If a circle passes through the points where the lines `3lambda x - 2y -1=0` and `4x - 3y +2 = 0` meet the coordinate axes then `lambda = `

Options

A. `-1`
B. `-1//2`
C. `1//2`
D. 1

Answer

To find the value of \( \lambda \) for which a circle passes through the points where the lines \( 3\lambda x - 2y - 1 = 0 \) and \( 4x - 3y + 2 = 0 \) meet the coordinate axes, we will follow these steps:

### Step 1: Find the intersection points of the first line with the axes
The first line is given by:
\[ 3\lambda x - 2y - 1 = 0 \]

**Finding the x-intercept:**
Set \( y = 0 \):
\[ 3\lambda x - 1 = 0 \implies x = \frac{1}{3\lambda} \]
So, the x-intercept is \( A\left(\frac{1}{3\lambda}, 0\right) \).

**Finding the y-intercept:**
Set \( x = 0 \):
\[ -2y - 1 = 0 \implies y = -\frac{1}{2} \]
So, the y-intercept is \( B\left(0, -\frac{1}{2}\right) \).

### Step 2: Find the intersection points of the second line with the axes
The second line is given by:
\[ 4x - 3y + 2 = 0 \]

**Finding the x-intercept:**
Set \( y = 0 \):
\[ 4x + 2 = 0 \implies x = -\frac{1}{2} \]
So, the x-intercept is \( C\left(-\frac{1}{2}, 0\right) \).

**Finding the y-intercept:**
Set \( x = 0 \):
\[ -3y + 2 = 0 \implies y = \frac{2}{3} \]
So, the y-intercept is \( D\left(0, \frac{2}{3}\right) \).

### Step 3: Verify the cyclic condition for the points
The points \( A\left(\frac{1}{3\lambda}, 0\right) \), \( B\left(0, -\frac{1}{2}\right) \), \( C\left(-\frac{1}{2}, 0\right) \), and \( D\left(0, \frac{2}{3}\right) \) must satisfy the condition for concyclic points:
\[ OA \cdot OC = OB \cdot OD \]
where \( O \) is the origin \( (0, 0) \).

### Step 4: Calculate the distances
- \( OA = \sqrt{\left(\frac{1}{3\lambda}\right)^2 + 0^2} = \frac{1}{3\lambda} \)
- \( OB = \sqrt{0^2 + \left(-\frac{1}{2}\right)^2} = \frac{1}{2} \)
- \( OC = \sqrt{\left(-\frac{1}{2}\right)^2 + 0^2} = \frac{1}{2} \)
- \( OD = \sqrt{0^2 + \left(\frac{2}{3}\right)^2} = \frac{2}{3} \)

### Step 5: Set up the equation
Substituting into the cyclic condition:
\[ \left(\frac{1}{3\lambda}\right) \cdot \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right) \cdot \left(\frac{2}{3}\right) \]
Simplifying:
\[ \frac{1}{6\lambda} = \frac{1}{3} \]

### Step 6: Solve for \( \lambda \)
Cross-multiplying gives:
\[ 1 = 2\lambda \implies \lambda = \frac{1}{2} \]

Thus, the value of \( \lambda \) is:
\[ \boxed{\frac{1}{2}} \]