A point P moves on line `2x-3y+4=0` If `Q(1,4)` and `R(3,-2)` are fixed points, then the locus of the centroid of `triangle PQR` is a line: (a) with slope `3/2` (b) parallel to y-axis (c) with slope `2/3` (d) parallel to x-axis

Options

A. parallel to x-axis
B. with slope `(2)/(3)`
C. with slope `(3)/(2)`
D. parallel to y-axis

Answer

To solve the problem, we need to find the locus of the centroid of triangle PQR where point P moves along the line \(2x - 3y + 4 = 0\) and points Q(1, 4) and R(3, -2) are fixed.

### Step-by-Step Solution:

1. **Express the line equation in terms of y**:
The line equation is given by:
\[ 2x - 3y + 4 = 0 \]
Rearranging this gives:
\[ 3y = 2x + 4 \implies y = \frac{2}{3}x + \frac{4}{3} \]
Thus, any point \(P\) on this line can be represented as \(P(h, \frac{2}{3}h + \frac{4}{3})\).

2. **Identify the coordinates of the fixed points**:
The coordinates of the fixed points are:
\[ Q(1, 4) \quad \text{and} \quad R(3, -2) \]

3. **Use the centroid formula**:
The centroid \(G\) of triangle \(PQR\) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of points \(P\), \(Q\), and \(R\):
\[ G\left(\frac{h + 1 + 3}{3}, \frac{\left(\frac{2}{3}h + \frac{4}{3}\right) + 4 - 2}{3}\right) \]
Simplifying this:
\[ G\left(\frac{h + 4}{3}, \frac{\frac{2}{3}h + \frac{4}{3} + 2}{3}\right) \]

4. **Simplify the y-coordinate**:
The y-coordinate simplifies to:
\[ \frac{\frac{2}{3}h + \frac{4}{3} + 2}{3} = \frac{\frac{2}{3}h + \frac{4}{3} + \frac{6}{3}}{3} = \frac{\frac{2}{3}h + \frac{10}{3}}{3} = \frac{2h + 10}{9} \]
Therefore, the coordinates of the centroid \(G\) are:
\[ G\left(\frac{h + 4}{3}, \frac{2h + 10}{9}\right) \]

5. **Express \(h\) in terms of \(x\) and \(y\)**:
Let \(x = \frac{h + 4}{3}\) and \(y = \frac{2h + 10}{9}\).
From the first equation, we can express \(h\):
\[ h = 3x - 4 \]
Substituting \(h\) into the second equation:
\[ y = \frac{2(3x - 4) + 10}{9} = \frac{6x - 8 + 10}{9} = \frac{6x + 2}{9} \]

6. **Rearranging to find the locus**:
Multiplying through by 9 gives:
\[ 9y = 6x + 2 \implies 6x - 9y + 2 = 0 \]
Rearranging gives:
\[ 6x = 9y - 2 \]

7. **Finding the slope**:
The equation can be rewritten in slope-intercept form:
\[ y = \frac{2}{3}x + \frac{2}{9} \]
Thus, the slope of the line is \(\frac{2}{3}\).

### Conclusion:
The locus of the centroid of triangle \(PQR\) is a line with slope \(\frac{2}{3}\).