Find the orthocentre of the triangle whose angular points are (0, 0), (2, -1), (-1, 3).
[Note. Orthocentre is the point of intersection of the altitudes]

Answer

To find the orthocenter of the triangle with vertices at points \( A(0, 0) \), \( B(2, -1) \), and \( C(-1, 3) \), we will follow these steps:

### Step 1: Find the slopes of the sides of the triangle.
1. **Slope of line \( BC \)**:
\[ \text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{2 - (-1)} = \frac{-4}{3} \]

2. **Slope of line \( AB \)**:
\[ \text{slope of } AB = \frac{-1 - 0}{2 - 0} = \frac{-1}{2} \]

3. **Slope of line \( AC \)**:
\[ \text{slope of } AC = \frac{3 - 0}{-1 - 0} = \frac{3}{-1} = -3 \]

### Step 2: Find the slopes of the altitudes.
1. **Slope of altitude from \( A \) (perpendicular to \( BC \))**:
\[ \text{slope of altitude from } A = -\frac{1}{\text{slope of } BC} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4} \]

2. **Slope of altitude from \( B \) (perpendicular to \( AC \))**:
\[ \text{slope of altitude from } B = -\frac{1}{\text{slope of } AC} = -\frac{1}{-3} = \frac{1}{3} \]

### Step 3: Write the equations of the altitudes.
1. **Equation of altitude from \( A(0, 0) \)**:
Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 0 = \frac{3}{4}(x - 0) \implies y = \frac{3}{4}x \]

2. **Equation of altitude from \( B(2, -1) \)**:
\[ y - (-1) = \frac{1}{3}(x - 2) \implies y + 1 = \frac{1}{3}x - \frac{2}{3} \]
Simplifying:
\[ y = \frac{1}{3}x - \frac{5}{3} \]

### Step 4: Find the intersection of the two altitudes.
Set the equations equal to each other:
\[ \frac{3}{4}x = \frac{1}{3}x - \frac{5}{3} \]

### Step 5: Solve for \( x \).
1. Multiply through by 12 to eliminate fractions:
\[ 12 \cdot \frac{3}{4}x = 12 \cdot \left(\frac{1}{3}x - \frac{5}{3}\right) \]
\[ 9x = 4x - 20 \]

2. Rearranging gives:
\[ 9x - 4x = -20 \implies 5x = -20 \implies x = -4 \]

### Step 6: Substitute \( x \) back to find \( y \).
Using \( y = \frac{3}{4}x \):
\[ y = \frac{3}{4}(-4) = -3 \]

### Conclusion
The orthocenter \( H \) of the triangle is at the point:
\[ H(-4, -3) \]