The equaiton of the lines representing the sides of a triangle are `3x - 4y =0 , x+y=0` and 2x - 3y = 7 . The line 3x + 2y = 0 always passes through the

Options

A. incentre
B. centrroid
C. circumcentre
D. orthocenter

Answer

To solve the problem, we need to determine the relationship between the given lines and the line \(3x + 2y = 0\). We will analyze the slopes of the lines and their intersections to find out where the line \(3x + 2y = 0\) passes through.

### Step-by-Step Solution:

1. **Identify the given lines:**
- Line 1: \(3x - 4y = 0\)
- Line 2: \(x + y = 0\)
- Line 3: \(2x - 3y = 7\)

2. **Convert the equations to slope-intercept form (y = mx + c):**
- For Line 1:
\[ 3x - 4y = 0 \implies 4y = 3x \implies y = \frac{3}{4}x \quad \text{(Slope \(m_1 = \frac{3}{4}\))} \]
- For Line 2:
\[ x + y = 0 \implies y = -x \quad \text{(Slope \(m_2 = -1\))} \]
- For Line 3:
\[ 2x - 3y = 7 \implies 3y = 2x - 7 \implies y = \frac{2}{3}x - \frac{7}{3} \quad \text{(Slope \(m_3 = \frac{2}{3}\))} \]

3. **Find the slopes of the lines:**
- \(m_1 = \frac{3}{4}\)
- \(m_2 = -1\)
- \(m_3 = \frac{2}{3}\)

4. **Check if any lines are perpendicular:**
- For two lines to be perpendicular, the product of their slopes should equal \(-1\).
- Check \(m_1 \cdot m_3\):
\[ m_1 \cdot m_3 = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2} \quad \text{(Not perpendicular)} \]
- Check \(m_1 \cdot m_2\):
\[ m_1 \cdot m_2 = \frac{3}{4} \cdot (-1) = -\frac{3}{4} \quad \text{(Not perpendicular)} \]
- Check \(m_2 \cdot m_3\):
\[ m_2 \cdot m_3 = (-1) \cdot \frac{2}{3} = -\frac{2}{3} \quad \text{(Not perpendicular)} \]

5. **Analyze the line \(3x + 2y = 0\):**
- Convert to slope-intercept form:
\[ 3x + 2y = 0 \implies 2y = -3x \implies y = -\frac{3}{2}x \quad \text{(Slope \(m_L = -\frac{3}{2}\))} \]

6. **Check the relationship between line \(3x + 2y = 0\) and Line 3:**
- Check if they are perpendicular:
\[ m_L \cdot m_3 = -\frac{3}{2} \cdot \frac{2}{3} = -1 \quad \text{(They are perpendicular)} \]

7. **Determine the intersection point (orthocenter):**
- Since the line \(3x + 2y = 0\) is perpendicular to Line 3 and passes through the origin (0,0), it indicates that the orthocenter of the triangle formed by the three lines is at the origin.

### Conclusion:
The line \(3x + 2y = 0\) always passes through the orthocenter of the triangle formed by the lines \(3x - 4y = 0\), \(x + y = 0\), and \(2x - 3y = 7\).