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Question #40370

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all ordered pairs (x,y)∈R2(x, y) \in \mathbb{R}^2 such that the complex number z=x+iyz = x + iy satisfies the following two equations: z2(3−4i)+zˉ2(3+4i)=0z^2 (3 - 4i) + \bar{z}^2 (3 + 4i) = 0 zzˉ−(3−i)z−(3+i)zˉ=0z \bar{z} - (3-i)z - (3+i)\bar{z} = 0 If the points in SS form the vertices of a triangle in the xyxy-plane, find the area of this triangle.

Options

  • A.

    485\frac{48}{5}

  • B.

    245\frac{24}{5}

  • C.

    10

  • D.

    365\frac{36}{5}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Ordered Pair Equality|Algebraic Manipulation of i|Geometrical Interpretation of Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to translate complex algebraic equations into geometric loci using ordered pair representations.

The primary trap lies in the first equation: students often attempt to take the modulus or use polar forms clumsily, missing that W+Wˉ=0W + \bar{W} = 0 simply means the real part is zero, which elegantly factors into two perpendicular lines.

The 'Lethal Collision' occurs when intersecting these lines with the circle from the second equation.

Students who mechanically substitute without recognizing the geometric implications (a right triangle inscribed in a circle) will waste time on tedious algebra.

Furthermore, failing to recognize the origin as a valid intersection point due to careless division by zz will lead to an incomplete set of vertices, destroying the polygon area calculation.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptOrdered Pair Equality
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree