JEE MainMathematics

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Locus of a Complex Number Defined by Differences of Squared Moduli

Let z_1=2+3i and z_2=3+4i. The set S={z ∈ C:|z-z_1|^2-|z-z_2|^2=|z_1-z_2|^2} represents a

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Ahyperbola with the length of the transverse axis 7Bhyperbola with eccentricity 2Cstraight line with the sum of its intercepts on the coordinate axes equals -18Dstraight line with the sum of its intercepts on the coordinate axes equals 14

Step-by-step solutionView

Correct answer

The equation reduces to the straight line $x + y = 7$ , whose axis intercepts are $7$ and $7$ , giving a sum of $14$ .

Option analysis

Why each option works or fails

A · hyperbola with the length of the transverse axis 7

Confusing the condition $∣ z - z_{1} ∣^{2} - ∣ z - z_{2} ∣^{2} = c$ with the definition of a hyperbola, which requires unsquared distances $∣ z - z_{1} ∣ - ∣ z - z_{2} ∣ = e x t c o n s t a n t$ . Recognize that the difference of squared Euclidean distances expands into linear terms in $x$ and $y$ because the quadratic terms $x^{2} + y^{2}$ cancel out, producing a straight line rather than a conic section.

B · hyperbola with eccentricity 2

Assuming that a difference between two distance terms involving fixed points must always define a hyperbola. Expand $z$ as $x + i y$ ; the terms $∣ z - z_{1} ∣^{2}$ and $∣ z - z_{2} ∣^{2}$ both contain $x^{2} + y^{2}$ , which cancel upon subtraction, leaving a degree-1 polynomial equation representing a straight line.

C · straight line with the sum of its intercepts on the coordinate axes equals -18

Making a sign error when expanding the squared terms or solving for the intercepts, leading to negative intercepts. Carefully expand $(x - 2)^{2} + (y - 3)^{2} - [(x - 3)^{2} + (y - 4)^{2}] = 2$ , yielding $2 x + 2 y - 12 = 2$ , or $x + y = 7$ , which has positive intercepts $7$ and $7$ .

D · straight line with the sum of its intercepts on the coordinate axes equals 14

This is the correct option. Substituting $z = x + i y$ and simplifying the squared distances yields $x + y = 7$ , with intercepts $a = 7$ and $b = 7$ , summing to $14$ .

Reviewed route

Solution

StepWorking

01Given

$z_{1} = 2 + 3 i$ , $z_{2} = 3 + 4 i$ , and the locus condition $∣ z - z_{1} ∣^{2} - ∣ z - z_{2} ∣^{2} = ∣ z_{1} - z_{2} ∣^{2}$ .

02Approach

Substitute $z = x + i y$ and express each squared modulus in terms of real coordinates $x$ and $y$ . Then simplify to determine the resulting curve.

03Execute

First, evaluate the right-hand side: $∣ z_{1} - z_{2} ∣^{2} = ∣ (2 - 3) + i (3 - 4) ∣^{2} = ∣ - 1 - i ∣^{2} = (- 1)^{2} + (- 1)^{2} = 2$ . Next, write the terms on the left: $∣ z - z_{1} ∣^{2} = (x - 2)^{2} + (y - 3)^{2} = x^{2} - 4 x + 4 + y^{2} - 6 y + 9$ , and $∣ z - z_{2} ∣^{2} = (x - 3)^{2} + (y - 4)^{2} = x^{2} - 6 x + 9 + y^{2} - 8 y + 16$ . Subtracting them gives: $[(x - 2)^{2} + (y - 3)^{2}] - [(x - 3)^{2} + (y - 4)^{2}] = (- 4 x + 6 x) + (- 6 y + 8 y) + (13 - 25) = 2 x + 2 y - 12$ . Setting this equal to 2: $2 x + 2 y - 12 = 2 ⟹ 2 x + 2 y = 14 ⟹ x + y = 7$ .

04Execute

The line is $x + y = 7$ , or in intercept form $\frac{x}{7} + \frac{y}{7} = 1$ . The $x$ -intercept is $7$ and the $y$ -intercept is $7$ . The sum of the intercepts on the coordinate axes is $7 + 7 = 14$ .

✓Verify

Since both $∣ z - z_{1} ∣^{2}$ and $∣ z - z_{2} ∣^{2}$ have quadratic terms $x^{2} + y^{2}$ with identical coefficients ( $1$ ), they cancel out completely under subtraction. Thus, the equation must be linear (a straight line), immediately ruling out hyperbolas (options 0 and 1). At $x = 0, y = 7$ and $y = 0, x = 7$ , sum of intercepts is $14$ .

Hints that build this answer step by step

Let $z = x + i y$ . What is the algebraic expression for $∣ z - z_{1} ∣^{2} - ∣ z - z_{2} ∣^{2}$ given $z_{1} = 2 + 3 i$ and $z_{2} = 3 + 4 i$ ?
[(x-2)^2 + (y-3)^2] - [(x-3)^2 + (y-4)^2] = 2x + 2y - 12
What is the value of $∣ z_{1} - z_{2} ∣^{2}$ ?
2
Equating $2 x + 2 y - 12 = 2$ , what is the resulting equation of the line and the sum of its intercepts?
x + y = 7, with sum of intercepts 14

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is this a straight line and not a hyperbola, even though it involves the difference between distances to two points?

A hyperbola is defined by the difference of unsquared distances, $∣ z - z_{1} ∣ - ∣ z - z_{2} ∣ = constant$ . When the distances are squared, the $x^{2} + y^{2}$ terms cancel out completely, yielding a first-degree equation in $x$ and $y$ , which is always a straight line.

Answer

The equation reduces to the straight line $x + y = 7$ , whose axis intercepts are $7$ and $7$ , giving a sum of $14$ .

Why each option works or fails

A: hyperbola with the length of the transverse axis 7 - Confusing the condition $∣ z - z_{1} ∣^{2} - ∣ z - z_{2} ∣^{2} = c$ with the definition of a hyperbola, which requires unsquared distances $∣ z - z_{1} ∣ - ∣ z - z_{2} ∣ = e x t c o n s t a n t$ . Recognize that the difference of squared Euclidean distances expands into linear terms in $x$ and $y$ because the quadratic terms $x^{2} + y^{2}$ cancel out, producing a straight line rather than a conic section.
B: hyperbola with eccentricity 2 - Assuming that a difference between two distance terms involving fixed points must always define a hyperbola. Expand $z$ as $x + i y$ ; the terms $∣ z - z_{1} ∣^{2}$ and $∣ z - z_{2} ∣^{2}$ both contain $x^{2} + y^{2}$ , which cancel upon subtraction, leaving a degree-1 polynomial equation representing a straight line.
C: straight line with the sum of its intercepts on the coordinate axes equals -18 - Making a sign error when expanding the squared terms or solving for the intercepts, leading to negative intercepts. Carefully expand $(x - 2)^{2} + (y - 3)^{2} - [(x - 3)^{2} + (y - 4)^{2}] = 2$ , yielding $2 x + 2 y - 12 = 2$ , or $x + y = 7$ , which has positive intercepts $7$ and $7$ .
D · correct: straight line with the sum of its intercepts on the coordinate axes equals 14 - This is the correct option. Substituting $z = x + i y$ and simplifying the squared distances yields $x + y = 7$ , with intercepts $a = 7$ and $b = 7$ , summing to $14$ .

Step-by-step solution

Given: $z_{1} = 2 + 3 i$ , $z_{2} = 3 + 4 i$ , and the locus condition $∣ z - z_{1} ∣^{2} - ∣ z - z_{2} ∣^{2} = ∣ z_{1} - z_{2} ∣^{2}$ .
Approach: Substitute $z = x + i y$ and express each squared modulus in terms of real coordinates $x$ and $y$ . Then simplify to determine the resulting curve.
Execute: First, evaluate the right-hand side: $∣ z_{1} - z_{2} ∣^{2} = ∣ (2 - 3) + i (3 - 4) ∣^{2} = ∣ - 1 - i ∣^{2} = (- 1)^{2} + (- 1)^{2} = 2$ . Next, write the terms on the left: $∣ z - z_{1} ∣^{2} = (x - 2)^{2} + (y - 3)^{2} = x^{2} - 4 x + 4 + y^{2} - 6 y + 9$ , and $∣ z - z_{2} ∣^{2} = (x - 3)^{2} + (y - 4)^{2} = x^{2} - 6 x + 9 + y^{2} - 8 y + 16$ . Subtracting them gives: $[(x - 2)^{2} + (y - 3)^{2}] - [(x - 3)^{2} + (y - 4)^{2}] = (- 4 x + 6 x) + (- 6 y + 8 y) + (13 - 25) = 2 x + 2 y - 12$ . Setting this equal to 2: $2 x + 2 y - 12 = 2 ⟹ 2 x + 2 y = 14 ⟹ x + y = 7$ .
Execute: The line is $x + y = 7$ , or in intercept form $\frac{x}{7} + \frac{y}{7} = 1$ . The $x$ -intercept is $7$ and the $y$ -intercept is $7$ . The sum of the intercepts on the coordinate axes is $7 + 7 = 14$ .
Verify: Since both $∣ z - z_{1} ∣^{2}$ and $∣ z - z_{2} ∣^{2}$ have quadratic terms $x^{2} + y^{2}$ with identical coefficients ( $1$ ), they cancel out completely under subtraction. Thus, the equation must be linear (a straight line), immediately ruling out hyperbolas (options 0 and 1). At $x = 0, y = 7$ and $y = 0, x = 7$ , sum of intercepts is $14$ .

Locus of a Complex Number Defined by Differences of Squared Moduli

What feels right?

The equation reduces to the straight line $x + y = 7$ , whose axis intercepts are $7$ and $7$ , giving a sum of $14$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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