Orthocenter of a triangle

The perpendiculars dropped from each vertex to the opposite side meet at a single point, called the orthocenter. We find it for the triangle with vertices $A(0, 0)$, $B(6, 3)$ and $C(2, 6)$.

The three sides as lines

Erecting the perpendiculars

Two perpendicular lines have slopes whose product is $-1$. The perpendicular from the vertex $C$ to the side $AB$ therefore has slope $-2$ against the slope $\dfrac{1}{2}$, and passing through $C(2, 6)$ it is $y = -2x + 10$.

The perpendicular from $B$ follows the same way. Against the slope $3$ of the side $AC$ its slope is $-\dfrac{1}{3}$, and passing through $B(6, 3)$ it is $y = -\dfrac{x}{3} + 5$. Solving the two together gives the following.

The orthocenter is $H(3, 4)$.

Checking with the third

The side $BC$ has slope $-\dfrac{3}{4}$, so the perpendicular from $A$ is $y = \dfrac{4}{3}x$. Putting $x = 3$ gives $y = 4$, so it too passes through $H$. That the three meet at one point is no accident; it holds for every triangle.

Where the orthocenter sits

All three angles of this triangle are acute, so $H$ lies inside.

The Euler line

The circumcenter $O$, the centroid $G$ and the orthocenter $H$ lie on one line¹.

Moreover $\overrightarrow{OH} = 3\,\overrightarrow{OG}$ holds. That line is called the Euler line.

The six lines on the graph are the three sides and the three perpendiculars, and the large dots are the three vertices together with the orthocenter $H(3, 4)$.

1. Euler line, Wikipedia