Area of Triangle & Distance Formulas in Complex Form – Quick Mastery

Introduction

Complex numbers offer elegant formulas for triangle area and distances that are faster than coordinate geometry for JEE Main problems.

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1. Area of Triangle – The Key Formula

For vertices $z_1, z_2, z_3$:

$\boxed{\text{Area} = \frac{1}{4i} \begin{vmatrix} z_1 & \bar{z}_1 & 1 \\ z_2 & \bar{z}_2 & 1 \\ z_3 & \bar{z}_3 & 1 \end{vmatrix}}$

Simpler version (use this!): $\text{Area} = \frac{1}{2}|\text{Im}(\overline{z_1}z_2 + \overline{z_2}z_3 + \overline{z_3}z_1)|$

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Memory Trick:

"Area = half |Imaginary part of (z1̄z2 + z2̄z3 + z3̄z1)|"

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2. Equilateral Triangle – Golden Identity

For $z_1, z_2, z_3$ to be equilateral:

$\boxed{z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1}$

Equivalent forms:

• $(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0$
• $\frac{1}{z_1-z_2} + \frac{1}{z_2-z_3} + \frac{1}{z_3-z_1} = 0$

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JEE PYQs – Lightning Solutions

PYQ 1: JEE Main 2021

$z_1, z_2$ are roots of $z^2 + az + 12 = 0$ forming equilateral △ with origin. Find $|a|$.

60-second solution: With vertices $0, z_1, z_2$, equilateral condition: $z_1^2 + z_2^2 = z_1z_2$

From equation: $z_1+z_2=-a$, $z_1z_2=12$

$z_1^2+z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = a^2 - 24$

Equate: $a^2 - 24 = 12$ $a^2 = 36$, $|a|=6$

Answer: $6$

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PYQ 2: Classic Problem

Find area of △ with $z, iz, z+iz$

Solution using shortcut: Let $z = x+iy$ Area = $\frac{1}{2}|z|^2$ (from rotation theorem)

Or calculate: $\text{Im}(\bar{z}(iz) + \overline{iz}(z+iz) + \overline{z+iz}(z))$ Simplify to $\frac{1}{2}(x^2+y^2) = \frac{1}{2}|z|^2$

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3. Distance from Point to Line

Line: $\bar{a}z + a\bar{z} + b = 0$ where $b$ is real

Distance from $z_0$: $\boxed{d = \frac{|\bar{a}z_0 + a\bar{z}_0 + b|}{2|a|}}$

Example: Distance from $2+3i$ to $\bar{z} + z + 2 = 0$

$a=1$, $z_0=2+3i$: $d = \frac{|1\cdot(2+3i)+1\cdot(2-3i)+2|}{2\cdot1} = \frac{|6|}{2}=3$

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4. Collinearity Condition

$z_1, z_2, z_3$ collinear if: $\frac{z_3-z_1}{z_2-z_1} \ \text{is real}$

Or equivalently: $\begin{vmatrix} z_1 & \bar{z}_1 & 1 \\ z_2 & \bar{z}_2 & 1 \\ z_3 & \bar{z}_3 & 1 \end{vmatrix} = 0$

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Quick Reference Table

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Special Cases

Case 1: Origin as vertex

For vertices $0, z_1, z_2$: Area = $\frac{1}{2}|\text{Im}(\overline{0}z_1 + \overline{z_1}z_2 + \overline{z_2}0)|$ = $\frac{1}{2}|\text{Im}(\overline{z_1}z_2)|$

Since $\overline{z_1}z_2$ is generally complex, its imaginary part gives area.

Case 2: Equilateral with circumcenter at origin

Then:

• $|z_1|=|z_2|=|z_3|$
• $z_1+z_2+z_3=0$
• $z_1^2+z_2^2+z_3^2=0$

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Practice Problems

Problem 1: Vertices $1+i, 2-i, -1+2i$. Find area.

Solution using shortcut: Area = $\frac{1}{2}|\text{Im}(\overline{(1+i)}(2-i) + \overline{(2-i)}(-1+2i) + \overline{(-1+2i)}(1+i))|$

Calculate:

1. $(1-i)(2-i) = 2-i-2i+i^2 = 1-3i$
2. $(2+i)(-1+2i) = -2+4i-i+2i^2 = -4+3i$
3. $(-1-2i)(1+i) = -1-i-2i-2i^2 = 1-3i$

Sum = $(1-3i) + (-4+3i) + (1-3i) = -2-3i$

Imaginary part = $-3$ Area = $\frac{1}{2}|-3| = 1.5$

Answer: $1.5$

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Problem 2: Show $z, \omega z, \omega^2 z$ form equilateral △.

Solution: Check identity: $z^2 + (\omega z)^2 + (\omega^2 z)^2 = z^2(1+\omega^2+\omega^4) = z^2(1+\omega^2+\omega)$ $= z^2(0) = 0$ (since $1+\omega+\omega^2=0$)

$z(\omega z) + (\omega z)(\omega^2 z) + (\omega^2 z)z = z^2(\omega + \omega^3 + \omega^2) = z^2(\omega+1+\omega^2)=0$

Both sides equal, so equilateral ✓

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Exam Strategy

1. For area: Use $\frac{1}{2}|\text{Im}(\bar{z_1}z_2+\bar{z_2}z_3+\bar{z_3}z_1)|$ - fastest method
2. For equilateral: Check $z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$
3. For distance: Formula works for any line in $\bar{a}z+a\bar{z}+b=0$ form
4. Time target: 1-2 minutes per problem

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Remember: Complex methods often avoid messy coordinate expansions, saving time in JEE.

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Related: Rotation Methods | Geometry Shortcuts