Find values of k if area of triangle is 4 sq. units and vertices are
(i) `(k, 0), (4, 0), (0, 2)`
(ii) `(–2, 0), (0, 4), (0, k)`

Options

A. 1)0
2) 8
B. 1)8
2)0
C. 1) 0 or 8
2) 0 or 8
D. 1) 0
2) 0

Answer

To solve the problem of finding the values of \( k \) such that the area of the triangle formed by the given vertices is 4 square units, we will break it down into two parts as stated in the question.

### Part (i): Vertices \( (k, 0), (4, 0), (0, 2) \)

1. **Area Formula**: The area \( A \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

2. **Substituting the vertices**: For our vertices \( (k, 0), (4, 0), (0, 2) \):
- \( (x_1, y_1) = (k, 0) \)
- \( (x_2, y_2) = (4, 0) \)
- \( (x_3, y_3) = (0, 2) \)

The area becomes:
\[ A = \frac{1}{2} \left| k(0 - 2) + 4(2 - 0) + 0(0 - 0) \right| \]
Simplifying this gives:
\[ A = \frac{1}{2} \left| -2k + 8 \right| \]

3. **Setting the area equal to 4**:
\[ \frac{1}{2} \left| -2k + 8 \right| = 4 \]
Multiplying both sides by 2:
\[ \left| -2k + 8 \right| = 8 \]

4. **Solving the absolute value equation**:
This gives us two cases:
- Case 1: \( -2k + 8 = 8 \)
- Case 2: \( -2k + 8 = -8 \)

**Case 1**:
\[ -2k + 8 = 8 \implies -2k = 0 \implies k = 0 \]

**Case 2**:
\[ -2k + 8 = -8 \implies -2k = -16 \implies k = 8 \]

5. **Final values for Part (i)**: The values of \( k \) are \( k = 0 \) and \( k = 8 \).

### Part (ii): Vertices \( (-2, 0), (0, 4), (0, k) \)

1. **Area Formula**: Using the same area formula as before:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

2. **Substituting the vertices**: For our vertices \( (-2, 0), (0, 4), (0, k) \):
- \( (x_1, y_1) = (-2, 0) \)
- \( (x_2, y_2) = (0, 4) \)
- \( (x_3, y_3) = (0, k) \)

The area becomes:
\[ A = \frac{1}{2} \left| -2(4 - k) + 0(k - 0) + 0(0 - 4) \right| \]
Simplifying this gives:
\[ A = \frac{1}{2} \left| -8 + 2k \right| \]

3. **Setting the area equal to 4**:
\[ \frac{1}{2} \left| -8 + 2k \right| = 4 \]
Multiplying both sides by 2:
\[ \left| -8 + 2k \right| = 8 \]

4. **Solving the absolute value equation**:
This gives us two cases:
- Case 1: \( -8 + 2k = 8 \)
- Case 2: \( -8 + 2k = -8 \)

**Case 1**:
\[ -8 + 2k = 8 \implies 2k = 16 \implies k = 8 \]

**Case 2**:
\[ -8 + 2k = -8 \implies 2k = 0 \implies k = 0 \]

5. **Final values for Part (ii)**: The values of \( k \) are \( k = 0 \) and \( k = 8 \).

### Conclusion:
For both parts of the question, the values of \( k \) that satisfy the area condition are:
- \( k = 0 \)
- \( k = 8 \)