Problem 72 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(A(2,-3)\) and \(B(-2,1)\) 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 (A) \(2 x+3 y=9\)004] (B) \(2 x-3 y=7\) (C) \(3 x+2 y=5\) (D) \(3 x-2 y=3\)

Step-by-Step Solution

Answer

Locus of vertex C is the line \(2x + 3y = 9\).

Step 1: Understand the Centroid Formula

The centroid of a triangle with vertices \(A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)\) is given by the formula: \(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\). Here, we have vertices \(A(2,-3)\) and \(B(-2,1)\). Let \(C(x, y)\) be the third vertex.

Step 2: Write Centroid with respects to C

Plug the coordinates of vertices \(A\) and \(B\) into the centroid formula: \[ \left(\frac{2 - 2 + x}{3}, \frac{-3 + 1 + y}{3}\right) = \left(\frac{x}{3}, \frac{-2 + y}{3}\right) \].

Step 3: Use Line Equation for Centroid

Since the centroid moves on the line \(2x + 3y = 1\), substitute the centroid coordinates into this equation: \(2\left(\frac{x}{3}\right) + 3\left(\frac{-2 + y}{3}\right) = 1\).

Step 4: Simplify the Equation

Simplify the equation: \(\frac{2x}{3} + \frac{3(-2 + y)}{3} = 1\) becomes \(\frac{2x - 6 + 3y}{3} = 1\). Multiply through by 3: \(2x - 6 + 3y = 3\).

Step 5: Rearrange to Find Line Locus

Rearrange the equation to find the locus of \(C\): \(2x + 3y = 9\).

Key Concepts

Centroid of a Triangle

The centroid of a triangle is a crucial concept in geometry. It is the point where the three medians of the triangle intersect. A median is a line segment connecting a vertex to the midpoint of the opposite side. The centroid is also known as the "center of mass" or "barycenter" of the triangle. The formula for finding the centroid is: \[\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\]This formula averages the x-coordinates and y-coordinates of the triangle's vertices, providing a balanced 'center'. This means that no matter the shape of the triangle (as long as it is not degenerate), this point is always inside the triangle.

Equation of a Line

Understanding the equation of a line is fundamental in coordinate geometry. The most common form of a line's equation in a Cartesian plane is the standard form: \[ ax + by = c \]where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. In this exercise, the line the centroid moves on is given by the equation \(2x + 3y = 1\). This allows you to determine the path of the centroid by examining the intersections and relationships of \(x\) and \(y\). Here, solving for the centroid's movement involves substituting its coordinate expression into the line equation, which helps in determining the locus of another point, such as vertex \(C\) in the triangle.

Triangle Vertices

Vertices are the corner points of a triangle. In coordinate geometry, they are essential for computations involving shapes. Each vertex is defined by its coordinates \((x, y)\) in the Cartesian plane, and the specific arrangement of these points defines the shape and size of the triangle. In this problem, the triangle's vertices are given partially, with \(A(2, -3)\) and \(B(-2, 1)\), while \(C(x, y)\) is undetermined and needs to be found based on other conditions. These vertices are crucial in determining other properties of the triangle, like its centroid or its area, and are always used as input variables in corresponding geometric equations.

Coordinate Geometry

Coordinate Geometry, also known as analytic geometry, is a powerful mathematical tool that combines algebra and geometry to study geometric figures. This branch of geometry allows us to use coordinate pairs and algebraic equations to define and understand properties of geometric shapes, including triangles. By representing points in a Cartesian plane, we can precisely calculate distances, slopes, and midpoints, and solve problems involving shapes' transformations and loci. In this task, using coordinate geometry transforms the geometric problem of a triangle's attributes into an algebraic equation. This allows for precise and reasoned solutions by simply manipulating equations and coordinates.