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

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

The locus of R is a straight line (option A).

Step 1: Set Up the Problem

Let the coordinates of point P be (0, k) on the OY axis and the coordinates of Q be (x, 0) on the OX axis. These points form the segment PQ.

Step 2: Define the Triangle Vertices

Considering segment PQ as the base of the equilateral triangle, point R will form an equilateral triangle with base PQ. The coordinates of point R can be represented as \((x_r, y_r)\).

Step 3: Use Properties of Equilateral Triangle

In an equilateral triangle, the height is perpendicular to the base and divides the base into two equal segments. The side of the triangle is equal to the distance PQ which is \(\sqrt{x^2 + k^2}\).

Step 4: Calculate the Coordinates of R

Using the rotation matrices and maintaining equal side lengths \(\sqrt{x^2 + k^2}\), the possible coordinates for R are obtained by rotating Q by 60 degrees around the midpoint of PQ and then translating it appropriately. This gives \(x_r = \frac{x}{2} - \frac{k\sqrt{3}}{2}, \; y_r = \frac{k}{2} + \frac{x\sqrt{3}}{2}\).

Step 5: Derive the Equation

The derived coordinates \((x_r, y_r)\) will form a relationship as \(y_r - \frac{k}{2} = \sqrt{3}\left(x_r - \frac{x}{2}\right)\). This is a linear equation representing a straight line.

Step 6: Confirm the Locus

Since the equation derived is a linear equation, the path traced by R, as Q moves along the OX axis, would be a straight line.

Key Concepts

Coordinates Geometry

In coordinate geometry, we work with points on a coordinate plane defined by ordered pairs of numbers, typically in the form \((x, y)\).The horizontal axis is labeled as OX and the vertical axis as OY.
Each point is defined by its distance from these axes. For example, a point like \((0, k)\) lies directly on the OY axis because its x-coordinate is zero.

The concept of a locus in coordinate geometry refers to the set of all points that satisfy certain conditions. Determining the coordinates of these points is essential in solving geometry problems on a plane.
By examining the situation described in any given problem, you can visualize and derive the path or shape that these points will trace across the plane. The goal is to identify these points and understand their geometric relationship with each other.

Equilateral Triangle Properties

An equilateral triangle is a special type of triangle where all three sides are equal in length, and each angle measures \(60^\circ\).This equality gives the equilateral triangle unique properties that are useful when solving geometry problems.
For instance, in an equilateral triangle, the height is perpendicular to the base and divides it into two equal parts.

When forming an equilateral triangle on a segment like PQ, the point R is such that it is the same distance from P as it is from Q.

• The height is a crucial element as it aids in calculating the exact position of the third vertex, R, using geometric methods.
• In our problem, using these properties allows us to derive the coordinates of R based on its equilateral and symmetrical nature, guaranteeing balanced distances among its vertices.

Equation of a Line

The equation of a line is a fundamental concept in coordinates geometry. It expresses a relationship between the x and y coordinates of points along a line. One common form is the slope-intercept form, expressed as \(y = mx + b\).
Here, \(m\) represents the line's slope, which indicates the steepness and direction, while \(b\) describes the y-intercept, the point where the line crosses the OY axis.

In the exercise, once we obtain the coordinates of the point R, we use these to derive a linear equation that describes the locus of R as Q varies.
The solution illuminates that the relationship between \((x_r, y_r)\) isn't just indicative of a random shape but forms a straight line because it fits a linear equation of the form \(y = mx + b\).
Understanding how to derive and interpret these equations is vital for determining the precise path or location of points defined by such equations.

Rotation Matrices

Rotation matrices provide a method to rotate points around an origin or a specified point in coordinate geometry. A rotation matrix in 2D is represented as:
\[\begin{bmatrix}\cos\theta & -\sin\theta \\sin\theta & \cos\theta\end{bmatrix}\]

Applying this to a point \((x, y)\), you obtain a new point \((x', y')\) after rotation by an angle \(\theta\).The concept is crucial for deriving the coordinates of point R as equilateral triangles involve rotations about points.

• In the given problem, R is derived by rotating the other vertex by \(60^\circ\) around a midpoint. This involves a rotation matrix with angle \(60^\circ\) being applied to the midpoint to get the new coordinates.
• Through this process, rotation matrices provide the systematic method to adjust the points within the geometric shape, ensuring accuracy and consistency.

Understanding and utilizing these matrices empower students to manage rotations in various geometric and analytic contexts efficiently.