Using determinant, find k if area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). a) 12 b) 2 c) -12,-2 d) 12,-2

Options

A. 12
B. -2
C. -12,-2
D. 12,-2

Answer

To find the value of \( k \) such that the area of the triangle formed by the vertices \( (2, -6) \), \( (5, 4) \), and \( (k, 4) \) is 35 square units, we can use the formula for the area of a triangle given by its vertices using determinants.

### Step-by-step Solution:

1. **Area Formula**: The area \( A \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

2. **Substituting the vertices**: For our triangle, the vertices are \( (2, -6) \), \( (5, 4) \), and \( (k, 4) \). Thus, we can substitute these values into the area formula:
\[ A = \frac{1}{2} \left| \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \right| \]

3. **Calculating the determinant**: We compute the determinant:
\[ \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = 2 \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} - (-6) \begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} + 1 \begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} \]

The determinants of the 2x2 matrices are calculated as follows:
- \( \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} = 4 \cdot 1 - 4 \cdot 1 = 0 \)
- \( \begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} = 5 \cdot 1 - k \cdot 1 = 5 - k \)
- \( \begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} = 5 \cdot 4 - k \cdot 4 = 20 - 4k \)

Plugging these back in:
\[ = 2(0) + 6(5 - k) + (20 - 4k) = 30 - 6k + 20 - 4k = 50 - 10k \]

4. **Setting the area**: The area is given as 35 square units, thus:
\[ \frac{1}{2} |50 - 10k| = 35 \]

Multiplying both sides by 2 gives:
\[ |50 - 10k| = 70 \]

5. **Solving the absolute value equation**: This leads to two cases:
- Case 1: \( 50 - 10k = 70 \)
\[ -10k = 70 - 50 \implies -10k = 20 \implies k = -2 \]
- Case 2: \( 50 - 10k = -70 \)
\[ -10k = -70 - 50 \implies -10k = -120 \implies k = 12 \]

6. **Final values of \( k \)**: Thus, the values of \( k \) are:
\[ k = -2 \quad \text{and} \quad k = 12 \]

### Conclusion:
The possible values of \( k \) are \( -2 \) and \( 12 \).