If the area of triangle is `35`sq.units with vertices `(2,-6), (5,4)` and `(k,4)`, find the values of 'k'

Answer

To find the values 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 the coordinates of its vertices.

### Step-by-Step Solution:

1. **Write the Area Formula**:
The area \( A \) of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our vertices \( (2, -6) \), \( (5, 4) \), and \( (k, 4) \):
\[ A = \frac{1}{2} \left| 2(4 - 4) + 5(4 - (-6)) + k(-6 - 4) \right| \]

2. **Substitute the Values**:
Substitute the coordinates into the area formula:
\[ A = \frac{1}{2} \left| 2(0) + 5(10) + k(-10) \right| \]
Simplifying this gives:
\[ A = \frac{1}{2} \left| 0 + 50 - 10k \right| = \frac{1}{2} \left| 50 - 10k \right| \]

3. **Set the Area Equal to 35**:
Since the area is given as \( 35 \) square units, we set up the equation:
\[ \frac{1}{2} \left| 50 - 10k \right| = 35 \]
Multiplying both sides by \( 2 \) gives:
\[ \left| 50 - 10k \right| = 70 \]

4. **Solve the Absolute Value Equation**:
The absolute value equation \( |x| = a \) gives two cases:
- Case 1: \( 50 - 10k = 70 \)
- Case 2: \( 50 - 10k = -70 \)

**Case 1**:
\[ 50 - 10k = 70 \]
Rearranging gives:
\[ -10k = 70 - 50 \implies -10k = 20 \implies k = -2 \]

**Case 2**:
\[ 50 - 10k = -70 \]
Rearranging gives:
\[ -10k = -70 - 50 \implies -10k = -120 \implies k = 12 \]

5. **Conclusion**:
The values of \( k \) that satisfy the area condition are:
\[ k = -2 \quad \text{and} \quad k = 12 \]