The line `L_1-=4x+3y-12=0` intersects the x-and y-axies at `Aa n dB ,` respectively. A variable line perpendicular to `L_1` intersects the x- and the y-axis at `P` and `Q` , respectively. Then the locus of the circumcenter of triangle `A B Q` is (a) `3x-4y+2=0` (b) `4x+3y+7=0` (c) `6x-8y+7=0` (d) none of these

Options

A. 3x-4y+2 = 0
B. 4x+3y+7 = 0
C. 6x-8y+7=0
D. none of these

Explanation

To solve the problem step by step, we will first identify the points where the line \( L_1: 4x + 3y - 12 = 0 \) intersects the x-axis and y-axis, then find the circumcenter of triangle \( ABQ \) where \( P \) and \( Q \) are points on the axes for a variable line perpendicular to \( L_1 \).

### Step 1: Find the points of intersection \( A \) and \( B \)

1. **Finding point \( A \) (x-intercept)**:
- Set \( y = 0 \) in the equation of the line:
\[ 4x + 3(0) - 12 = 0 \implies 4x - 12 = 0 \implies 4x = 12 \implies x = 3 \]
- Thus, point \( A \) is \( (3, 0) \).

2. **Finding point \( B \) (y-intercept)**:
- Set \( x = 0 \) in the equation of the line:
\[ 4(0) + 3y - 12 = 0 \implies 3y - 12 = 0 \implies 3y = 12 \implies y = 4 \]
- Thus, point \( B \) is \( (0, 4) \).

### Step 2: Determine the slope of line \( L_1 \)

- The slope \( m_1 \) of line \( L_1 \) can be found by rewriting the equation in slope-intercept form \( y = mx + c \):
\[ 3y = -4x + 12 \implies y = -\frac{4}{3}x + 4 \]
- Therefore, the slope \( m_1 = -\frac{4}{3} \).

### Step 3: Find the slope of the perpendicular line

- The slope \( m_2 \) of a line perpendicular to \( L_1 \) is given by:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4} \]

### Step 4: Equation of the variable line perpendicular to \( L_1 \)

- The equation of the line passing through point \( P \) (x-intercept) and point \( Q \) (y-intercept) can be expressed as:
\[ y - y_1 = m(x - x_1) \]
- Let \( P = (p, 0) \) and \( Q = (0, q) \). Using the slope \( m_2 = \frac{3}{4} \):
\[ y - 0 = \frac{3}{4}(x - p) \implies y = \frac{3}{4}x - \frac{3}{4}p \]

### Step 5: Find the circumcenter of triangle \( ABQ \)

- The circumcenter of triangle \( ABQ \) lies on the perpendicular bisector of segment \( AB \).
- The midpoint \( M \) of segment \( AB \) is:
\[ M = \left( \frac{3 + 0}{2}, \frac{0 + 4}{2} \right) = \left( \frac{3}{2}, 2 \right) \]

### Step 6: Equation of the perpendicular bisector of \( AB \)

- The slope of \( AB \) is:
\[ m_{AB} = \frac{4 - 0}{0 - 3} = -\frac{4}{3} \]
- The slope of the perpendicular bisector is:
\[ m_{perpendicular} = \frac{3}{4} \]
- Using point-slope form at point \( M \):
\[ y - 2 = \frac{3}{4}\left(x - \frac{3}{2}\right) \]
- Rearranging gives:
\[ 4y - 8 = 3x - \frac{9}{2} \implies 8y - 16 = 6x - 9 \implies 6x - 8y + 7 = 0 \]

### Final Answer

The locus of the circumcenter of triangle \( ABQ \) is given by:
\[ 6x - 8y + 7 = 0 \]