If the equations of three sides of a triangle are `x+y=1, 3x + 5y = 2 and x - y = 0` then the orthocentre of the triangle lies on the line/lines

Options

A. `5x-3y=1`
B. `5y-3x=1`
C. `2x-3y=1`
D. `5x-3y=2`

Answer

To find the orthocenter of the triangle formed by the lines \(x + y = 1\), \(3x + 5y = 2\), and \(x - y = 0\), we will follow these steps:

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

1. **Intersection of \(x + y = 1\) and \(x - y = 0\)**:
- From \(x - y = 0\), we have \(x = y\).
- Substitute \(y\) in \(x + y = 1\):
\[ x + x = 1 \implies 2x = 1 \implies x = \frac{1}{2}, \, y = \frac{1}{2} \]
- Vertex \(A\) is \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

2. **Intersection of \(x + y = 1\) and \(3x + 5y = 2\)**:
- Substitute \(y = 1 - x\) into \(3x + 5y = 2\):
\[ 3x + 5(1 - x) = 2 \implies 3x + 5 - 5x = 2 \implies -2x + 5 = 2 \implies -2x = -3 \implies x = \frac{3}{2} \]
- Substitute \(x = \frac{3}{2}\) back into \(y = 1 - x\):
\[ y = 1 - \frac{3}{2} = -\frac{1}{2} \]
- Vertex \(B\) is \(\left(\frac{3}{2}, -\frac{1}{2}\right)\).

3. **Intersection of \(3x + 5y = 2\) and \(x - y = 0\)**:
- From \(x - y = 0\), we have \(x = y\).
- Substitute \(y\) in \(3x + 5y = 2\):
\[ 3x + 5x = 2 \implies 8x = 2 \implies x = \frac{1}{4}, \, y = \frac{1}{4} \]
- Vertex \(C\) is \(\left(\frac{1}{4}, \frac{1}{4}\right)\).

### Step 2: Determine the slopes of the sides of the triangle.

1. **Slope of line \(x + y = 1\)**:
- Rearranging gives \(y = -x + 1\) (slope = -1).

2. **Slope of line \(3x + 5y = 2\)**:
- Rearranging gives \(y = -\frac{3}{5}x + \frac{2}{5}\) (slope = -\(\frac{3}{5}\)).

3. **Slope of line \(x - y = 0\)**:
- Rearranging gives \(y = x\) (slope = 1).

### Step 3: Find the slopes of the altitudes.

1. **Altitude from vertex \(A\) to line \(3x + 5y = 2\)**:
- The slope of the altitude is the negative reciprocal of the slope of the line:
\[ \text{slope} = \frac{5}{3} \]

2. **Altitude from vertex \(B\) to line \(x + y = 1\)**:
- The slope of the altitude is the negative reciprocal of the slope of the line:
\[ \text{slope} = 1 \]

3. **Altitude from vertex \(C\) to line \(x - y = 0\)**:
- The slope of the altitude is the negative reciprocal of the slope of the line:
\[ \text{slope} = -1 \]

### Step 4: Find the equations of the altitudes.

1. **Equation of altitude from \(A\) \(\left(\frac{1}{2}, \frac{1}{2}\right)\)**:
\[ y - \frac{1}{2} = \frac{5}{3}\left(x - \frac{1}{2}\right) \]

2. **Equation of altitude from \(B\) \(\left(\frac{3}{2}, -\frac{1}{2}\right)\)**:
\[ y + \frac{1}{2} = 1\left(x - \frac{3}{2}\right) \]

3. **Equation of altitude from \(C\) \(\left(\frac{1}{4}, \frac{1}{4}\right)\)**:
\[ y - \frac{1}{4} = -1\left(x - \frac{1}{4}\right) \]

### Step 5: Find the orthocenter.

- The orthocenter is the intersection of any two altitudes. We can solve the equations of the altitudes to find the orthocenter.

### Final Answer:
The orthocenter of the triangle lies on the lines formed by the altitudes.