Perpendicular lines — product of slopes = −1

2.3.9 · Maths › Coordinate Geometry

Slope HOTA KYA hai, actually?

Slope $2$ ka matlab hai: $1$ right jao, $2$ upar jao.
Slope literally inclination angle ka tangent hota hai — yahi poore proof ki key hai.

Product $- 1$ kyun hota hai? (Derivation from scratch)

Hum ise teen alag-alag taaron se prove karte hain taaki yeh kabhi magic formula na lage.

Tarika 1 — Line ko $90°$ rotate karo (sabse clean idea)

Ek line lo jo direction $(Δ x, Δ y)$ mein ja rahi hai. Uska slope hai $m_{1} = \frac{Δ y}{Δ x}$ .

Is vector ko $90°$ anticlockwise rotate karo. Rotation rule hai: $(x, y) \mapsto (- y, x)$

Yeh rule kyun? $90°$ turn $x$ -axis direction $(1, 0)$ ko $(0, 1)$ par bhejta hai (upar point karta hai), aur $y$ -axis direction $(0, 1)$ ko $(- 1, 0)$ par (left point karta hai). Isse $(Δ x, Δ y)$ par apply karne par milta hai $(- Δ y, Δ x)$ .

Rotated (perpendicular) line ka slope: $m_{2} = \frac{new rise}{new run} = \frac{Δ x}{- Δ y} = - \frac{Δ x}{Δ y}$

Ab multiply karo: $m_{1} m_{2} = \frac{Δ y}{Δ x} \cdot (- \frac{Δ x}{Δ y}) = - 1 ■$

Yeh step kyun? $Δ x$ aur $Δ y$ cancel ho jaate hain, exactly $- 1$ bachta hai. Yeh dikhata hai ki yeh result inevitable hai, koi coincidence nahi.

Tarika 2 — Angles aur tangent identity

Maan lo do lines $x$ -axis ke saath angles $θ_{1}$ aur $θ_{2}$ banati hain, toh $m_{1} = tan θ_{1}$ , $m_{2} = tan θ_{2}$ .

Perpendicular ka matlab hai $θ_{2} = θ_{1} + 90°$ . Isliye: $m_{2} = tan (θ_{1} + 90°) = - cot θ_{1} = - \frac{1}{t a n θ _{1}} = - \frac{1}{m _{1}}$

Kyun? Identity $tan (θ + 90°) = - cot θ$ quarter turn shift se aati hai jo sine aur cosine ko sign flip ke saath swap karta hai. Rearrange karne par: $m_{1} m_{2} = - 1 ■$

Tarika 3 — Pythagoras (koi trig nahi, koi vectors nahi)

Origin se do lines lo: $y = m_{1} x$ aur $y = m_{2} x$ . Pehli line par point $A = (1, m_{1})$ lo aur doosri par $B = (1, m_{2})$ .

Origin $O$ par right angle $⟺$ triangle $O A B$ mein Pythagoras satisfy ho: $∣ O A ∣^{2} + ∣ O B ∣^{2} = ∣ A B ∣^{2}$

Har ek compute karo:

$∣ O A ∣^{2} = 1 + m_{1}^{2}$
$∣ O B ∣^{2} = 1 + m_{2}^{2}$
$∣ A B ∣^{2} = (1 - 1)^{2} + (m_{1} - m_{2})^{2} = (m_{1} - m_{2})^{2}$

Substitute karo: $(1 + m_{1}^{2}) + (1 + m_{2}^{2}) = (m_{1} - m_{2})^{2}$ $2 + m_{1}^{2} + m_{2}^{2} = m_{1}^{2} - 2 m_{1} m_{2} + m_{2}^{2}$

Yeh step kyun? $2$ aur $- 2 m_{1} m_{2}$ ke alawa sab kuch cancel ho jaata hai: $2 = - 2 m_{1} m_{2} ⟹ m_{1} m_{2} = - 1 ■$

Teen raaste, ek manzil — rule bilkul bulletproof hai.

USE kaise karein

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Cover karo aur answer karo

Perpendicularity condition ek line mein batao.
$m = - \frac{3}{4}$ ka perpendicular slope batao.
Pythagoras proof mein $2 = - 2 m_{1} m_{2}$ tak kyun cancel hota hai?
Product rule kis case mein fail karta hai, aur kyun?

Answers: 1) $m_{1} m_{2} = - 1$ . 2) $+ \frac{4}{3}$ . 3) $m_{1}^{2}$ aur $m_{2}^{2}$ terms dono sides par cancel ho jaate hain. 4) Horizontal ⊥ vertical, kyunki vertical slope undefined hota hai.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tum ek pahaadi par chal rahe ho. Slope batata hai tumhara raasta kitna steep hai. Ab agar tumhe ek aisa raasta chahiye jo tumhare raaste ko perfect square corner par kaate, toh tumhe quarter-circle ghoomna hoga. Jab tum aisa ghoomte ho, toh "seedha upar chadna" ban jaata hai "dheere se side mein neeche jaana" — steepness palat jaati hai AUR neeche ki taraf jhuk jaati hai. Yahi "fraction flip karo aur minus sign lagao" ka matlab hai. Aur jab tum dono steepnesses ko multiply karte ho, woh hamesha ek seedha $- 1$ mein tidy ho jaate hain, jaise do puzzle pieces jo sirf us ek tarike se fit hote hain.

Connections

Slope of a Line — definition $m = tan θ$ jis par yeh note rely karta hai.
Parallel Lines — equal slopes — twin rule: $m_{1} = m_{2}$ .
Equation of a Line — point-slope form — Example 3 mein use hua.
Angle Between Two Lines — perpendicular special case hai $θ = 90°$ .
Rotation of Vectors by 90° — Tarika 1 of the proof ko power deta hai.
Distance Formula & Pythagoras Theorem — Tarika 3 ko power dete hain.

Do lines ke perpendicular hone ki condition (slopes mein)

m_{1} m_{2} = - 1

Slope

m

wali line ka perpendicular slope

- \frac{1}{m}

(negative reciprocal)

m = \frac{2}{3}

ka perpendicular slope

- \frac{3}{2}

Direction ko

90°

rotate karne par perpendicular slope kyun milta hai?

(Δ x, Δ y) \mapsto (- Δ y, Δ x)

, toh slope

\frac{Δ y}{Δ x} \to - \frac{Δ x}{Δ y}

, aur unka product

- 1

hai

tan (θ + 90°)

kya hota hai?

- cot θ = - 1/ tan θ

Product rule kis perpendicular pair ko cover NAHI karta?

Horizontal line (

m = 0

) aur vertical line (undefined slope)

m

ke perpendicular ke liye common wrong answer, aur fix

Wrong:

1/ m

(minus bhool gaye); Fix:

- 1/ m

(1, 2)

aur

(4, 3)

se guzarti line ka slope, phir uska perpendicular slope

m = \frac{1}{3}

; perpendicular

= - 3

y = m x + c

mein

m

kya hai?

Slope (

x

ka coefficient)

PARALLEL lines vs PERPENDICULAR lines ke slopes ka product

Parallel:

m_{1} = m_{2}

; Perpendicular:

m_{1} m_{2} = - 1

Perpendicular lines — product of slopes = −1

Slope HOTA KYA hai, actually?

Product $- 1$ kyun hota hai? (Derivation from scratch)

Tarika 1 — Line ko $90°$ rotate karo (sabse clean idea)

Tarika 2 — Angles aur tangent identity

Tarika 3 — Pythagoras (koi trig nahi, koi vectors nahi)

USE kaise karein

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

Concept Map

Deep Dive