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 analyze the equations of the lines representing the sides of a triangle and determine whether the line \(3x + 2y = 0\) passes through a specific point related to the triangle.

### Step 1: Identify the equations of the lines
The equations of the lines representing the sides of the triangle are:
1. \(L_1: 3x - 4y = 0\)
2. \(L_2: x + y = 0\)
3. \(L_3: 2x - 3y = 7\)

### Step 2: Find the slopes of the lines
To find the slopes, we can rearrange each equation into the slope-intercept form \(y = mx + c\), where \(m\) is the slope.

1. For \(L_1: 3x - 4y = 0\):
\[ 4y = 3x \implies y = \frac{3}{4}x \quad \text{(Slope, } m_1 = \frac{3}{4}\text{)} \]

2. For \(L_2: x + y = 0\):
\[ y = -x \quad \text{(Slope, } m_2 = -1\text{)} \]

3. For \(L_3: 2x - 3y = 7\):
\[ 3y = 2x - 7 \implies y = \frac{2}{3}x - \frac{7}{3} \quad \text{(Slope, } m_3 = \frac{2}{3}\text{)} \]

### Step 3: Check if the lines are perpendicular
Two lines are perpendicular if the product of their slopes is \(-1\).

- Check \(L_1\) and \(L_2\):
\[ m_1 \cdot m_2 = \frac{3}{4} \cdot (-1) = -\frac{3}{4} \quad (\text{Not perpendicular}) \]

- Check \(L_2\) and \(L_3\):
\[ m_2 \cdot m_3 = (-1) \cdot \frac{2}{3} = -\frac{2}{3} \quad (\text{Not perpendicular}) \]

- Check \(L_1\) and \(L_3\):
\[ m_1 \cdot m_3 = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2} \quad (\text{Not perpendicular}) \]

### Step 4: Find the intersection points of the lines
To find the vertices of the triangle, we need to find the intersection points of the lines.

1. **Intersection of \(L_1\) and \(L_2\)**:
\[ 3x - 4(-x) = 0 \implies 3x + 4x = 0 \implies 7x = 0 \implies x = 0, y = 0 \]
Intersection point: \( (0, 0) \)

2. **Intersection of \(L_2\) and \(L_3\)**:
\[ x + y = 0 \quad \text{and} \quad 2x - 3y = 7 \]
Substitute \(y = -x\) into \(2x - 3(-x) = 7\):
\[ 2x + 3x = 7 \implies 5x = 7 \implies x = \frac{7}{5}, y = -\frac{7}{5} \]
Intersection point: \( \left( \frac{7}{5}, -\frac{7}{5} \right) \)

3. **Intersection of \(L_1\) and \(L_3\)**:
\[ 3x - 4y = 0 \quad \text{and} \quad 2x - 3y = 7 \]
From \(L_1\), \(y = \frac{3}{4}x\):
\[ 2x - 3\left(\frac{3}{4}x\right) = 7 \implies 2x - \frac{9}{4}x = 7 \implies \frac{8}{4}x - \frac{9}{4}x = 7 \]
\[ -\frac{1}{4}x = 7 \implies x = -28, y = \frac{3}{4}(-28) = -21 \]
Intersection point: \( (-28, -21) \)

### Step 5: Check if the line \(3x + 2y = 0\) passes through the orthocenter
The orthocenter of a triangle formed by three lines is the point where the altitudes intersect. We can check if the line \(3x + 2y = 0\) passes through the point \( (0, 0) \) (the intersection of \(L_1\) and \(L_2\)).

Substituting \(x = 0\) and \(y = 0\) into \(3x + 2y = 0\):
\[ 3(0) + 2(0) = 0 \quad \text{(True)} \]

### Conclusion
The line \(3x + 2y = 0\) passes through the orthocenter of the triangle formed by the given lines.