Find the values of k if area of tringle is 4 sq. units and dvertices are : (i) (k,0), (4,0), (0,2)

Explanation

To find the values of \( k \) such that the area of the triangle formed by the vertices \( (k, 0) \), \( (4, 0) \), and \( (0, 2) \) is 4 square units, we can use the determinant method for calculating the area of a triangle.

### Step 1: Set up the area formula using determinants

The area \( A \) of a triangle given its vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:

\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

For our triangle, the vertices are \( (k, 0) \), \( (4, 0) \), and \( (0, 2) \). Thus, we can substitute these values into the formula:

\[ A = \frac{1}{2} \left| \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \right| \]

### Step 2: Calculate the determinant

Now, we need to calculate the determinant:

\[ \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \]

We can expand this determinant along the second column:

\[ = 0 \cdot \begin{vmatrix} 4 & 1 \\ 0 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} k & 1 \\ 0 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} \]

Calculating the \( 2 \times 2 \) determinant:

\[ \begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} = k \cdot 1 - 4 \cdot 1 = k - 4 \]

Thus, the area becomes:

\[ A = \frac{1}{2} \left| 2(k - 4) \right| = |k - 4| \]

### Step 3: Set the area equal to 4 square units

According to the problem, the area of the triangle is 4 square units:

\[ |k - 4| = 4 \]

### Step 4: Solve the absolute value equation

This absolute value equation gives us two cases to solve:

1. \( k - 4 = 4 \)
2. \( k - 4 = -4 \)

**Case 1:**

\[ k - 4 = 4 \implies k = 8 \]

**Case 2:**

\[ k - 4 = -4 \implies k = 0 \]

### Conclusion

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

\[ k = 8 \quad \text{or} \quad k = 0 \]