Problem 44 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(O X\) and \(O Y\) are two coordinate axes. On \(O Y\) is taken a fixed point \(P\) and on \(O X\) any point \(Q .\) On \(P Q\) an equilateral triangle is described, its vertex \(R\) being on the side of \(P Q\) away from \(O\), then the locus of \(R\) will be (A) straight line (B) circle (C) ellipse (D) parabola

Step-by-Step Solution

Answer

(B) circle

Step 1: Understanding the Setup

First, we define the coordinate axes with the origin at point \( O \). The line \( OY \) is vertical, and \( OX \) is horizontal. Point \( P \) is fixed on \( OY \), and point \( Q \) can be any point on \( OX \). The line \( PQ \) is then a segment connecting points \( P \) and \( Q \).

Step 2: Forming the Triangle

An equilateral triangle \( PQR \) is described using the line segment \( PQ \) such that the vertex \( R \) is directed away from the origin. The triangle's properties include equal side lengths \( PQ = PR = QR \).

Step 3: Determining the Coordinates of P and Q

Take \( P \) as \((0, b)\) since it is on \( OY \), where \( b \) is constant. Let \( Q \) be \((a, 0)\) on \( OX \), where \( a \) is variable.

Step 4: Finding the Locus of R

Since \( PQR \) is an equilateral triangle, \( R \) lies on a perpendicular bisector of \( PQ \). Assuming \( R \)'s coordinates are \((x, y)\), because \( R \) is the same distance from \( P \) and \( Q \), we can form equations: 1. \((x - 0)^2 + (y - b)^2 = (x - a)^2 + y^2\).2. Use symmetry and rotation in the complex plane method, solving for the point \( R \) ensures it traces a circle as \( Q \) moves.

Key Concepts

Equilateral Triangle

An equilateral triangle is a special type of polygon where all three sides are of equal length. In addition to the equal sides, all internal angles of an equilateral triangle are equal, each measuring 60 degrees. This results in a perfectly symmetrical shape, making it a favorite in geometry.
One important aspect of an equilateral triangle is that the altitude, angle bisector, median, and perpendicular bisector all converge at the same point, known as the centroid. This unique property can be used to solve various geometric problems involving symmetry and balance.
In coordinate geometry, when forming an equilateral triangle on a plane, the positioning of its vertices can be critical. Here, the equilateral triangle is described using points on two perpendicular lines, where one point is fixed and the other can move. This setup leads to interesting conclusions regarding the geometric paths the vertices can trace under movement.

Locus

The term "locus" refers to the set of all points that satisfy a given condition or a set of conditions. In the context of this exercise, we are interested in the path that the vertex, denoted as point \( R \), takes as point \( Q \) moves along the \( OX \) axis while \( P \) remains fixed on \( OY \).
The locus of a point is essentially a geometric representation of all positions where the point can exist under specified constraints. Understanding the locus is crucial for visualizing and solving complex geometric problems. For example, the exercise shows that as the point \( Q \) travels horizontally, the path of \( R \) creates a circle.

• This circle is due to the equilateral triangle's balance and the equal distances maintained by point \( R \) as it relates to both \( P \) and \( Q \).
• By analyzing the symmetry and coordinate setup, we determine the locus by ensuring that point \( R \) is always equidistant from both \( P \) and \( Q \).

This powerful concept allows one to predict the movement and final positioning of geometric figures based on initial conditions.

Complex Plane

The complex plane is an extension of the traditional Cartesian coordinate system and is used to explore numbers with both real and imaginary components. In the complex plane, each point represents a complex number, with the horizontal axis as the real component and the vertical axis as the imaginary component. Instead of traditional coordinates like (x, y), points are expressed as \( z = a + bi \), where \( a \) and \( b \) are real numbers.
In this exercise, using the complex plane helps leverage symmetry and rotational properties of complex numbers to find the locus of \( R \). Often, rotations and symmetries in the complex plane correspond to geometric transformations in Euclidean space, offering a more algebraic approach to geometry.

• By considering \( PQ \) as a complex number and using the geometric property that transforms \( P \) and \( Q \) into \( R \), the circle's path is easily derived.
• This method simplifies finding loci as it incorporates rotation, symmetry, and distance much more effectively than working purely in Cartesian coordinates.

The complex plane is a powerful tool for solving geometric problems, especially when symmetry and rotations are involved.