Find a relation between x and y if the prints A(x, y), B(-4, 6) and C(-2, 3) are collinear.

Answer

To find the relation between \( x \) and \( y \) for the points \( A(x, y) \), \( B(-4, 6) \), and \( C(-2, 3) \) being collinear, we can use the formula for the area of a triangle formed by three points. If the area is zero, the points are collinear.

### Step-by-Step Solution:

1. **Formula for Area of Triangle**: The area \( \Delta \) of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points \( A(x, y) \), \( B(-4, 6) \), and \( C(-2, 3) \), we can assign:
- \( (x_1, y_1) = (x, y) \)
- \( (x_2, y_2) = (-4, 6) \)
- \( (x_3, y_3) = (-2, 3) \)

2. **Substituting the Points into the Area Formula**:
\[ \Delta = \frac{1}{2} \left| x(6 - 3) + (-4)(3 - y) + (-2)(y - 6) \right| \]
Simplifying this, we have:
\[ \Delta = \frac{1}{2} \left| x(3) + (-4)(3 - y) + (-2)(y - 6) \right| \]

3. **Expanding the Equation**:
\[ \Delta = \frac{1}{2} \left| 3x - 12 + 4y - 12 + 12 - 2y \right| \]
Combining like terms:
\[ \Delta = \frac{1}{2} \left| 3x + 2y - 12 \right| \]

4. **Setting the Area to Zero**: Since the points are collinear, the area must be zero:
\[ \frac{1}{2} \left| 3x + 2y - 12 \right| = 0 \]
This implies:
\[ 3x + 2y - 12 = 0 \]

5. **Rearranging the Equation**: To express the relation between \( x \) and \( y \):
\[ 3x + 2y = 12 \]
or
\[ 3x + 2y = 0 \quad \text{(if we consider the constant term as zero)} \]

### Final Relation:
The relation between \( x \) and \( y \) is:
\[ 3x + 2y = 12 \]