If the lines `x-2y+3=0, 3x+ky+7=0` cut the coordinate axes in concylic points, then k=

Options

A. a) 1.5
B. b) 0.5
C. c) `-3//2`
D. d) `-4`

Answer

To solve the problem, we need to find the value of \( k \) such that the points where the lines \( x - 2y + 3 = 0 \) and \( 3x + ky + 7 = 0 \) intersect the coordinate axes are concyclic.

### Step-by-Step Solution:

1. **Identify the equations of the lines**:
- The first line is \( x - 2y + 3 = 0 \).
- The second line is \( 3x + ky + 7 = 0 \).

2. **Find the intercepts of the first line**:
- To find the x-intercept, set \( y = 0 \):
\[ x + 3 = 0 \implies x = -3 \quad \text{(x-intercept)} \]
- To find the y-intercept, set \( x = 0 \):
\[ -2y + 3 = 0 \implies y = \frac{3}{2} \quad \text{(y-intercept)} \]
- So, the intercepts of the first line are \( A(-3, 0) \) and \( B(0, \frac{3}{2}) \).

3. **Find the intercepts of the second line**:
- To find the x-intercept, set \( y = 0 \):
\[ 3x + 7 = 0 \implies x = -\frac{7}{3} \quad \text{(x-intercept)} \]
- To find the y-intercept, set \( x = 0 \):
\[ ky + 7 = 0 \implies y = -\frac{7}{k} \quad \text{(y-intercept)} \]
- So, the intercepts of the second line are \( C(-\frac{7}{3}, 0) \) and \( D(0, -\frac{7}{k}) \).

4. **Use the condition for concyclic points**:
- The points \( A, B, C, D \) are concyclic if the following condition holds:
\[ a_1 a_2 = b_1 b_2 \]
- Here, \( a_1 = 1 \), \( b_1 = -2 \) (from the first line), \( a_2 = 3 \), and \( b_2 = k \) (from the second line).
- Therefore, we have:
\[ 1 \cdot 3 = (-2) \cdot k \]
- Simplifying this gives:
\[ 3 = -2k \]

5. **Solve for \( k \)**:
- Rearranging the equation:
\[ k = -\frac{3}{2} \]

### Final Answer:
Thus, the value of \( k \) is \( -\frac{3}{2} \).