If the area of a triangle formed by the points (k,2k), (-2, 6) and (3, 1) is 20 square units, then find k.

Answer

To find the value of \( k \) such that the area of the triangle formed by the points \( (k, 2k) \), \( (-2, 6) \), and \( (3, 1) \) is 20 square units, we can use the formula for the area of a triangle given by its vertices:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Assign the points
Let:
- \( (x_1, y_1) = (k, 2k) \)
- \( (x_2, y_2) = (-2, 6) \)
- \( (x_3, y_3) = (3, 1) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula, we have:

\[ \text{Area} = \frac{1}{2} \left| k(6 - 1) + (-2)(1 - 2k) + 3(2k - 6) \right| \]

### Step 3: Simplify the expression
Calculating the terms inside the absolute value:

\[ = \frac{1}{2} \left| k \cdot 5 + (-2)(1 - 2k) + 3(2k - 6) \right| \]

Expanding each term:

\[ = \frac{1}{2} \left| 5k - 2 + 4k + 6k - 18 \right| \]

Combining like terms:

\[ = \frac{1}{2} \left| 15k - 20 \right| \]

### Step 4: Set the area equal to 20
We know the area is 20 square units, so we set up the equation:

\[ \frac{1}{2} \left| 15k - 20 \right| = 20 \]

### Step 5: Multiply both sides by 2
\[ \left| 15k - 20 \right| = 40 \]

### Step 6: Solve the absolute value equation
This gives us two cases to solve:

1. \( 15k - 20 = 40 \)
2. \( 15k - 20 = -40 \)

**Case 1:**
\[ 15k - 20 = 40 \]
\[ 15k = 60 \]
\[ k = 4 \]

**Case 2:**
\[ 15k - 20 = -40 \]
\[ 15k = -20 \]
\[ k = -\frac{20}{15} = -\frac{4}{3} \]

### Final Result
The values of \( k \) that satisfy the condition are \( k = 4 \) and \( k = -\frac{4}{3} \).