Straight Lines πŸ“

Ek point se ek line draw karo β€” infinite lines possible hain. Do points do toh exactly ek line banti hai. Yahi hai geometry ka beauty!

πŸ—ΊοΈ GPS aur Lines

Jab GPS tumhe kisi jagah ka rasta batata hai β€” woh straight roads actually coordinate geometry ki lines hain! Ek city map par har road ek line equation se represent hoti hai. Ek road ki steepness (slope) batata hai woh kitni chadhai hai ya dhalaaan hai. Signals pe perpendicular roads (90Β° angle wali) β€” unke slopes ka product βˆ’1 hota hai. Yahi saari cheezein hum is chapter mein seekhte hain!

1. Slope (Dhalaaan) β€” m kya hai?

Slope batata hai line kitni "tilted" hai β€” kitni steep ya flat hai. Do points (x₁,y₁) aur (xβ‚‚,yβ‚‚) given ho toh slope:

m = (yβ‚‚ βˆ’ y₁) / (xβ‚‚ βˆ’ x₁) = Rise / Run

Angle ΞΈ se slope: m = tan ΞΈ (jahan ΞΈ x-axis ke saath angle hai)

Alag alag slopes ka visual

m > 0 (upar jaati) m = 0 (horizontal) m < 0 (neeche jaati) m = ∞ (vertical)

βœ… A(2,3) aur B(5,9) ke beech slope nikalo

m = (9βˆ’3)/(5βˆ’2) = 6/3 = 2

matlab: x mein 1 unit badhne par y mein 2 unit badhti hai β€” yeh moderate slope hai.

2. Parallel aur Perpendicular Lines

Parallel lines: Dono lines ka slope equal hota hai. Woh kabhi nahi milti.

m₁ = mβ‚‚ (parallel lines)

Perpendicular lines: Dono slopes ka product = βˆ’1. Yeh 90Β° par milti hain.

m₁ Γ— mβ‚‚ = βˆ’1 β†’ mβ‚‚ = βˆ’1/m₁ (perpendicular)

Trick: Perpendicular slope = original ka negative reciprocal. Agar m=2 hai toh perpendicular slope = βˆ’1/2.

3. Line ke Different Forms

Ek straight line ko alag alag situations mein alag alag equations se likha ja sakta hai. Yeh saari forms same line ko represent karti hain β€” bas starting information alag hoti hai:

β‘  Slope-Intercept Form

Slope aur y-intercept given. Sabse common aur useful form.

y = mx + c

m = slope, c = y-intercept (jahan line y-axis ko kaati hai)

β‘‘ Point-Slope Form

Ek point (x₁,y₁) aur slope m given ho toh use karo.

y βˆ’ y₁ = m(x βˆ’ x₁)

Real life: direction aur ek location know ho toh path nikalna.

β‘’ Two-Point Form

Do points given ho aur slope calculate karke point-slope use karo.

(yβˆ’y₁)/(yβ‚‚βˆ’y₁) = (xβˆ’x₁)/(xβ‚‚βˆ’x₁)

Do locations pata ho toh line ka equation nikalna.

β‘£ Intercept Form

x-intercept (a) aur y-intercept (b) dono given ho.

x/a + y/b = 1

Jab line dono axes par kahan milti hai yeh pata ho.

β‘€ Normal Form

Origin se perpendicular distance p aur angle Ο‰ given.

x·cosω + y·sinω = p

Advanced form β€” physics mein projection problems mein aata hai.

β‘₯ General Form

Ax + By + C = 0. Kisi bhi line ko is form mein likha ja sakta hai.

Ax + By + C = 0

Slope = βˆ’A/B, y-intercept = βˆ’C/B

βœ… Point-Slope: (3,βˆ’2) se guzarne wali line jiska slope 4 hai

y βˆ’ (βˆ’2) = 4(x βˆ’ 3)
y + 2 = 4x βˆ’ 12
y = 4x βˆ’ 14 ya 4x βˆ’ y βˆ’ 14 = 0

βœ… Two Points: A(1,2) aur B(4,8) se line ka equation

m = (8βˆ’2)/(4βˆ’1) = 6/3 = 2
y βˆ’ 2 = 2(x βˆ’ 1) β†’ y = 2x (y-intercept = 0, line passes through origin!)

4. Distance Formulas

Point se line ki perpendicular distance: Point P(xβ‚€,yβ‚€) aur line Ax+By+C=0 ke beech:

d = |Axβ‚€ + Byβ‚€ + C| / √(AΒ² + BΒ²)

Do parallel lines ke beech distance: Ax+By+C₁=0 aur Ax+By+Cβ‚‚=0 ke beech:

d = |C₁ βˆ’ Cβ‚‚| / √(AΒ² + BΒ²)

βœ… Point (3,4) ki line 3x βˆ’ 4y + 5 = 0 se distance nikalo

A=3, B=βˆ’4, C=5, xβ‚€=3, yβ‚€=4
d = |3(3) + (βˆ’4)(4) + 5| / √(9+16)
= |9 βˆ’ 16 + 5| / √25 = |βˆ’2| / 5 = 2/5

5. Two Lines ka Angle

Jab do lines milti hain unke beech ka angle ΞΈ nikalna ho:

tan ΞΈ = |(m₁ βˆ’ mβ‚‚) / (1 + m₁mβ‚‚)|

Agar m₁mβ‚‚ = βˆ’1 β†’ lines perpendicular (ΞΈ = 90Β°). Agar m₁ = mβ‚‚ β†’ lines parallel (ΞΈ = 0Β°).

6. Practice Questions

Q1. A(βˆ’1,3) aur B(4,βˆ’2) se guzarne wali line ka slope aur equation nikalo.

Solution

m = (βˆ’2βˆ’3)/(4+1) = βˆ’1. yβˆ’3=βˆ’1(x+1) β†’ y=βˆ’x+2 β†’ x+yβˆ’2=0

Q2. 3x + 4y = 12 ki slope aur intercepts nikalo.

Solution

x/4 + y/3 = 1 β†’ x-intercept=4, y-intercept=3. Slope = βˆ’3/4

Q3. y = 3x + 5 ke parallel aur perpendicular line jo (2,1) se guzre β€” equations nikalo.

Solution

Parallel (m=3): yβˆ’1=3(xβˆ’2) β†’ y=3xβˆ’5. Perpendicular (m=βˆ’1/3): yβˆ’1=βˆ’1/3(xβˆ’2) β†’ x+3y=5

Q4. Point (5,6) aur line 2x βˆ’ y + 4 = 0 ke beech perpendicular distance nikalo.

Solution

d = |2(5)βˆ’6+4|/√(4+1) = |8|/√5 = 8/√5 = 8√5/5

Q5. Lines y=2x+3 aur y=βˆ’x+1 ke beech angle nikalo.

Solution

tan ΞΈ = |(2βˆ’(βˆ’1))/(1+2Γ—(βˆ’1))| = |3/(1βˆ’2)| = |3/βˆ’1| = 3 β†’ ΞΈ = arctan(3) β‰ˆ 71.6Β°

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