If the lines `2x-3y+7=0, 3x+ky+5=0` cut the coordinate axes in concyclic points then k=

Options

A. 2
B. 3
C. `-2`
D. `-3`

Answer

To solve the problem, we need to determine the value of \( k \) such that the points where the lines \( 2x - 3y + 7 = 0 \) and \( 3x + ky + 5 = 0 \) intersect the coordinate axes are concyclic.

### Step 1: Find the intercepts of the first line \( 2x - 3y + 7 = 0 \).

- **X-intercept**: Set \( y = 0 \):
\[ 2x + 7 = 0 \implies 2x = -7 \implies x = -\frac{7}{2} \]
So, the x-intercept is \( A\left(-\frac{7}{2}, 0\right) \).

- **Y-intercept**: Set \( x = 0 \):
\[ -3y + 7 = 0 \implies 3y = 7 \implies y = \frac{7}{3} \]
So, the y-intercept is \( B\left(0, \frac{7}{3}\right) \).

### Step 2: Find the intercepts of the second line \( 3x + ky + 5 = 0 \).

- **X-intercept**: Set \( y = 0 \):
\[ 3x + 5 = 0 \implies 3x = -5 \implies x = -\frac{5}{3} \]
So, the x-intercept is \( C\left(-\frac{5}{3}, 0\right) \).

- **Y-intercept**: Set \( x = 0 \):
\[ ky + 5 = 0 \implies ky = -5 \implies y = -\frac{5}{k} \]
So, the y-intercept is \( D\left(0, -\frac{5}{k}\right) \).

### Step 3: Use the concyclic condition.

For the points \( A \), \( B \), \( C \), and \( D \) to be concyclic, the following condition must hold:
\[ a_1 a_2 = b_1 b_2 \]
where \( a_1 \) and \( a_2 \) are the coefficients of \( x \) in the two lines, and \( b_1 \) and \( b_2 \) are the coefficients of \( y \).

From our lines:
- For the first line, \( a_1 = 2 \) and \( b_1 = -3 \).
- For the second line, \( a_2 = 3 \) and \( b_2 = k \).

### Step 4: Substitute into the concyclic condition.

Substituting into the condition:
\[ 2 \cdot 3 = (-3) \cdot k \]
This simplifies to:
\[ 6 = -3k \]

### Step 5: Solve for \( k \).

Dividing both sides by -3:
\[ k = -2 \]

### Conclusion

Thus, the value of \( k \) is \( -2 \).