Key Concepts and Formulas

• Centroid of a Triangle: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
• Equation of a Line: A linear equation in the form $ax + by + c = 0$ represents a straight line.
• Locus: The set of all points satisfying a given condition.

Step-by-Step Solution

Step 1: Define Coordinates and the Centroid

We are given that point $P(\alpha, \beta)$ moves on the line $2x - 3y + 4 = 0$. This means that the coordinates of P satisfy the equation of the line, so we have:

$2\alpha - 3\beta + 4 = 0$

We are also given the fixed points $Q(1, 4)$ and $R(3, -2)$. Let the centroid of $\Delta PQR$ be $G(h, k)$. Our goal is to find the relationship between $h$ and $k$, which will give us the equation of the locus of the centroid.

Step 2: Apply the Centroid Formula

The coordinates of the centroid $G(h, k)$ are given by:

$h = \frac{\alpha + 1 + 3}{3} = \frac{\alpha + 4}{3}$ $k = \frac{\beta + 4 + (-2)}{3} = \frac{\beta + 2}{3}$

Step 3: Express $\alpha$ and $\beta$ in terms of $h$ and $k$

We need to eliminate $\alpha$ and $\beta$ to find the relationship between $h$ and $k$. From the centroid equations, we can express $\alpha$ and $\beta$ as:

$\alpha = 3h - 4$ $\beta = 3k - 2$

Step 4: Substitute into the Equation of the Line

Since $P(\alpha, \beta)$ lies on the line $2x - 3y + 4 = 0$, we can substitute the expressions for $\alpha$ and $\beta$ in terms of $h$ and $k$ into the equation of the line:

$2(3h - 4) - 3(3k - 2) + 4 = 0$

Step 5: Simplify the Equation

Simplify the equation to find the locus of the centroid:

$6h - 8 - 9k + 6 + 4 = 0$ $6h - 9k + 2 = 0$

Step 6: Rewrite in terms of x and y

Replace $h$ with $x$ and $k$ with $y$ to represent the locus in standard coordinate form:

$6x - 9y + 2 = 0$

Step 7: Analyze the Locus

The equation $6x - 9y + 2 = 0$ represents a straight line. We want to determine its properties. We can rewrite it as:

$9y = 6x + 2$ $y = \frac{6}{9}x + \frac{2}{9}$ $y = \frac{2}{3}x + \frac{2}{9}$

This is in the slope-intercept form, $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. Thus, the slope of the line representing the locus is $\frac{2}{3}$.

Common Mistakes & Tips

• Using (x, y) for both the moving point and the centroid: Avoid this by using different variables like $(\alpha, \beta)$ for the moving point and $(h, k)$ for the centroid, converting to $(x,y)$ only at the final step.
• Forgetting to substitute back into the original equation: The condition that P lies on the given line is crucial to eliminating the parameters $\alpha$ and $\beta$.
• Careless Arithmetic: Double-check your calculations, especially when simplifying the equation of the locus.

Summary

We found the locus of the centroid of $\Delta PQR$ by expressing the coordinates of the centroid in terms of the coordinates of the moving point $P(\alpha, \beta)$ and the fixed points $Q$ and $R$. We then used the fact that $P$ lies on the line $2x - 3y + 4 = 0$ to eliminate $\alpha$ and $\beta$ and obtain the equation of the locus. The resulting equation, $6x - 9y + 2 = 0$, represents a straight line with a slope of $\frac{2}{3}$.

Final Answer

The final answer is \boxed{B}, which corresponds to option (B).