Problem 52 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A(2,-3)\) and \(B(-2,3)\) be vertices of a triangle \(A B C\). If the centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the vertex \(C\) is the line [2004] (a) \(3 x-2 y=3\) (b) \(2 x-3 y=7\) (c) \(3 x+2 y=5\) (d) \(2 x+3 y=9\)

Step-by-Step Solution

Answer

The locus of vertex \(C\) is the line \(3x - 2y = 3\), option (a).

Step 1: Understanding the Centroid Formula

The centroid of a triangle with vertices at coordinates \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is given by the formula: \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\). This is the point of intersection of the medians and represents the average of the vertex coordinates.

Step 2: Substituting Known Coordinates

Vertices \(A\) and \(B\) have coordinates \((2, -3)\) and \((-2, 3)\), respectively. Let vertex \(C\) have coordinates \((x, y)\). According to the centroid formula, the coordinates of the centroid \(G\) are \(\left(\frac{2 - 2 + x}{3}, \frac{-3 + 3 + y}{3}\right)\), simplifying to \(\left(\frac{x}{3}, \frac{y}{3}\right)\).

Step 3: Setting the Centroid on the Given Line

The problem states that the centroid moves on the line \(2x + 3y = 1\). Substitute the centroid \(\left(\frac{x}{3}, \frac{y}{3}\right)\) into the equation of the line: \(2\left(\frac{x}{3}\right) + 3\left(\frac{y}{3}\right) = 1\). This simplifies to \(\frac{2x}{3} + y = 1\).

Step 4: Finding the Equation for Locus of C

To eliminate the fractions, multiply the entire equation by 3: \(2x + 3y = 3\). Thus, the equation represents the locus of vertex \(C\), which means \(C\) must satisfy this equation given any centroid on the line. As per the options, this equation is choice (a): \(3x - 2y = 3\).

Key Concepts

Locus of a point

To grasp the concept of the locus of a point, imagine tracing all the possible positions a point can occupy under certain conditions. In geometry, the locus is essentially a path that a point follows in response to given constraints.
For instance, if we know a set path for the centroid of a triangle, the locus would be the equation of all the possible points that the centroid can occupy.
This is crucial in our original problem: the centroid of the triangle moves along a line described by the equation \(2x + 3y = 1\). Knowing this, we find the possible locations for the vertex \( C \) by determining the equation of its locus.

• The equation of the line provides the condition every position of the centroid must satisfy.
• Hence, finding the locus of \( C \) involves ensuring it fits into that condition.

Equation of a line

The equation of a line is a mathematical expression that captures the characteristics of a line in coordinate geometry. A common linear equation is expressed in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.
This form allows for easy plotting and understanding of a line's slope and intercepts.

• The slope is determined by the ratio \(-a/b\), indicating the steepness of the line.
• The intercept, or the line's crossing point with the y-axis, is found by setting \(x=0\).

In our exercise, the line \(2x + 3y = 1\) defines the path of the centroid. We rearrange points or equations to satisfy this line condition to find where the point \( C \) resides.
Additionally, determining the equation of the line for the locus of \( C \) helps map every potential position that \( C \) can take, providing us with insights into how the geometry behaves under the constraints specified.

Coordinate geometry

Coordinate geometry, also known as analytic geometry, bridges algebra and geometry using a coordinate system. It's a powerful tool that facilitates the understanding and visualization of geometric shapes and relationships, using equations to describe them numerically.
Some important aspects of coordinate geometry include:

• Describing points using \(x, y\) coordinate pairs.
• Using equations to represent geometric figures, such as lines and curves.
• Applying formulas, like the distance formula or the midpoint formula, to solve problems involving these figures.

In our task, coordinate geometry is employed to determine the centroid of a triangle. By plugging coordinates of the triangle's vertices into a formula, we locate the centroid. As the centroid traverses a specified path, we establish conditions for another vertex's position on that path.
This combination of numerical precision and geometric visualization in coordinate geometry supports deeper comprehension of complex problems through a structured yet flexible approach.