Orthocenter — Slope Condition

The altitude from A is perpendicular to BC. Orthocenter H is found by intersecting any two altitudes.

The orthocenter $H$ is the intersection of the altitudes of a triangle.

Construction

For triangle $ABC$ with $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$:

The altitude from $A$ is perpendicular to $BC$. Slope of $BC$:

$$m_{BC} = \frac{y_3-y_2}{x_3-x_2}$$

Slope of altitude from $A$: $m_{AH} = -\dfrac{1}{m_{BC}}$ (perpendicularity condition).

Equation of altitude from $A$:

$$y - y_1 = -\frac{1}{m_{BC}}(x - x_1)$$

Write the altitude from $B$ similarly (perpendicular to $AC$). Solve the two altitudes simultaneously — the intersection is $H$.

Key facts

• All three altitudes are concurrent at $H$ — this is a theorem, not assumed.
• For an acute triangle, $H$ lies inside; for an obtuse triangle, $H$ lies outside; for a right triangle, $H$ is at the right-angle vertex.
• $O$ (circumcenter), $G$ (centroid), $H$ (orthocenter) are collinear — the Euler line — with $G$ dividing $OH$ in ratio $1:2$.