If A is (1,-2), B (3,k), C(-3,1) and D(k,4) where lines AB`bot` CD then :k=

Options

A. `-5/12`
B. `5//12`
C. `-12//5`
D. `12//5`

Answer

To solve the problem, we need to find the value of \( k \) such that the lines \( AB \) and \( CD \) are perpendicular. We will use the concept of slopes of lines.

### Step-by-Step Solution:

1. **Identify the Points**:
- Let \( A = (1, -2) \)
- Let \( B = (3, k) \)
- Let \( C = (-3, 1) \)
- Let \( D = (k, 4) \)

2. **Calculate the Slope of Line AB**:
- The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
- For line \( AB \):
\[ m_1 = \frac{k - (-2)}{3 - 1} = \frac{k + 2}{2} \]

3. **Calculate the Slope of Line CD**:
- For line \( CD \):
\[ m_2 = \frac{4 - 1}{k - (-3)} = \frac{3}{k + 3} \]

4. **Set Up the Perpendicularity Condition**:
- Two lines are perpendicular if the product of their slopes is -1:
\[ m_1 \cdot m_2 = -1 \]
- Substituting the slopes:
\[ \left(\frac{k + 2}{2}\right) \cdot \left(\frac{3}{k + 3}\right) = -1 \]

5. **Cross-Multiply to Eliminate the Fractions**:
- This gives:
\[ 3(k + 2) = -2(k + 3) \]

6. **Expand Both Sides**:
- Expanding the left side:
\[ 3k + 6 \]
- Expanding the right side:
\[ -2k - 6 \]

7. **Combine Like Terms**:
- Set the equation:
\[ 3k + 6 = -2k - 6 \]
- Rearranging gives:
\[ 3k + 2k = -6 - 6 \]
\[ 5k = -12 \]

8. **Solve for k**:
- Dividing both sides by 5:
\[ k = -\frac{12}{5} \]

### Final Answer:
The value of \( k \) is \( -\frac{12}{5} \).

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