The equations of the sides of a triangle are `x-3y=0,4x+3y=5,3x+y=0.` The line `3x-4y=0` passes through (A) Incentre (B) Centroid (C) Orthocentre (D) Circumcentre

Options

A. the incentre
B. the centroid
C. the orthocentre
D. the circumcentre

Explanation

To solve the problem, we need to find the orthocenter of the triangle formed by the given lines and check if the line \(3x - 4y = 0\) passes through this point.

### Step 1: Find the intersection points of the lines to determine the vertices of the triangle.

1. **Intersection of Line 1 and Line 2:**
- Line 1: \(x - 3y = 0 \Rightarrow x = 3y\)
- Line 2: \(4x + 3y = 5\)

Substitute \(x = 3y\) into Line 2:
\[ 4(3y) + 3y = 5 \Rightarrow 12y + 3y = 5 \Rightarrow 15y = 5 \Rightarrow y = \frac{1}{3} \]
Now substitute \(y\) back to find \(x\):
\[ x = 3\left(\frac{1}{3}\right) = 1 \]
So, the intersection point (vertex A) is \(A(1, \frac{1}{3})\).

2. **Intersection of Line 2 and Line 3:**
- Line 2: \(4x + 3y = 5\)
- Line 3: \(3x + y = 0 \Rightarrow y = -3x\)

Substitute \(y = -3x\) into Line 2:
\[ 4x + 3(-3x) = 5 \Rightarrow 4x - 9x = 5 \Rightarrow -5x = 5 \Rightarrow x = -1 \]
Now substitute \(x\) back to find \(y\):
\[ y = -3(-1) = 3 \]
So, the intersection point (vertex B) is \(B(-1, 3)\).

3. **Intersection of Line 1 and Line 3:**
- Line 1: \(x - 3y = 0 \Rightarrow x = 3y\)
- Line 3: \(3x + y = 0 \Rightarrow y = -3x\)

Substitute \(x = 3y\) into Line 3:
\[ 3(3y) + y = 0 \Rightarrow 9y + y = 0 \Rightarrow 10y = 0 \Rightarrow y = 0 \]
Now substitute \(y\) back to find \(x\):
\[ x = 3(0) = 0 \]
So, the intersection point (vertex C) is \(C(0, 0)\).

### Step 2: Find the orthocenter of the triangle.

The orthocenter of a triangle is the point where the altitudes intersect. Since we have a right triangle (as shown by the perpendicularity of the sides), the orthocenter is located at the vertex where the right angle is formed.

From the vertices:
- \(A(1, \frac{1}{3})\)
- \(B(-1, 3)\)
- \(C(0, 0)\)

We can see that the right angle is at vertex \(C(0, 0)\).

### Step 3: Check if the line \(3x - 4y = 0\) passes through the orthocenter.

To check if the line \(3x - 4y = 0\) passes through the orthocenter \(C(0, 0)\):
\[ 3(0) - 4(0) = 0 \]
This is true, so the line passes through the orthocenter.

### Conclusion:
The line \(3x - 4y = 0\) passes through the orthocenter of the triangle formed by the given lines.