The equations to the sides of a triangle are `x-3y=0, 4x+3y=5` and `3x+y=0`. The line `3x-4y=0` passes through

Options

A. the incentre
B. the centoid
C. the circumcentre
D. the orthocentre of the triangle

Answer

To solve the problem, we need to analyze the given equations of the sides of the triangle and determine if the line \(3x - 4y = 0\) passes through any significant point related to the triangle, such as the orthocenter.

### Step-by-Step Solution:

1. **Identify the equations of the sides of the triangle:**
The equations given are:
- \(x - 3y = 0\) (Equation 1)
- \(4x + 3y = 5\) (Equation 2)
- \(3x + y = 0\) (Equation 3)

2. **Find the intersection points of the sides to determine the vertices of the triangle:**
- **Intersection of Equation 1 and Equation 2:**
\[ x - 3y = 0 \implies x = 3y \]
Substitute \(x = 3y\) into Equation 2:
\[ 4(3y) + 3y = 5 \implies 12y + 3y = 5 \implies 15y = 5 \implies y = \frac{1}{3} \]
Then, substituting \(y = \frac{1}{3}\) back to find \(x\):
\[ x = 3\left(\frac{1}{3}\right) = 1 \]
So, one vertex is \(A(1, \frac{1}{3})\).

- **Intersection of Equation 2 and Equation 3:**
\[ 4x + 3y = 5 \quad \text{and} \quad 3x + y = 0 \implies y = -3x \]
Substitute \(y = -3x\) into Equation 2:
\[ 4x + 3(-3x) = 5 \implies 4x - 9x = 5 \implies -5x = 5 \implies x = -1 \]
Then, substituting \(x = -1\) back to find \(y\):
\[ y = -3(-1) = 3 \]
So, another vertex is \(B(-1, 3)\).

- **Intersection of Equation 1 and Equation 3:**
\[ x - 3y = 0 \quad \text{and} \quad 3x + y = 0 \implies y = -3x \]
Substitute \(y = -3x\) into Equation 1:
\[ x - 3(-3x) = 0 \implies x + 9x = 0 \implies 10x = 0 \implies x = 0 \]
Then, substituting \(x = 0\) back to find \(y\):
\[ y = -3(0) = 0 \]
So, the third vertex is \(C(0, 0)\).

3. **Determine the orthocenter of the triangle:**
The orthocenter is the intersection of the altitudes of the triangle. Since we have the vertices \(A(1, \frac{1}{3})\), \(B(-1, 3)\), and \(C(0, 0)\), we can find the slopes of the sides and then the slopes of the altitudes.

- **Slope of side AB:**
\[ \text{slope} = \frac{3 - \frac{1}{3}}{-1 - 1} = \frac{\frac{9}{3} - \frac{1}{3}}{-2} = \frac{\frac{8}{3}}{-2} = -\frac{4}{3} \]
The slope of the altitude from \(C\) (perpendicular to AB) is \(\frac{3}{4}\).

- **Equation of altitude from C(0, 0):**
\[ y = \frac{3}{4}x \]

- **Slope of side BC:**
\[ \text{slope} = \frac{3 - 0}{-1 - 0} = -3 \]
The slope of the altitude from \(A\) (perpendicular to BC) is \(\frac{1}{3}\).

- **Equation of altitude from A(1, \frac{1}{3}):**
\[ y - \frac{1}{3} = \frac{1}{3}(x - 1) \implies y = \frac{1}{3}x + \frac{1}{3} - \frac{1}{3} = \frac{1}{3}x \]

4. **Find the intersection of the altitudes:**
Set the equations of the altitudes equal to find the orthocenter:
\[ \frac{3}{4}x = \frac{1}{3}x \implies 3x = 4x \implies x = 0 \]
Substitute \(x = 0\) into either altitude equation:
\[ y = \frac{3}{4}(0) = 0 \]
Thus, the orthocenter is at \(H(0, 0)\).

5. **Check if the line \(3x - 4y = 0\) passes through the orthocenter:**
Substitute \(H(0, 0)\) into the line equation:
\[ 3(0) - 4(0) = 0 \implies 0 = 0 \]
Therefore, the line \(3x - 4y = 0\) passes through the orthocenter.

### Final Answer:
The line \(3x - 4y = 0\) passes through the orthocenter of the triangle.