Area of triangle using coordinate formula

2.3.11 · Maths › Coordinate Geometry

The Core Formula

Derivation from First Principles

YEH KAAM KYUN KARTA HAI? Chaliye ise trapezoid method use karke scratch se banate hain.

Step 1: Triangle ko trapezoids se x-axis tak enclose karo

Har vertex se x-axis tak vertical lines girao. Triangle ki har edge, uske saath do vertical lines aur neeche x-axis segment milke ek trapezoid banate hain. Inhi trapezoids ka signed sum triangle ka area deta hai.

Step 2: Signed trapezoid sum set up karo (A→B→C→A ke order mein ghoomte hue)

Hum triangle ki boundary ko order mein traverse karte hain. Har directed edge ke liye $(x_a,y_a)$ se $(x_b,y_b)$ tak, trapezoid ek signed area contribute karta hai: $\frac{1}{2}(y_a+y_b)(x_b-x_a)$

SIGNED KYUN, aur YEH KISI BHI ORDERING KO KYUN HANDLE KARTA HAI? Factor $(x_b-x_a)$ automatically negative hota hai jab hum left move karte hain aur positive jab hum right move karte hain. Toh hume yeh assume nahi karna padta ki $x_1\le x_2\le x_3$. Jab hum closed loop A→B→C→A par chalte hain, rightward edges area add karte hain aur leftward edges subtract karte hain, aur jo bachta hai woh exactly enclosed triangle hota hai — kisi bhi vertex arrangement ke liye. Agar hum counterclockwise chalein toh total positive hoga; clockwise negative dega. Isliye hum end mein $|\cdot |$ lete hain.

Step 3: Teeno directed-edge contributions likho

${\text{Edge}}_{A\to B}=\frac{1}{2}(y_1+y_2)(x_2-x_1)$ ${\text{Edge}}_{B\to C}=\frac{1}{2}(y_2+y_3)(x_3-x_2)$ ${\text{Edge}}_{C\to A}=\frac{1}{2}(y_3+y_1)(x_1-x_3)$

Step 4: Teeno ko sum karo (har ek apna sign already carry kar raha hai)

$\text{Signed Area}=\frac{1}{2}[(y_1+y_2)(x_2-x_1)+(y_2+y_3)(x_3-x_2)+(y_3+y_1)(x_1-x_3)]$

Note: teeno terms ADD hoti hain — signs already $(x_b-x_a)$ factors mein built-in hain. (Yahi crucial fix hai: yeh loop ke around ek sum hai, "do add karo, ek subtract karo" nahi.)

Step 5: Multiply out karo aur simplify karo

Har product expand karo: $(y_1+y_2)(x_2-x_1)=x_2{y}_1+x_2{y}_2-x_1{y}_1-x_1{y}_2$ $(y_2+y_3)(x_3-x_2)=x_3{y}_2+x_3{y}_3-x_2{y}_2-x_2{y}_3$ $(y_3+y_1)(x_1-x_3)=x_1{y}_3+x_1{y}_1-x_3{y}_3-x_3{y}_1$

Inhe add karo. $\pm x_1{y}_1$, $\pm x_2{y}_2$, $\pm x_3{y}_3$ diagonal terms pairs mein cancel ho jaate hain. Jo bachta hai: $x_2{y}_1-x_1{y}_2+x_3{y}_2-x_2{y}_3+x_1{y}_3-x_3{y}_1$

Ab har $x_i$ ke hisaab se group karo: $=x_1(y_3-y_2)+x_2(y_1-y_3)+x_3(y_2-y_1)$

Yeh target expression ka negative hai, yaani yeh $-[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$ ke barabar hai. Kyunki hum absolute value lete hain, sign irrelevant hai: $\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|✓$

Absolute value lo taaki vertex order ki parwah kiye bina area positive rahe. ✓

Worked Examples

Common Mistakes and Fixes

Memory Aids

Connections

• Coordinate Geometry Basics - points plot karne ka prerequisite
• Distance Formula - coordinates se related calculation
• Section Formula - coordinates se line segments ko divide karna
• Determinants - 3×3 determinant is area ka "linear algebra" view hai
• Colinearity Test - jab area = 0, points collinear hain
• Shoelace Theorem - n vertices wale polygons tak generalization
• Vector Cross Product - area as ½|AB × AC| vector version hai
• Heron's Formula - side lengths use karke area ka alternative method

#flashcards/maths

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
• D5 · Question bank — concept traps