Distance from a point to a line

2.3.10 · Maths › Coordinate Geometry

Core Understanding: Hum Dhoondh KYA Rahe Hain?

Hum point $P(x_1,y_1)$ se line $L:ax+by+c=0$ tak ki perpendicular distance nikalna chahte hain.

Key insight: $P$ se line $L$ ke kisi bhi point tak ke saare possible raston mein, perpendicular raasta sabse chota hota hai. Kyun? Kyunki koi bhi tirchha raasta ek right triangle ka hypotenuse banata hai, jo hamesha perpendicular leg se lamba hota hai.

Formula ko First Principles se Derive Karna

Method 1: Triangle ke Area ka Approach

Maan lo ek line $ax+by+c=0$ hai aur ek point $P(x_1,y_1)$ jo line par nahi hai.

Step 1: Line par DO points dhoondho.

• $x=0$ rakhne par: $by+c=0\Rightarrow y=-\frac{c}{b}$ (agar $b\ne 0$), jisse point $A(0,-\frac{c}{b})$ milta hai
• $y=0$ rakhne par: $ax+c=0\Rightarrow x=-\frac{c}{a}$ (agar $a\ne 0$), jisse point $B(-\frac{c}{a},0)$ milta hai

Step 2: Distance $AB$ calculate karo (hamare triangle ka base). $AB=\sqrt{{\left(-\frac{c}{a}\right)}^2+{\left(-\frac{c}{b}\right)}^2}=\sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}=|c|\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{|c|}{\sqrt{a^2+b^2}}\cdot \frac{\sqrt{a^2+b^2}}{\sqrt{a^2{b}^2}}\cdot \sqrt{a^2+b^2}$

Ruko, dobara calculate karte hain: $AB=\sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}=|c|\sqrt{\frac{b^2+a^2}{a^2{b}^2}}=\frac{|c|\sqrt{a^2+b^2}}{|ab|}$

Step 3: Triangle $PAB$ ka area coordinate formula se nikalo: $\text{Area}=\frac{1}{2}|x_1(y_A-y_B)+x_A(y_B-y_1)+x_B(y_1-y_A)|$ $=\frac{1}{2}\left|x_1\left(-\frac{c}{b}-0\right)+0\left(0-y_1\right)+\left(-\frac{c}{a}\right)\left(y_1+\frac{c}{b}\right)\right|$ $=\frac{1}{2}\left|-\frac{c{x}_1}{b}-\frac{c{y}_1}{a}-\frac{c^2}{ab}\right|=\frac{|c|}{2|ab|}\left|a{x}_1+b{y}_1+c\right|$

Step 4: Area = $\frac{1}{2}\times \text{base}\times \text{height}$ ka relationship use karo.

Agar $d$ perpendicular distance (height) hai, toh: $\frac{|c|}{2|ab|}|a{x}_1+b{y}_1+c|=\frac{1}{2}\times \frac{|c|\sqrt{a^2+b^2}}{|ab|}\times d$

Simplify karne par: $|a{x}_1+b{y}_1+c|=\sqrt{a^2+b^2}\cdot d$

Method 2: Point se Perpendicular (Vector Approach)

Yeh kyun kaam karta hai: Perpendicular distance kisi bhi position vector ka woh component hoti hai jo line ke normal ki direction mein hoti hai.

Line $ax+by+c=0$ ka normal vector $\vec{n}=(a,b)$ hota hai.

Line par koi bhi point $Q(x_0,y_0)$ lo (toh $a{x}_0+b{y}_0+c=0$).

$Q$ se $P$ tak ka vector $\overrightarrow{QP}=(x_1-x_0,y_1-y_0)$ hai.

Perpendicular distance $\overrightarrow{QP}$ ka unit normal par projection hai: $d=\left|\overrightarrow{QP}\cdot \frac{\vec{n}}{|\vec{n}|}\right|=\frac{|a(x_1-x_0)+b(y_1-y_0)|}{\sqrt{a^2+b^2}}$

Kyunki $a{x}_0+b{y}_0=-c$: $d=\frac{|a{x}_1+b{y}_1-(a{x}_0+b{y}_0)|}{\sqrt{a^2+b^2}}=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Worked Examples

Solution: Yahan $a=3,b=4,c=-10$, aur $(x_1,y_1)=(3,4)$ hai.

$d=\frac{|3(3)+4(4)+(-10)|}{\sqrt{3^2+4^2}}=\frac{|9+16-10|}{\sqrt{9+16}}=\frac{|15|}{\sqrt{25}}=\frac{15}{5}=3$

Har step kyun?

• Hum seedha formula mein values substitute kar rahe hain.
• $\sqrt{3^2+4^2}=5$ ek 3-4-5 Pythagorean triple hai—speed ke liye inhe pehchano!
• Distance 3 units hai.

Solution: Yahan $a=1,b=-2,c=4$, aur $(x_1,y_1)=(-1,2)$ hai.

$d=\frac{|1(-1)+(-2)(2)+4|}{\sqrt{1^2+(-2{)}^2}}=\frac{|-1-4+4|}{\sqrt{1+4}}=\frac{|-1|}{\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

Har step kyun?

• Numerator negative ho jaata hai ($-1$), lekin absolute value use positive bana deti hai.
• $\frac{1}{\sqrt{5}}$ ko rationalize karte hain $\frac{\sqrt{5}}{\sqrt{5}}$ se multiply karke.
• Distance $\frac{\sqrt{5}}{5}\approx 0.447$ units hai.

Solution: Step 1: Line ki equation nikalo. Slope $m=\frac{5-1}{4-1}=\frac{4}{3}$.

Point-slope form use karke: $y-1=\frac{4}{3}(x-1)$ $3(y-1)=4(x-1)$ $3y-3=4x-4$ $4x-3y-1=0$

Step 2: Distance formula apply karo. Yahan $a=4,b=-3,c=-1$, aur $(x_1,y_1)=(2,-3)$ hai.

$d=\frac{|4(2)+(-3)(-3)+(-1)|}{\sqrt{4^2+(-3{)}^2}}=\frac{|8+9-1|}{\sqrt{16+9}}=\frac{16}{\sqrt{25}}=\frac{16}{5}$

Har step kyun?

• Pehle $ax+by+c=0$ form mein convert karna zaroori hai—slope-intercept seedha kaam nahi karega.
• Distance $\frac{16}{5}=3.2$ units hai.

Common Mistakes

Kyun sahi lagta hai: Tumhe lagta hai distance negative ho sakti hai (jaise "signed distance"), lekin distance hamesha non-negative hoti hai.

Steel-man: Kuch contexts mein (jaise yeh pata karna ki point line ke kis taraf hai), signed distance useful hoti hai. Lekin jab "distance" puchhi jaaye, hum magnitude chahte hain.

Fix: Numerator ke around hamesha $|\cdots |$ lagao.

Kyun sahi lagta hai: $y=mx+c$ sabse familiar form hai.

Problem: Formula ko $ax+by+c=0$ form chahiye. Agar tum $y=2x+3$ use karte ho, toh ise $-2x+y-3=0$ (ya $2x-y+3=0$) mein likhna padega.

Fix: Pehle hamesha $ax+by+c=0$ mein convert karo. Yaad rakho: $y=mx+k\Rightarrow mx-y+k=0$.

Kyun sahi lagta hai: Square root bhool jaate ho, socha $a^2+b^2$ hi hai.

Fix: Denominator $\sqrt{a^2+b^2}$ hai, $a^2+b^2$ nahi. Yeh normal vector $(a,b)$ ki magnitude se aata hai.

Special Cases

Jab point line PAR ho: $a{x}_1+b{y}_1+c=0$, toh $d=0$. ✓

Parallel lines: Do parallel lines $ax+by+c_1=0$ aur $ax+by+c_2=0$ ke beech ki doori nikalne ke liye, ek line par koi bhi point lo aur doosri line se uski distance nikalo: $d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Deep Connections

• Connections:
  • Equation of a Line: $ax+by+c=0$ form mein convert karna zaroori hai
  • Perpendicular Lines: Sabse choti distance perpendicular ke along hoti hai
  • Projection of Vectors: Method 2 vector projection use karta hai
  • Area of Triangle Using Coordinates: Method 1 area se derive hota hai
  • Distance Formula: Saari distance calculations ki foundation
  • Normal Vector to a Line: $(a,b)$, $ax+by+c=0$ par perpendicular hai
  • Linear Programming: Distance constraints is formula ko use karti hain
  • Least Squares Regression: Perpendicular distances ko minimize karna

Ya socho: "Point ko line equation mein plug karo, normal vector $(a,b)$ ki length se divide karo, absolute value lo."

Denominator $\sqrt{a^2+b^2}$ bas normal vector ki magnitude hai—agar vectors jaante ho toh yaad rakhna aasaan hai!

Tum kitna chalna padega yeh pata karne ke liye, humare paas ek magic formula hai. Tum raaste ki equation lete ho (jo $ax+by+c=0$ jaisi dikhti hai), jahan tum khade ho $(x_1,y_1)$ woh plug karte ho, aur ek aisi number se divide karte ho jo raaste ki "steepness" par depend karta hai ($\sqrt{a^2+b^2}$). Answer tumhe sabse choti distance batata hai!

Absolute value bars $|\cdots |$ ka matlab hai "ignore karo ki number positive hai ya negative—bas batao kitna bada hai," kyunki distance negative nahi ho sakti!

#flashcards/maths

What is the formula for the distance from a point $(x_1,y_1)$ to a line $ax+by+c=0$? :: $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions ke saath
• D5 · Question bank — concept traps