Let $$A(2, -3)$$ and $$B(-2, 1)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line

Let the coordinates of the third vertex C of the triangle be given by:

$$ (h, k) $$

The coordinates of the given vertices are A(2, -3) and B(-2, 1). Let the coordinates of the centroid of the triangle ABC be denoted as G(X, Y).

The formula for the coordinates of the centroid of a triangle is:

$$ X = \frac{x_1 + x_2 + x_3}{3} $$

$$ Y = \frac{y_1 + y_2 + y_3}{3} $$

Substitute the given values into the centroid formula:

$$ X = \frac{2 + (-2) + h}{3} $$

$$ Y = \frac{-3 + 1 + k}{3} $$

Simplify the numerator expressions for both coordinates:

$$ X = \frac{h}{3} $$

$$ Y = \frac{k - 2}{3} $$

The problem states that the centroid G(X, Y) moves along a specific straight line:

$$ 2x + 3y = 1 $$

Since the point G(X, Y) lies on this line, its coordinates must satisfy the line equation:

$$ 2X + 3Y = 1 $$

Substitute the expressions for X and Y in terms of h and k back into this equation where $$ X = \frac{h}{3} $$ and $$ Y = \frac{k - 2}{3} $$:

$$ 2\left(\frac{h}{3}\right) + 3\left(\frac{k - 2}{3}\right) = 1 $$

Multiply the entire equation by 3 to clear the denominators:

$$ 2h + 3(k - 2) = 3 $$

Expand the terms inside the parentheses:

$$ 2h + 3k - 6 = 3 $$

Isolate the constant terms on the right side of the equation:

$$ 2h + 3k = 3 + 6 $$

$$ 2h + 3k = 9 $$

To find the locus of the moving vertex C, replace the coordinates (h, k) with the general coordinate variables (x, y):

$$ 2x + 3y = 9 $$

Final Answer:

The locus of the vertex C is the line $$ 2x + 3y = 9 $$.