Problem 150 - A Complete Resource Book in Mathematics for JEE Main

Question

\(A B C D\) is a rhombus. Its diagonals \(A C\) and \(B D\) intersect at the point \(M\) and satisfy \(B D=2 A C\). If the points \(D\) and \(M\) represent the complex numbers \(1+i\) and \(2-i\), respectively, then \(A\) represents the complex number (A) \(3-\frac{i}{2}\) or \(3+\frac{i}{2}\) (B) \(3+\frac{i}{2}\) or \(1+\frac{3}{2} i\) (C) \(3-i\) or \(1-3 i\) (D) None of these

Step-by-Step Solution

Answer

Option (A) is correct; solutions for \(A\) are \(3-\frac{i}{2}\) and \(3+\frac{i}{2}\).

Step 1: Understand the Problem

We have a rhombus \(ABCD\) where the diagonals \(AC\) and \(BD\) intersect at \(M\). We know \(BD = 2AC\) and the complex numbers for \(D\) and \(M\) are \(1+i\) and \(2-i\), respectively. We need the complex number for \(A\).

Step 2: Properties of Rhombus Diagonals

In a rhombus, diagonals bisect each other at right angles, and bisect each other into two equal parts. Thus, \(D\) and \(B\) are equidistant from \(M\), and so are \(A\) and \(C\). So, \(D - M = M - B\) and \(A - M = M - C\).

Step 3: Express Diagonals in Terms of Complex Numbers

For \(BD\), since \(M\) is the midpoint: \(BD = 2\times (D - M)\). Thus, \(B = M - (D - M) = 2M - D\). Substituting \(D = 1+i\) and \(M = 2-i\): \(B = (2-i) - (1+i) = 1 - 2i\).

Step 4: Applying Diagonal Length Relationship

Given \(BD = 2AC\), we have \(|BD| = 2 |AC|\). We already established \(|BD| = 2|D - M| = 2\times |AC|\). Thus \(|BD| = 4 |A - M|\).

Step 5: Use Complex Midpoint Property

Since \(M\) is the midpoint of \(AC\), \(2M = A + C\). So \(A + C = 2(2-i) = 4-2i\).

Step 6: Solve for Potential Positions of \(A\)

Using \(|A-M| = |C-M|\) and \(A + C = 4-2i\), solve by substituting potential options for position (A) \((x, y) = (3, -\frac{1}{2})\) and evaluate if these satisfy the midpoint condition. Check all alternatives the same way.

Step 7: Verification with Provided Options

Verify each answer against the above conditions. The solutions for \(A\) need to satisfy both the real and imaginary midpoint conditions and equality of distances.

Key Concepts

Geometry of Rhombus

A rhombus is a fascinating geometric figure with special properties that distinguish it from other quadrilaterals. It is a type of parallelogram where all sides are of equal length. This property is quite useful in various geometric problems, especially those related to coordinate geometry and algebraic calculations.

The key characteristics of a rhombus include:

• All sides are congruent.
• Opposite angles are equal.
• Diagonals intersect at right angles, forming perpendicular lines.
• Diagonals bisect each other into two equal segments.

These properties make the rhombus an interesting subject when it comes to solving problems involving complex numbers, where the geometric conditions are translated into algebraic expressions using complex coordinates.

Properties of Diagonals

The diagonals of a rhombus hold unique properties that often come into play during problem-solving. One major property is that the diagonals bisect each other at right angles, which provides a framework for solving the locations of vertices in specific coordinate planes, such as the complex plane.

A critical feature of a rhombus is that its diagonals not only intersect at a right angle but also divide each other into equal halves. This implies:

• For diagonal AC, point M is the midpoint such that AM = MC.
• For diagonal BD, point M is also the midpoint such that BM = MD.

In problems involving the intersection of diagonals, these properties are pivotal. For example, if we know the lengths of diagonals in terms of complex numbers, such as in our exercise, we can use the property that one diagonal is twice the length of the other to find unknown vertices using relationships established by the midpoint and equality of diagonal segments.

Midpoint in Complex Plane

In complex geometry, the concept of midpoints plays a crucial role when working with shapes like rhombuses. The midpoint formula in the complex plane is used to find the center point between two complex numbers. Given the points as complex numbers, the midpoint can be found using an average of the real and imaginary parts.

For two complex numbers, say Z1 (a + bi) and Z2 (c + di), the midpoint, Zm, is obtained by:\[Z_m = \frac{Z_1 + Z_2}{2} = \frac{(a+c) + (b+d)i}{2}\]

This concept is useful for problems involving symmetrical properties or where the midpoint corresponds to a specific balance point, such as in a rhombus. In the exercise, for example, we used M as the midpoint of diagonals to establish the relationships between various points. If given specific coordinates or conditions, one can solve for one vertex using the midpoint condition combined with other properties, like the equal length of sides or specific diagonal relationships.