Find the value of k so that the area of the triangle with vertices `(1,-1),(-4,2k)and (-k,-5)` is 24 square units.

Answer

To find the value of \( k \) such that the area of the triangle with vertices \( (1, -1) \), \( (-4, 2k) \), and \( (-k, -5) \) is 24 square units, we can use the formula for the area of a triangle given its vertices.

### Step-by-Step Solution:

1. **Identify the vertices**:
Let \( A(1, -1) \), \( B(-4, 2k) \), and \( C(-k, -5) \).

2. **Use the 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:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points \( A \), \( B \), and \( C \):
\[ A = \frac{1}{2} \left| 1(2k - (-5)) + (-4)(-5 - (-1)) + (-k)(-1 - 2k) \right| \]

3. **Simplify the expression**:
\[ A = \frac{1}{2} \left| 1(2k + 5) + (-4)(-5 + 1) + (-k)(-1 - 2k) \right| \]
\[ = \frac{1}{2} \left| 2k + 5 + 4(4) + k(1 + 2k) \right| \]
\[ = \frac{1}{2} \left| 2k + 5 + 16 + k + 2k^2 \right| \]
\[ = \frac{1}{2} \left| 2k^2 + 3k + 21 \right| \]

4. **Set the area equal to 24**:
Since the area is given as 24 square units, we have:
\[ \frac{1}{2} \left| 2k^2 + 3k + 21 \right| = 24 \]
Multiplying both sides by 2:
\[ \left| 2k^2 + 3k + 21 \right| = 48 \]

5. **Solve the absolute value equation**:
This gives us two equations:
\[ 2k^2 + 3k + 21 = 48 \quad \text{(1)} \]
\[ 2k^2 + 3k + 21 = -48 \quad \text{(2)} \]

6. **Solve equation (1)**:
\[ 2k^2 + 3k + 21 - 48 = 0 \]
\[ 2k^2 + 3k - 27 = 0 \]

Now we can factor or use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2} \]
\[ = \frac{-3 \pm \sqrt{9 + 216}}{4} = \frac{-3 \pm \sqrt{225}}{4} = \frac{-3 \pm 15}{4} \]
This gives us:
\[ k = \frac{12}{4} = 3 \quad \text{and} \quad k = \frac{-18}{4} = -\frac{9}{2} \]

7. **Solve equation (2)**:
\[ 2k^2 + 3k + 21 + 48 = 0 \]
\[ 2k^2 + 3k + 69 = 0 \]
The discriminant \( b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 69 = 9 - 552 < 0 \), so this equation has no real solutions.

### Final Values of \( k \):
Thus, the values of \( k \) that satisfy the area condition are:
\[ k = 3 \quad \text{and} \quad k = -\frac{9}{2} \]