Show that the orthocentre of ΔA B C having vertices A(𝑥₁, 𝑦₁), B(𝑥₂, 𝑦₂), C(𝑥₃, 𝑦₃) is :

• 𝑥₁ tan A + 𝑥₂ tan B + 𝑥₃ tan C
• tan A + tan B + tan C

To find the orthocenter of triangle ΔABC with vertices A(𝑥₁, 𝑦₁), B(𝑥₂, 𝑦₂), and C(𝑥₃, 𝑦₃), we need to understand how the altitudes of the triangle intersect. The orthocenter is the point where these altitudes meet.

Finding the Slopes of the Sides

The slopes of the sides of the triangle can be calculated as follows:

• Slope of AB: \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \)
• Slope of BC: \( m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} \)
• Slope of CA: \( m_{CA} = \frac{y_1 - y_3}{x_1 - x_3} \)

Calculating the Altitudes

The altitudes of the triangle can be derived from the slopes of the sides. The altitude from vertex A is perpendicular to side BC, and its slope is the negative reciprocal of \( m_{BC} \). Similarly, we can find the altitudes from vertices B and C.

Using Trigonometric Relationships

Let angles A, B, and C be the angles at vertices A, B, and C respectively. The tangent of these angles can be expressed in terms of the sides of the triangle:

• tan A = \( \frac{\text{opposite}}{\text{adjacent}} \)
• tan B = \( \frac{\text{opposite}}{\text{adjacent}} \)
• tan C = \( \frac{\text{opposite}}{\text{adjacent}} \)

Finding the Orthocenter Coordinates

The coordinates of the orthocenter (H) can be derived using the formula:

H = (x₁ tan A + x₂ tan B + x₃ tan C) / (tan A + tan B + tan C)

Thus, the x-coordinate of the orthocenter can be expressed as:

x_H = x₁ tan A + x₂ tan B + x₃ tan C / (tan A + tan B + tan C)

Conclusion

In summary, the orthocenter of triangle ΔABC can be calculated using the coordinates of the vertices and the tangents of the angles at those vertices. This provides a clear method to find the orthocenter based on the triangle's geometry.