To solve the problem, we need to find the value of \( x \) such that the ratio of the area of triangle \( DBC \) to the area of triangle \( ABC \) is \( \frac{1}{2} \).
### Step 1: Calculate the Area of Triangle ABC
The formula for the area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
For triangle \( ABC \) with points \( A(6, 3) \), \( B(-3, 5) \), and \( C(4, -2) \):
- \( x_1 = 6, y_1 = 3 \)
- \( x_2 = -3, y_2 = 5 \)
- \( x_3 = 4, y_3 = -2 \)
Substituting these values into the area formula:
\[
\text{Area}_{ABC} = \frac{1}{2} \left| 6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5) \right|
\]
Calculating each term:
1. \( 6(5 + 2) = 6 \times 7 = 42 \)
2. \( -3(-2 - 3) = -3 \times -5 = 15 \)
3. \( 4(3 - 5) = 4 \times -2 = -8 \)
Now, summing these:
\[
\text{Area}_{ABC} = \frac{1}{2} \left| 42 + 15 - 8 \right| = \frac{1}{2} \left| 49 \right| = \frac{49}{2}
\]
### Step 2: Calculate the Area of Triangle DBC
For triangle \( DBC \) with point \( D(x, 2x) \), \( B(-3, 5) \), and \( C(4, -2) \):
- \( x_1 = x, y_1 = 2x \)
- \( x_2 = -3, y_2 = 5 \)
- \( x_3 = 4, y_3 = -2 \)
Substituting these values into the area formula:
\[
\text{Area}_{DBC} = \frac{1}{2} \left| x(5 - (-2)) + (-3)(-2 - 2x) + 4(2x - 5) \right|
\]
Calculating each term:
1. \( x(5 + 2) = 7x \)
2. \( -3(-2 - 2x) = -3(-2 - 2x) = 6 + 6x \)
3. \( 4(2x - 5) = 8x - 20 \)
Now, summing these:
\[
\text{Area}_{DBC} = \frac{1}{2} \left| 7x + 6 + 6x + 8x - 20 \right| = \frac{1}{2} \left| 21x - 14 \right|
\]
### Step 3: Set Up the Ratio
According to the problem, we have:
\[
\frac{\text{Area}_{DBC}}{\text{Area}_{ABC}} = \frac{1}{2}
\]
Substituting the areas we calculated:
\[
\frac{\frac{1}{2} |21x - 14|}{\frac{49}{2}} = \frac{1}{2}
\]
This simplifies to:
\[
\frac{|21x - 14|}{49} = \frac{1}{2}
\]
Cross-multiplying gives:
\[
|21x - 14| = \frac{49}{2}
\]
### Step 4: Solve the Absolute Value Equation
This leads to two cases:
1. \( 21x - 14 = \frac{49}{2} \)
2. \( 21x - 14 = -\frac{49}{2} \)
**Case 1:**
\[
21x - 14 = \frac{49}{2}
\]
\[
21x = \frac{49}{2} + 14 = \frac{49}{2} + \frac{28}{2} = \frac{77}{2}
\]
\[
x = \frac{77}{42} = \frac{11}{6}
\]
**Case 2:**
\[
21x - 14 = -\frac{49}{2}
\]
\[
21x = -\frac{49}{2} + 14 = -\frac{49}{2} + \frac{28}{2} = -\frac{21}{2}
\]
\[
x = -\frac{21}{42} = -\frac{1}{2}
\]
### Final Solution
The possible values of \( x \) are \( \frac{11}{6} \) and \( -\frac{1}{2} \).
