Orthocenter Calculator

What Is the Orthocenter of a Triangle?

The orthocenter (denoted by $H$) of a triangle is the single point where all three of its altitudes intersect. An altitude is a straight line passing through a triangle vertex that is perpendicular to the opposite side (or the line containing the opposite side).

In Euclidean geometry, the three altitudes of any non-degenerate triangle are always concurrent, which guarantees that they meet at exactly one unique point $H$.

Location of the Orthocenter

The position of the orthocenter relative to the triangle depends entirely on the type of triangle:

• Acute Triangle: All three angles are strictly less than 90 degrees. The orthocenter lies inside the boundary of the triangle.
• Right Triangle: One angle equals 90 degrees. The orthocenter coincides exactly with the vertex containing the right angle.
• Obtuse Triangle: One angle is greater than 90 degrees. The orthocenter lies outside the triangle, opposite the obtuse vertex.
• Equilateral Triangle: The orthocenter, centroid ($G$), incenter ($I$), and circumcenter ($O$) all coincide at the exact same geometric point.

How to Calculate the Orthocenter from Coordinates

Given three vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$:

1. Find the Slope of Side $BC$: $$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2}$$
2. Determine the Perpendicular Slope for Altitude $h_A$: $$m_{\perp A} = -\frac{1}{m_{BC}} = -\frac{x_3 - x_2}{y_3 - y_2}$$ The equation of the altitude line passing through $A(x_1, y_1)$ is: $$(x_3 - x_2)(x - x_1) + (y_3 - y_2)(y - y_1) = 0$$
3. Determine the Equation for Altitude $h_B$: Passes through $B(x_2, y_2)$ and is perpendicular to side $AC$: $$(x_3 - x_1)(x - x_2) + (y_3 - y_1)(y - y_2) = 0$$
4. Solve the System of Equations: Solving these two linear equations simultaneously gives the exact orthocenter coordinates $(x_H, y_H)$.

The Euler Line Connection

In any non-equilateral triangle, the orthocenter ($H$), centroid ($G$), and circumcenter ($O$) are strictly collinear, forming the famous Euler Line.

Moreover, the centroid $G$ divides the line segment connecting the circumcenter $O$ and orthocenter $H$ in a fixed $2:1$ ratio:

$$GH = 2 \cdot GO \quad \text{and} \quad \vec{H} = 3\vec{G} - 2\vec{O}$$

Worked Example

Let a triangle have vertices $A(0, 0)$, $B(6, 0)$, and $C(2, 4)$.

1. Altitude from $C$ to side $AB$: Side $AB$ lies on the line $y = 0$. The perpendicular line through $C(2, 4)$ is the vertical line $x = 2$.
2. Altitude from $A(0, 0)$ to side $BC$: The slope of $BC$ from $(6, 0)$ to $(2, 4)$ is $m_{BC} = \frac{4 - 0}{2 - 6} = \frac{4}{-4} = -1$. The perpendicular slope is $m = 1$. The line through $(0, 0)$ with slope 1 is $y = x$.
3. Find Intersection: Substituting $x = 2$ into $y = x$ gives $y = 2$.
4. Result: The orthocenter is $H(2, 2)$.