To find the value of \( k \) for the triangle with vertices \( (2, -6) \), \( (5, 4) \), and \( (k, 4) \) such that the area is \( 35 \) square units, we can use the formula for the area of a triangle given by its vertices using determinants.
### Step 1: Set up the area formula
The area \( A \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant:
\[
A = \frac{1}{2} \left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \right|
\]
For our triangle, the vertices are:
- \( (x_1, y_1) = (2, -6) \)
- \( (x_2, y_2) = (5, 4) \)
- \( (x_3, y_3) = (k, 4) \)
### Step 2: Substitute the vertices into the determinant
Substituting the coordinates into the determinant gives:
\[
A = \frac{1}{2} \left| \begin{vmatrix}
2 & -6 & 1 \\
5 & 4 & 1 \\
k & 4 & 1
\end{vmatrix} \right|
\]
### Step 3: Calculate the determinant
We can calculate the determinant as follows:
\[
\begin{vmatrix}
2 & -6 & 1 \\
5 & 4 & 1 \\
k & 4 & 1
\end{vmatrix} = 2 \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} - (-6) \begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} + 1 \begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} = 4 \cdot 1 - 4 \cdot 1 = 0 \)
2. \( \begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} = 5 \cdot 1 - k \cdot 1 = 5 - k \)
3. \( \begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} = 5 \cdot 4 - k \cdot 4 = 20 - 4k \)
Substituting these back into the determinant gives:
\[
= 2(0) + 6(5 - k) + (20 - 4k) = 30 - 6k + 20 - 4k = 50 - 10k
\]
### Step 4: Set the area equal to 35
Now we set the area equal to 35:
\[
\frac{1}{2} |50 - 10k| = 35
\]
Multiplying both sides by 2:
\[
|50 - 10k| = 70
\]
### Step 5: Solve the absolute value equation
This gives us two cases to solve:
1. \( 50 - 10k = 70 \)
2. \( 50 - 10k = -70 \)
**Case 1:**
\[
50 - 10k = 70 \implies -10k = 20 \implies k = -2
\]
**Case 2:**
\[
50 - 10k = -70 \implies -10k = -120 \implies k = 12
\]
### Conclusion
The values of \( k \) are \( -2 \) and \( 12 \). Therefore, the answer is:
**Option D: \( k = 12, -2 \)**
