The vertex C of a triangle ABC moves on the line `L-=3x+4y+5=0`. The co-ordinates of the points A and B are `(2,7)` and `(5,8)` The locus of centroid of `DeltaABC` is a line parallel to:

Options

A. AB
B. BC
C. CA
D. L

Answer

To solve the problem step by step, we will find the locus of the centroid of triangle ABC as point C moves along the line \( L: 3x + 4y + 5 = 0 \).

### Step 1: Identify the coordinates of points A and B
The coordinates of points A and B are given as:
- \( A(2, 7) \)
- \( B(5, 8) \)

### Step 2: Use the formula for the centroid
The centroid \( G \) of triangle \( ABC \) with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

### Step 3: Substitute the coordinates of A and B
Let the coordinates of point C be \( C(x, y) \). Then, substituting the coordinates of points A and B into the centroid formula gives:
\[ G\left( \frac{2 + 5 + x}{3}, \frac{7 + 8 + y}{3} \right) = G\left( \frac{7 + x}{3}, \frac{15 + y}{3} \right) \]

### Step 4: Express the coordinates of the centroid in terms of \( x \) and \( y \)
Let \( G(h, k) \) represent the coordinates of the centroid:
\[ h = \frac{7 + x}{3} \quad \text{and} \quad k = \frac{15 + y}{3} \]

### Step 5: Solve for \( x \) and \( y \)
From the equations for \( h \) and \( k \), we can express \( x \) and \( y \) in terms of \( h \) and \( k \):
\[ x = 3h - 7 \quad \text{and} \quad y = 3k - 15 \]

### Step 6: Substitute \( y \) into the line equation
Since point C lies on the line \( L: 3x + 4y + 5 = 0 \), we substitute \( x \) and \( y \) into this equation:
\[ 3(3h - 7) + 4(3k - 15) + 5 = 0 \]

### Step 7: Simplify the equation
Expanding the equation:
\[ 9h - 21 + 12k - 60 + 5 = 0 \]
Combining like terms:
\[ 9h + 12k - 76 = 0 \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ 9h + 12k = 76 \]

### Step 9: Find the slope of the locus
To find the slope of the locus of the centroid, we can express this in slope-intercept form:
\[ 12k = -9h + 76 \quad \Rightarrow \quad k = -\frac{9}{12}h + \frac{76}{12} \]
This simplifies to:
\[ k = -\frac{3}{4}h + \frac{19}{3} \]

### Step 10: Identify the slope of the given line
The line \( L: 3x + 4y + 5 = 0 \) can be rewritten in slope-intercept form:
\[ 4y = -3x - 5 \quad \Rightarrow \quad y = -\frac{3}{4}x - \frac{5}{4} \]
The slope of this line is \( -\frac{3}{4} \).

### Conclusion
Since the slope of the locus of the centroid is also \( -\frac{3}{4} \), it is parallel to the line \( L \).