The combined equation of lines passing through the point `(2, 3)` and perpendicular to the lines `3x+2y-1=0 and x-3y+2=0` is

Options

A. `6x^(2)-7xy+3y^(2)-3x-32y-45=0`
B. `6x^(2)-7xy-3y^(2)-3x+32y-45=0`
C. `6x^(2)-7xy-3y^(2)+3x+32y-45=0`
D. `6x^(2)-7xy+3y^(2)-3x+32y-45=0`

Answer

To find the combined equation of the lines passing through the point (2, 3) and perpendicular to the lines given by the equations \(3x + 2y - 1 = 0\) and \(x - 3y + 2 = 0\), we can follow these steps:

### Step 1: Find the slopes of the given lines

1. **For the line \(3x + 2y - 1 = 0\)**:
- Rearranging the equation to slope-intercept form \(y = mx + c\):
\[ 2y = -3x + 1 \implies y = -\frac{3}{2}x + \frac{1}{2} \]
- The slope \(m_1\) of this line is \(-\frac{3}{2}\).

2. **For the line \(x - 3y + 2 = 0\)**:
- Rearranging the equation:
\[ -3y = -x - 2 \implies y = \frac{1}{3}x + \frac{2}{3} \]
- The slope \(m_2\) of this line is \(\frac{1}{3}\).

### Step 2: Find the slopes of the required lines

Since the required lines are perpendicular to the given lines, we can find their slopes using the negative reciprocal of the slopes of the given lines.

1. **For the first line**:
- The slope of the required line \(m_1'\) is:
\[ m_1' = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3} \]

2. **For the second line**:
- The slope of the required line \(m_2'\) is:
\[ m_2' = -\frac{1}{m_2} = -\frac{1}{\frac{1}{3}} = -3 \]

### Step 3: Write the equations of the required lines

Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point (2, 3):

1. **For the first line with slope \(\frac{2}{3}\)**:
\[ y - 3 = \frac{2}{3}(x - 2) \]
Rearranging gives:
\[ y - 3 = \frac{2}{3}x - \frac{4}{3} \implies y = \frac{2}{3}x + \frac{5}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y = 2x + 5 \implies 2x - 3y + 5 = 0 \]

2. **For the second line with slope \(-3\)**:
\[ y - 3 = -3(x - 2) \]
Rearranging gives:
\[ y - 3 = -3x + 6 \implies y = -3x + 9 \]
Rearranging gives:
\[ 3x + y - 9 = 0 \]

### Step 4: Find the combined equation

Now we have the two equations:
1. \(2x - 3y + 5 = 0\)
2. \(3x + y - 9 = 0\)

To find the combined equation, we multiply these two equations:
\[ (2x - 3y + 5)(3x + y - 9) = 0 \]

Expanding this:
\[ 6x^2 + 2xy - 18x - 9xy - 3y^2 + 27y + 15x - 45 = 0 \]
Combining like terms gives:
\[ 6x^2 - 7xy - 3y^2 - 3x + 32y - 45 = 0 \]

### Final Answer

The combined equation of the lines is:
\[ 6x^2 - 7xy - 3y^2 - 3x + 32y - 45 = 0 \]