Find the value of k, if the area of the triangle formed by the points (k,0),(3,4) and (5,-2) is 10.

Answer

To find the value of \( k \) such that the area of the triangle formed by the points \( (k, 0) \), \( (3, 4) \), and \( (5, -2) \) is equal to 10, we can use the formula for the area of a triangle given by three vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### Step 1: Assign the points
Let:
- \( (x_1, y_1) = (k, 0) \)
- \( (x_2, y_2) = (3, 4) \)
- \( (x_3, y_3) = (5, -2) \)

### Step 2: Substitute the points into the area formula
Substituting the points into the area formula gives:

\[ \text{Area} = \frac{1}{2} \left| k(4 - (-2)) + 3(-2 - 0) + 5(0 - 4) \right| \]

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

\[ = \frac{1}{2} \left| k(4 + 2) + 3(-2) + 5(0 - 4) \right| \]

\[ = \frac{1}{2} \left| 6k - 6 - 20 \right| \]

\[ = \frac{1}{2} \left| 6k - 26 \right| \]

### Step 4: Set the area equal to 10
Since the area is given as 10, we set up the equation:

\[ \frac{1}{2} \left| 6k - 26 \right| = 10 \]

### Step 5: Multiply both sides by 2
Multiplying both sides by 2 to eliminate the fraction:

\[ \left| 6k - 26 \right| = 20 \]

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

1. \( 6k - 26 = 20 \)
2. \( 6k - 26 = -20 \)

#### Case 1: \( 6k - 26 = 20 \)
\[ 6k = 20 + 26 \]
\[ 6k = 46 \]
\[ k = \frac{46}{6} = \frac{23}{3} \]

#### Case 2: \( 6k - 26 = -20 \)
\[ 6k = -20 + 26 \]
\[ 6k = 6 \]
\[ k = 1 \]

### Final Answer
The values of \( k \) that satisfy the condition are:

\[ k = \frac{23}{3} \quad \text{and} \quad k = 1 \]

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