IfA(2,-3) and B(-2, 1) are two vertices of a triangle and third vertex moves on the line 2x +3y 9,th the locus of the centroid of the triangle is 21. AIEEE-2011 (1) x-y=1 (2) 2x + 3y = 1 (3) 2x + 3y = 3 (4) 2x-3y = 1

Options

A. 2x + 3y = 1
B. 2x + y = 3
C. 2x - 3y = 1
D. x - y = 1

Explanation

To find the locus of the centroid of the triangle formed by the points A(2, -3), B(-2, 1), and a third vertex C(x, y) that moves along the line 2x + 3y = 9, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**:
The given vertices are A(2, -3) and B(-2, 1). Let the coordinates of the third vertex C be (x, y).

2. **Formula for the centroid**:
The coordinates of the centroid (G) of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) are given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
For our triangle, we have:
\[ G\left(\frac{2 + (-2) + x}{3}, \frac{-3 + 1 + y}{3}\right) \]
Simplifying this gives:
\[ G\left(\frac{x}{3}, \frac{-2 + y}{3}\right) \]

3. **Express x and y in terms of the centroid coordinates**:
Let the coordinates of the centroid be (h, k). Then we have:
\[ h = \frac{x}{3} \quad \Rightarrow \quad x = 3h \]
\[ k = \frac{-2 + y}{3} \quad \Rightarrow \quad y = 3k + 2 \]

4. **Substitute x and y into the line equation**:
Since the point C(x, y) lies on the line \(2x + 3y = 9\), we substitute x and y:
\[ 2(3h) + 3(3k + 2) = 9 \]
Simplifying this gives:
\[ 6h + 9k + 6 = 9 \]
\[ 6h + 9k = 3 \]
Dividing the entire equation by 3:
\[ 2h + 3k = 1 \]

5. **Write the locus equation**:
The locus of the centroid (h, k) is given by the equation:
\[ 2x + 3y = 1 \]

### Conclusion:
The locus of the centroid of the triangle is given by the equation:
\[ \boxed{2x + 3y = 1} \]