ko right angle par cross karti hai aur uska y-intercept −2 hai. Use nikaalo.

    Solution dikhao

    Diya gaya slope −23 hai, to chahiye wala slope 32 hai, aur y=32x−2.

    Equation: 3x−2y−4=0.

  8. Short answerMediumNCERT Exercise 9.3, Q8

    Lines x−3y+1=0 aur x−y=3 ke beech ke angles nikaalo.

    Solution dikhao

    Inclinations 30∘ aur 45∘ hain, to acute angle 15∘ hai.

    (9.6) se: tan⁡θ=1−1/31+1/3=3−13+1=2−3=tan⁡15∘, jo match karta hai.

    Angles 15∘ aur 165∘ hain.

  9. Quick answerMediumNCERT Exercise 9.3, Q9

    (2,k) aur (−1,5) se guzarti line 3x+2y−8=0 ke perpendicular hai. k nikaalo.

    Answer dekho

    Answer 7

    Diya gaya slope −32 hai, to chahiye wala slope 23 hai. Phir k−52−(−1)=23 se k−5=2 milta hai, to k=7.
  10. ProofMediumNCERT Exercise 9.3, Q10

    Prove karo ki (x1,y1) se guzarti aur Ax+By+C=0 ke perpendicular line B(x−x1)−A(y−y1)=0 hai.

    Solution dikhao

    Agar A,B≠0: diya gaya slope −AB hai, to chahiye wala slope BA hai. Phir y−y1=BA(x−x1), jo rearrange hokar B(x−x1)−A(y−y1)=0 ban jaata hai.

    Agar B=0 to di gayi line vertical hai, chahiye wali line y=y1 hai, aur formula −A(y−y1)=0 ban jaata hai. Agar A=0 to di gayi line horizontal hai, chahiye wali line x=x1 hai, aur formula B(x−x1)=0 ban jaata hai.

  11. Long answerMediumNCERT Exercise 9.3, Q11

    (1,−2) se do lines guzarti hain jo 60∘ par kaatti hain. Agar ek ka slope 3 hai, to doosri ka kya ho sakta hai?

    Solution dikhao

    Pehli line ka inclination 60∘ hai. Doosri usse 60∘ alag hai, to uska inclination 0∘ ya 120∘ hai, aur slope 0 ya −3.

    Slope 0: y=−2. Slope −3: y+2=−3(x−1), yaani 3x+y+2−3=0.

    (9.6) se check karo: dono se tan⁡θ=3 milta hai.

  12. Short answerMediumNCERT Exercise 9.3, Q12

    Ek line ka har point (−2,5) aur (4,1) se barabar door hai. Line nikaalo.

    Solution dikhao

    Do diye gaye points se barabar door wale points unhe jodne wale segment ka perpendicular bisector banaate hain. Midpoint (1,3). Segment ka slope 1−54+2=−23 hai, to bisector ka slope 32 hai.

    y−3=32(x−1), to 3x−2y+3=0.

  13. Short answerMediumNCERT Exercise 9.3, Q13

    (4,1) se line 2x−y+3=0 par perpendicular daalo. Woh kahan girta hai?

    Solution dikhao

    (4,1) se guzarte perpendicular ka slope −12 hai: x+2y−6=0.

    2x−y+3=0 ke saath solve karo: pehli se x=6−2y, to 12−5y+3=0, y=3, x=0.

    Foot (0,3) hai.

  14. Short answerMediumNCERT Exercise 9.3, Q14

    Line y=mx+c ka origin ke sabse paas wala point (3,−1) hai. m aur c nikaalo.

    Solution dikhao

    O se (3,−1) ka slope −13 hai, to m=3.

    Point line par hai: −1=3(3)+c, to c=−10.

  15. ProofHardNCERT Exercise 9.3, Q15

    Maan lo p origin se (a,0) aur (0,atan⁡θ) se guzarti line ki distance hai, aur q origin se (a,0) aur (0,acot⁡θ) se guzarti line ki distance hai, jahan 0∘<θ<90∘ aur a>0. Prove karo ki p2+q2=a2.

    Solution dikhao

    Pehli line: xa+yatan⁡θ=1, yaani xtan⁡θ+y−atan⁡θ=0. To p=atan⁡θtan⁡2θ+1=atan⁡θsec⁡θ=asin⁡θ.

    Isi tarah, doosri line xcot⁡θ+y−acot⁡θ=0 hai, aur q=acot⁡θcsc⁡θ=acos⁡θ.

    Isliye p2+q2=a2(sin⁡2θ+cos⁡2θ)=a2.

  16. Long answerMediumNCERT Exercise 9.3, Q16

    A(1,4), B(−2,−1) aur C(4,1) ke liye A se guzarti aur BC ke perpendicular line nikaalo, aur A se BC ki distance bhi.

    Solution dikhao

    BC ka slope 1+14+2=13 hai: y+1=13(x+2), yaani x−3y−1=0.

    Altitude ka slope −3 hai: y−4=−3(x−1), yaani 3x+y−7=0.

    Length =∣1−12−1∣10=1210=6105≈3.79.

  17. ProofMediumNCERT Exercise 9.3, Q17

    Maan lo ek line ke intercepts a aur b hain (dono non-zero), aur p origin se uski distance hai. Prove karo ki 1p2=1a2+1b2, phir a=6 aur b=8 ke liye p nikaalo.

    Solution dikhao

    Line bx+ay−ab=0 hai, to p=∣ab∣a2+b2 aur 1p2=a2+b2a2b2=1b2+1a2.

    a=6, b=8 ke liye: p=4810=4.8.

Miscellaneous Examples

Yeh examples chapter ke tools ko milate hain: intersections, slopes, midpoints aur distances. Jo lines ek common point share karti hain unhe concurrent kehte hain. Teen lines ke liye jaldi wala test hai: do ko ek saath solve karo aur woh point teesri mein substitute karo.

Example 9.11NCERT Example 11 · p. 168

Teen concurrent lines

Lines x+2y−5=0, 3x−y−1=0 aur kx+3y−10=0 ek hi point se guzarti hain. k nikaalo.

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

Example 9.12NCERT Example 12 · p. 169

Di gayi direction mein distance

P(2,1) se shuru karke tum inclination 45∘ wali line ke saath tab tak chalte ho jab tak line 2x+y−11=0 tak na pahunch jao. Tum kitna chale?

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

Example 9.13NCERT Example 13 · p. 169

Line mirror mein image

Line x+2y−4=0 ek mirror ka kaam karti hai. P(3,−2) ka reflection kahan dikhega?

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

3/3N ke aage utni hi distance jao: Q(5,2).

Fig. 9.14Mirror PQ ka perpendicular bisector hai.NCERT Fig. 9.17 · p. 169

Mirror line x + 2y − 4 = 0 left se right neeche utarti hai. Step 1 mein sirf mirror ke neeche point P(3, −2) dikhta hai. Step 2 mein P se guzarta slope 2 wala dashed perpendicular judta hai, jo mirror ko N(4, 0) par right-angle mark ke saath milta hai. Step 3 mein dashed line N ke aage utni hi distance tak image Q(5, 2) tak badhti hai, aur PN aur NQ par barabar tick marks hain.

Example 9.14NCERT Example 14 · p. 170

Do lines aur ek axis se bana triangle

Lines y=m1x+c1 aur y=m2x+c2, jahan m1≠m2 aur m1,m2≠0, x-axis ke saath ek triangle banati hain. Dikhao ki uska area (m1c2−m2c1)22∣m1m2(m1−m2)∣ hai.

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

Example 9.15NCERT Example 15 · p. 171

Diye gaye point par bisect hua segment

Ek line x−y+1=0 ko A par aur 3x+y−15=0 ko B par milti hai, aur AB ka midpoint (2,4) hai. Line ki equation nikaalo.

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

Example 9.16NCERT Example 16 · p. 172

Do lines se barabar door

Ek point aise chalta hai ki woh hamesha x+2y−3=0 se utna hi door rehta hai jitna 2x+y+1=0 se. Dikhao ki uska path straight lines se bana hai, aur unhe nikaalo.

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

Quick check
3 mein se 1
Lines x+y=4, x−y=2 aur 2x+ky=7 concurrent hain. To k hai
Answer dekho

Answer (D) 1

Pehli do (3,1) par milti hain (jodo: 2x=6). Phir 6+k=7, to k=1.
Practice · 28 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

  1. Short answerMediumNCERT Miscellaneous Exercise, Q1

    Line (p−2)x+(p2−9)y+p2−5p+4=0 lo. p ki kin values ke liye yeh

    1. x-axis ke parallel hai?

      Solution dikhao

      x-coefficient 0 hona chahiye (y-coefficient non-zero ke saath): p=2, jisse −5y−2=0 milta hai.

    2. y-axis ke parallel hai?

      Solution dikhao

      y-coefficient 0 hona chahiye: p2=9, to p=3 ya p=−3 (x-coefficients 1 aur −5 non-zero hain).

    3. origin se guzarti line hai?

      Solution dikhao

      Constant 0 hona chahiye: p2−5p+4=0, to p=1 ya p=4.

  2. Short answerMediumNCERT Miscellaneous Exercise, Q2

    Ek line ke do intercepts ka sum −1 aur product −12 hai. Aisi har line nikaalo.

    Solution dikhao

    Intercepts 3 aur −4 hain, kisi bhi order mein.

    a=3, b=−4: x3−y4=1, yaani 4x−3y=12.

    a=−4, b=3: −x4+y3=1, yaani 3x−4y+12=0.

  3. Short answerMediumNCERT Miscellaneous Exercise, Q3

    y-axis ke kaun se points line x5+y12=1 se 5 units door hain?

    Solution dikhao

    Line 12x+5y−60=0 hai. (0,k) ke liye: ∣5k−60∣13=5, to 5k−60=±65.

    k=25 ya k=−1: points (0,25) aur (0,−1).

  4. ProofHardNCERT Miscellaneous Exercise, Q4

    Dikhao ki (acos⁡α,asin⁡α) aur (acos⁡β,asin⁡β) ko jodne wali line se origin ki distance ∣acos⁡α−β2∣ hai, jahan a>0 aur α≠β (dono 0∘ aur 360∘ ke beech). a=6, α=20∘ aur β=140∘ ke liye ise nikaalo.

    Solution dikhao

    Maan lo s=α+β2. (acos⁡α,asin⁡α) par xcos⁡s+ysin⁡s=acos⁡α−β2 ki left side acos⁡(α−s)=acos⁡α−β2 hai. Doosra point bhi isi tarah kaam karta hai, to yahi line hai.

    Iske coefficients cos⁡2s+sin⁡2s=1 satisfy karte hain, to (9.13) se origin se distance ∣acos⁡α−β2∣ hai.

    Numbers ke saath: 6cos⁡60∘=3.

  5. Short answerEasyNCERT Miscellaneous Exercise, Q5

    Jahan 2x+3y−7=0 aur x−4y+2=0 milti hain, us point se x-axis ke parallel ek line kheenchi jaati hai. Uski equation kya hai?

    Solution dikhao

    Doosri line se x=4y−2. Phir 8y−4+3y−7=0, to y=1 aur x=2.

    Line y=1 hai.

  6. Short answerMediumNCERT Miscellaneous Exercise, Q6

    Line x3−y2=1 x-axis ko K par kaatti hai. K se guzarti aur is line ke right angle par line nikaalo.

    Solution dikhao

    Line 2x−3y=6 hai, slope 23 ke saath, jo x-axis ko (3,0) par milti hai.

    Perpendicular ka slope −32 hai: y=−32(x−3), yaani 3x+2y−9=0.

  7. Short answerMediumNCERT Miscellaneous Exercise, Q7

    Lines y=2x aur y=−2x aur horizontal line y=k (k≠0) ek triangle gherti hain. Woh kitna bada hai?

    Solution dikhao

    Vertices (0,0), (k2,k) aur (−k2,k) hain.

    Base =∣k∣ (y=k ke saath), height =∣k∣, to area k22 hai.

  8. Quick answerEasyNCERT Miscellaneous Exercise, Q8

    p ki kis value ke liye lines 2x−y−1=0, x+3y−11=0 aur px−2y+4=0 ka ek common point hota hai?

    Answer dekho

    Answer 1

    Pehli se y=2x−1; phir x+6x−3−11=0 se (2,3) milta hai. To 2p−6+4=0 aur p=1.
  9. ProofHardNCERT Miscellaneous Exercise, Q9

    Teen lines y=mix+ci (i=1,2,3) ke slopes alag-alag hain.

    1. Dikhao ki woh tabhi aur sirf tabhi concurrent hain jab c2−c1m1−m2=c3−c2m2−m3.

      Solution dikhao

      Lines 1 aur 2 wahan milti hain jahan x=c2−c1m1−m2; lines 2 aur 3 wahan jahan x=c3−c2m2−m3.

      Dono points line 2 par hain, isliye woh tabhi ek hote hain jab unke x-coordinates barabar hon.

    2. Iska use karke check karo ki y=x+1, y=2x−1 aur y=3x−3 concurrent hain ya nahi.

      Solution dikhao

      −1−11−2=2 aur −3+12−3=2. Yeh concurrent hain, (2,3) par.

  10. Long answerMediumNCERT Miscellaneous Exercise, Q10

    (−1,4) se guzarti do lines line 3x−y+2=0 ko 45∘ par kaatti hain. Dono nikaalo.

    Solution dikhao

    Diya gaya slope 3 hai. m−31+3m=±1 se: m=−2 ya m=12.

    m=−2: y−4=−2(x+1), yaani 2x+y−2=0.

    m=12: y−4=12(x+1), yaani x−2y+9=0.

  11. Short answerMediumNCERT Miscellaneous Exercise, Q11

    Ek line x−y+2=0 aur 2x+y−8=0 ke milne wale point se guzarti hai aur dono axes par barabar intercepts kaatti hai. Use nikaalo.

    Solution dikhao

    Jodne par: 3x−6=0, to x=2 aur y=4.

    Barabar non-zero intercepts wali line x+y=a hai; (2,4) se guzarne par a=6. To x+y=6.

    (Agar zero intercepts allowed hon, to origin se guzarti y=2x bhi chalegi.)

  12. ProofHardNCERT Miscellaneous Exercise, Q12

    Dikhao ki (h,k) se guzarti jo lines y=mx+c ke saath angle θ (0∘<θ<90∘) banati hain, unke slopes m+tan⁡θ1−mtan⁡θ aur m−tan⁡θ1+mtan⁡θ hain. m=1 aur θ=45∘ hone par yeh lines kya hain?

    Solution dikhao

    Slope μ ke liye (9.6) se μ−m=±tan⁡θ (1+mμ) milta hai, to μ(1∓mtan⁡θ)=m±tan⁡θ, jisse do slopes milte hain. (Agar koi denominator 0 ho, to woh line vertical hai.)

    m=1, θ=45∘ ke saath: pehla denominator 0 hai, to ek line vertical hai, x=h; doosra slope 0 hai, to doosri line y=k hai. Sach mein y=x+c dono axes ke saath 45∘ banati hai.

  13. Quick answerMediumNCERT Miscellaneous Exercise, Q13

    Line 2x+y−6=0, (−2,1) se (4,4) tak ke segment ko (−2,1) se gin kar ratio λ:1 mein kaatti hai. λ nikaalo.

    Answer dekho

    Answer 1.5

    Dividing point substitute karo: 2(4λ−2)+(4λ+1)−6(λ+1)=0, to 6λ=9 aur λ=32. Ratio 3:2 hai; point (1.6,2.8) hai.
  14. Short answerMediumNCERT Miscellaneous Exercise, Q14

    (−1,3) se tum line x+2y−5=0 ke saath tab tak chalte ho jab tak 3x+y+5=0 se na mil jao. Tum kitna chale?

    Solution dikhao

    (−1,3) line x+2y−5=0 par hai. Isse x=5−2y; phir 15−6y+y+5=0 se y=4, x=−3 milta hai.

    (−1,3) se (−3,4) tak distance: 4+1=5.

  15. Long answerHardNCERT Miscellaneous Exercise, Q15

    P(1,2) se guzarti ek line, line x=4 ko P se 23 units door ek point par milti hai. Uske possible inclinations aur equations nikaalo.

    Solution dikhao

    Agar inclination θ hai, to horizontal gap 23 ∣cos⁡θ∣=3 hai, to ∣cos⁡θ∣=32 aur θ=30∘ ya 150∘.

    θ=30∘: y−2=13(x−1), yaani x−3y+23−1=0.

    θ=150∘: y−2=−13(x−1), yaani x+3y−23−1=0.

  16. Short answerEasyNCERT Miscellaneous Exercise, Q16

    Ek right triangle ke legs axes ke parallel hain, aur uske hypotenuse ke sire (−2,5) aur (3,−1) hain. Legs ki possible equations nikaalo.

    Solution dikhao

    Agar right angle (−2,−1) par hai: legs x=−2 aur y=−1 hain.

    Agar yeh (3,5) par hai: legs x=3 aur y=5 hain.

  17. Short answerMediumNCERT Miscellaneous Exercise, Q17

    Point (−2,4) ko line 2x−y+3=0 mein reflect karo.

    Solution dikhao

    Maan lo image (h,k) hai. Join ka slope −12 hai: k−4h+2=−12, to h+2k=6.

    Midpoint mirror par hai: 2⋅h−22−k+42+3=0, to 2h−k−2=0.

    k=2h−2 ko h+2k=6 mein substitute karne par: 5h=10, to h=2 aur k=2. Image (2,2) hai.

  18. Long answerHardNCERT Miscellaneous Exercise, Q18

    Lines y=7x+2 aur y=−x+5, line y=mx+1 ke saath barabar angles banati hain. m nikaalo.

    Solution dikhao

    Barabar inclination ka matlab 7−m1+7m=−−1−m1−m=1+m1−m.

    Cross-multiply karne par: 7−8m+m2=1+8m+7m2, to 3m2+8m−3=0, jisse m=13 ya m=−3 milta hai.

    m=13 check karo: dono angles ke liye tan⁡θ=2.

  19. ProofHardNCERT Miscellaneous Exercise, Q19

    Ek chalte point P(x,y) ki 3x−4y+2=0 aur 12x+5y−6=0 se distances ka sum hamesha 3 hai. Samjhao ki kisi bhi aise region mein jahan koi bhi expression sign nahi badalta, P ek straight line ke saath kyun chalta hai.

    Solution dikhao

    Condition hai ∣3x−4y+2∣5+∣12x+5y−6∣13=3.

    Jahan dono expressions positive hain, wahan 65 se multiply karne par 13(3x−4y+2)+5(12x+5y−6)=195 milta hai, yaani 99x−27y−199=0: ek straight line.

    Baaki har region mein bhi signs isi tarah fixed hote hain, aur condition phir se x aur y mein linear hoti hai.

  20. Short answerMediumNCERT Miscellaneous Exercise, Q20

    Parallel lines 4x−6y+5=0 aur 2x−3y−7=0 ke theek beech wali line nikaalo.

    Solution dikhao

    Doosri line ko 4x−6y−14=0 likho. Beech wali line ka constant 5+(−14)2=−4.5 hai: 4x−6y−4.5=0.

    Yaani 8x−12y−9=0.

  21. Short answerMediumNCERT Miscellaneous Exercise, Q21

    (3,2) par rakhe lamp ki light y-axis ke saath padi mirror se takra kar laut-ti hai aur phir (6,8) se guzarti hai. Mirror ke kis point B par woh takrati hai?

    Solution dikhao

    y-axis mein (3,2) ki image (−3,2) hai. Reflected ray (−3,2) aur (6,8) se guzarti line par hai, jiska slope 23 hai.

    x=0 par: y=2+23⋅3=4. To B=(0,4).

  22. ProofHardNCERT Miscellaneous Exercise, Q22

    Maan lo c=a2+b2. Prove karo ki (c,0) aur (−c,0) se line xasec⁡θ−ybtan⁡θ=1 par daale gaye perpendiculars ka product b2 hai.

    Solution dikhao

    Product ∣(casec⁡θ−1)(−casec⁡θ−1)∣sec⁡2θa2+tan⁡2θb2=∣c2a2sec⁡2θ−1∣sec⁡2θa2+tan⁡2θb2 hai.

    Kyunki c2=a2+b2, numerator sec⁡2θ−1+b2a2sec⁡2θ=tan⁡2θ+b2sec⁡2θa2 hai.

    Yeh denominator ka b2 guna hai, to product b2 hai.

  23. Long answerMediumNCERT Miscellaneous Exercise, Q23

    Ek cyclist wahan khadi hai jahan sadakein x+2y−4=0 aur 2x−y−3=0 kaatti hain. Woh canal road 3x+4y−30=0 tak jitna jaldi ho sake pahunchna chahti hai. Use kis line ke saath chalana chahiye, aur woh kitni door hai?

    Solution dikhao

    Sadakein (2,1) par kaatti hain (pehli se x=4−2y; phir 8−5y−3=0).

    Canal road ka slope −34 hai, to woh slope 43 ke saath chalati hai: y−1=43(x−2), yaani 4x−3y−5=0.

    Distance =∣6+4−30∣5=4 units.

  24. Quick answerMedium

    (Entrance-style) Lines 3x+4y=9 aur 6x+8y=15 ke beech ki distance hai

    Answer dekho

    Answer (D) 310

    Doosri ko aadha karo: 3x+4y=7.5. Phir d=∣9−7.5∣5=310. 9 aur 15 ko bina badle use karne se trap answer 65 milta hai.
  25. Quick answerMedium

    (Assertion–reason) A: Lines 2x+3y=5 aur 4x+6y=7 intersect nahi karti. R: Barabar slopes wali do alag lines parallel hoti hain.

    Answer dekho

    Answer (A) Dono true hain, aur R, A ko explain karta hai.

    Dono slopes −23 hain, aur lines alag hain (4x+6y=10 versus 7). R true hai aur theek isi wajah se A true hai.
  26. Quick answerMedium

    (Entrance-style) Origin se 3x+4y=25 par daale gaye perpendicular ka foot hai

    Answer dekho

    Answer (B) (3,4)

    O se perpendicular y=43x hai. Substitute karne par: 3x+163x=25, to x=3, y=4. (4,3) line par to hai, lekin O se uska join perpendicular nahi hai.
  27. Quick answerEasy

    (Board-style) Dono axes ke saath line 3x+4y=24 first quadrant mein ek triangle gherti hai. Uska area kya hai?

    Answer dekho

    Answer 24

    Intercepts 8 aur 6 hain, to area 12⋅8⋅6=24 square units hai.
  28. Quick answerEasy

    (Assertion–reason) A: Slopes 2 aur −12 wali lines perpendicular hain. R: Do lines tab perpendicular hoti hain jab unke slopes ka sum zero ho.

    Answer dekho

    Answer (C) A true hai, R false hai.

    A true hai kyunki 2×(−12)=−1. R false hai: slopes 2 aur −2 ka sum zero hai, phir bhi aisi lines lagbhag 53∘ par milti hain, 90∘ par nahi. Condition yeh hai ki product −1 ho.
Recap

Summary aur key points

  1. Distance PQ2=(x2−x1)2+(y2−y1)2, section aur midpoint formulas, aur triangle-area formula; zero area ka matlab collinear points (Section 9.1).
  2. Inclination θ positive x-axis se anticlockwise naapa jaata hai, 0∘≤θ<180∘, aur slope m=tan⁡θ hai (Section 9.2).
  3. Horizontal lines ka slope 0 hota hai; vertical lines ka koi slope nahi hota.
  4. (x1,y1) aur (x2,y2) se guzarti line ke liye, x1≠x2 hone par m=y2−y1x2−x1 ((9.5)).
  5. Jo lines vertical nahi hain unke liye: barabar slopes ka matlab parallel, aur m1m2=−1 ka matlab perpendicular. Teen points tabhi ek line par hote hain jab AB aur BC ka slope same ho (Section 9.2.2).
  6. Do lines ke beech ka acute angle tan⁡θ=∣m2−m11+m1m2∣ satisfy karta hai; doosra angle 180∘−θ hai ((9.6)).
  7. Horizontal lines y=±a aur vertical lines x=±b hoti hain; axes y=0 aur x=0 hain (Section 9.3.1).
  8. Point-slope form y−y0=m(x−x0) ((9.7)) aur two-point form ((9.8)).
  9. Slope-intercept forms y=mx+c aur y=m(x−d) ((9.9)); intercept form xa+yb=1 ((9.11)).
  10. Har line Ax+By+C=0 hai (A, B dono 0 nahi), jiska slope −AB, x-intercept −CA aur y-intercept −CB hai ((9.12)).
  11. Ax+By+C=0 se (x1,y1) ki distance ∣Ax1+By1+C∣A2+B2 hai ((9.13)).
  12. Same A aur B aur constants C1, C2 wali do parallel lines ke beech ∣C1−C2∣A2+B2 ki doori hai; pehle coefficients match karo ((9.15)).

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 meinYahan practice question
NCERT Exercise 9.1, Q1Practice 9.1 · Q1
NCERT Exercise 9.1, Q2Practice 9.1 · Q2
NCERT Exercise 9.1, Q3Practice 9.1 · Q3
NCERT Exercise 9.1, Q4Practice 9.1 · Q4
NCERT Exercise 9.1, Q5Practice 9.1 · Q5
NCERT Exercise 9.1, Q6Practice 9.1 · Q6
NCERT Exercise 9.1, Q7Practice 9.1 · Q7
NCERT Exercise 9.1, Q8Practice 9.1 · Q8
NCERT Exercise 9.1, Q9Practice 9.1 · Q9
NCERT Exercise 9.1, Q10Practice 9.1 · Q10
NCERT Exercise 9.1, Q11Practice 9.1 · Q11
NCERT Exercise 9.2, Q1Practice 9.2 · Q1
NCERT Exercise 9.2, Q2Practice 9.2 · Q2
NCERT Exercise 9.2, Q3Practice 9.2 · Q3
NCERT Exercise 9.2, Q4Practice 9.2 · Q4
NCERT Exercise 9.2, Q5Practice 9.2 · Q5
NCERT Exercise 9.2, Q6Practice 9.2 · Q6
NCERT Exercise 9.2, Q7Practice 9.2 · Q7
NCERT Exercise 9.2, Q8Practice 9.2 · Q8
NCERT Exercise 9.2, Q9Practice 9.2 · Q9
NCERT Exercise 9.2, Q10Practice 9.2 · Q10
NCERT Exercise 9.2, Q11Practice 9.2 · Q11
NCERT Exercise 9.2, Q12Practice 9.2 · Q12
NCERT Exercise 9.2, Q13Practice 9.2 · Q13
NCERT Exercise 9.2, Q14Practice 9.2 · Q14
NCERT Exercise 9.2, Q15Practice 9.2 · Q15
NCERT Exercise 9.2, Q16Practice 9.2 · Q16
NCERT Exercise 9.2, Q17Practice 9.2 · Q17
NCERT Exercise 9.2, Q18Practice 9.2 · Q18
NCERT Exercise 9.2, Q19Practice 9.2 · Q19
NCERT Exercise 9.3, Q1Practice 9.3 · Q1
NCERT Exercise 9.3, Q2Practice 9.3 · Q2
NCERT Exercise 9.3, Q3Practice 9.3 · Q3
NCERT Exercise 9.3, Q4Practice 9.3 · Q4
NCERT Exercise 9.3, Q5Practice 9.3 · Q5
NCERT Exercise 9.3, Q6Practice 9.3 · Q6
NCERT Exercise 9.3, Q7Practice 9.3 · Q7
NCERT Exercise 9.3, Q8Practice 9.3 · Q8
NCERT Exercise 9.3, Q9Practice 9.3 · Q9
NCERT Exercise 9.3, Q10Practice 9.3 · Q10
NCERT Exercise 9.3, Q11Practice 9.3 · Q11
NCERT Exercise 9.3, Q12Practice 9.3 · Q12
NCERT Exercise 9.3, Q13Practice 9.3 · Q13
NCERT Exercise 9.3, Q14Practice 9.3 · Q14
NCERT Exercise 9.3, Q15Practice 9.3 · Q15
NCERT Exercise 9.3, Q16Practice 9.3 · Q16
NCERT Exercise 9.3, Q17Practice 9.3 · Q17
NCERT Miscellaneous Exercise, Q1Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8Miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9Miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10Miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11Miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12Miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13Miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14Miscellaneous exercises · Q14
NCERT Miscellaneous Exercise, Q15Miscellaneous exercises · Q15
NCERT Miscellaneous Exercise, Q16Miscellaneous exercises · Q16
NCERT Miscellaneous Exercise, Q17Miscellaneous exercises · Q17
NCERT Miscellaneous Exercise, Q18Miscellaneous exercises · Q18
NCERT Miscellaneous Exercise, Q19Miscellaneous exercises · Q19
NCERT Miscellaneous Exercise, Q20Miscellaneous exercises · Q20
NCERT Miscellaneous Exercise, Q21Miscellaneous exercises · Q21
NCERT Miscellaneous Exercise, Q22Miscellaneous exercises · Q22
NCERT Miscellaneous Exercise, Q23Miscellaneous exercises · Q23

Quick check: MCQs

  1. Pichhli classes ke tools3 questions
  2. 9.2.2 Parallel aur perpendicular lines ki conditions4 questions
  3. 9.2.3 Do lines ke beech ka angle3 questions
  4. Line ki general equation4 questions
  5. 9.4.1 Do parallel lines ke beech ki distance3 questions
  6. Miscellaneous Examples3 questions
Glossary · 7

Key terms (English + Hindi)

collinearसंरेखsanrekh
Ek hi straight line par hona. Teen points tab collinear hote hain jab unse bane triangle ka area zero ho.Chapter mein pehli baar yahan aaya →
concurrent linesसंगामी रेखाएँsangami rekhaen
Teen ya zyada lines jo sab ek hi common point se guzarti hain.Chapter mein pehli baar yahan aaya →
general linear equationव्यापक रैखिक समीकरणvyapak rekhik samikaran
Ax+By+C=0 jaisi equation, jahan A, B dono zero nahi hain. Har line ko is tarah likha ja sakta hai.Chapter mein pehli baar yahan aaya →
inclinationझुकावjhukav
Positive x-axis se anticlockwise ghoomte hue kisi line tak ka angle θ; 0∘≤θ<180∘.Chapter mein pehli baar yahan aaya →
interceptअंतःखंडantahkhand
Woh coordinate jahan line kisi axis ko kaatti hai: agar woh y-axis ko (0,c) par milti hai to c, agar x-axis ko (d,0) par milti hai to d.Chapter mein pehli baar yahan aaya →
perpendicular distanceलंबवत दूरीlambvat doori
Ek point line se kitna door hai, line par daale gaye perpendicular ke saath naapa gaya, jo sabse chhota raasta hai.Chapter mein pehli baar yahan aaya →
slopeढालdhaal
Inclination θ≠90∘ wali line ke liye m=tan⁡θ; line ke kisi bhi do points ke beech rise upon run ke barabar.

Tum chapter ke end tak pahunch gaye