If the area of the quadrilateral whose angular points taken in order are (1,2) , (-5,6) (7,-4) and (h,-2) be zero, find the value of h :

Options

A. a. 2
B. b. 3
C. c. 5
D. d. 6

Answer

To find the value of \( h \) such that the area of the quadrilateral formed by the points \( A(1, 2) \), \( B(-5, 6) \), \( C(7, -4) \), and \( D(h, -2) \) is zero, we can use the formula for the area of a polygon based on its vertices.

### Step-by-step Solution:

1. **Understanding the Area Condition**:
The area of the quadrilateral is zero when the points are collinear. We can use the area formula for triangles to find the condition for collinearity.

2. **Using the Area Formula**:
The area \( A \) of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our quadrilateral, we can split it into two triangles, \( ABC \) and \( ACD \).

3. **Calculating Area of Triangle \( ABC \)**:
- Let \( A(1, 2) \), \( B(-5, 6) \), \( C(7, -4) \).
- Plugging into the formula:
\[ A_{ABC} = \frac{1}{2} \left| 1(6 + 4) + (-5)(-4 - 2) + 7(2 - 6) \right| \]
\[ = \frac{1}{2} \left| 1 \cdot 10 + (-5)(-6) + 7(-4) \right| \]
\[ = \frac{1}{2} \left| 10 + 30 - 28 \right| = \frac{1}{2} \left| 12 \right| = 6 \]

4. **Calculating Area of Triangle \( ACD \)**:
- Let \( D(h, -2) \).
- Plugging into the formula:
\[ A_{ACD} = \frac{1}{2} \left| 1(-2 + 4) + 7(-2 - 2) + h(2 + 4) \right| \]
\[ = \frac{1}{2} \left| 1 \cdot 2 + 7(-4) + h(6) \right| \]
\[ = \frac{1}{2} \left| 2 - 28 + 6h \right| = \frac{1}{2} \left| 6h - 26 \right| \]

5. **Setting the Total Area to Zero**:
Since the total area of the quadrilateral is zero, we have:
\[ A_{ABC} + A_{ACD} = 0 \implies 6 + \frac{1}{2} \left| 6h - 26 \right| = 0 \]
This implies:
\[ \frac{1}{2} \left| 6h - 26 \right| = -6 \]
Since area cannot be negative, we set:
\[ \left| 6h - 26 \right| = 12 \]

6. **Solving the Absolute Value Equation**:
This gives us two cases:
- Case 1: \( 6h - 26 = 12 \)
\[ 6h = 38 \implies h = \frac{38}{6} = \frac{19}{3} \]
- Case 2: \( 6h - 26 = -12 \)
\[ 6h = 14 \implies h = \frac{14}{6} = \frac{7}{3} \]

7. **Conclusion**:
The values of \( h \) that make the area of the quadrilateral zero are \( h = \frac{19}{3} \) and \( h = \frac{7}{3} \).