Altitudes and Orthocenter of a Triangle

The altitude of a triangle is the line that passes through a vertex and is perpendicular to the side (or to the extension of that side) opposite to that vertex.

The problem presented here is about the point of concurrency of the altitudes of a triangle. This point is referred to as the orthocenter of the triangle.

For a right triangle, this point of concurrency of the altitudes is at one of its vertices. This is so because two of its altitudes coincide with its legs. See illustration below.

For an obtuse triangle, this point of concurrency of the altitudes is outside of the triangle. Two of its altitudes are drawn to the respective extensions of its opposite sides. See illustration below.

The Problem

Given triangle ABC with its vertices at A(-5, 0), B(3, 4) and C(0, -5), find the point of concurrency of its altitudes.

Discussion

The altitudes of a triangle are concurrent at a point.This point of concurrency is called the orthocenter of a triangle.

In the illustrated triangle above, triangle ABC is a scalene triangle. Line segments AT, BU and CS are its altitudes. Point O is its orthocenter.

Solution

Point of Concurrency: O (-2, -1)

Distance Between Two Points

• distance_AB = √ 80
• distance_BC = √ 90
• distance_AC = √ 50
• distance_AT = √ 40
• distance_BU = √ 72
• distance_CS = √ 45

Equations of the Line Segments

• side AB: x - 2y + 5 = 0; x-intercept = - 5; y-intercept = 5/2
• side BC: 3x - y - 5 = 0; x-intercept = 5/3; y-intercept = - 5
• side AC: x + y + 5 = 0; x-intercept = - 5; y-intercept = - 5
• altitude AT: x + 3y + 5 = 0; x-intercept = - 5; y-intercept = - 5/3
• altitude BU: x - y + 1 = 0; x-intercept = -1; y-intercept = 1
• altitude CS: 2x + y + 5 = 0; x-intercept = - 5/2; y-intercept = - 5

Slopes of the Line Segments

• m_AB = 1/2
• m_BC = 3
• m_AC = - 1
• m_AT = - 1/3
• m_BU = 1
• m_CS = - 2