Orthocenter of a Triangle – Definition, Formula and Examples

The orthocenter of a triangle is a point that represents the intersection of the three heights of the triangle. In turn, the heights are the perpendicular lines that connect the vertices with their opposite sides.

In this article, we will learn about the orthocenter of a triangle in more detail. We will learn how to determine its position and solve some practical examples.

GEOMETRY

Relevant for

Learning about the orthocenter of a triangle.

See definition

GEOMETRY

Relevant for

Learning about the orthocenter of a triangle.

See definition

Definition of the orthocenter of a triangle

The orthocenter of a triangle is the point where the three heights of the triangle intersect. In turn, let us remember that the altitudes of the triangle are the perpendicular lines that connect a vertex with the opposite side. The following is a diagram of the orthocenter in a triangle:

The location of the orthocenter varies depending on the type of triangle we have. For example, for equilateral triangles, the orthocenter is located in the same position as the centroid. However, for obtuse triangles, the orthocenter is outside the triangle.

Orthocenter of an acute triangle

For all acute triangles, the orthocenter is located inside the triangle. Remember that an acute triangle is a triangle that has all its interior angles with a size less than 90°.

Orthocenter of an obtuse triangle

For all obtuse triangles, the orthocenter lies outside the triangle. Remember that an obtuse triangle is a triangle that has an internal angle that is greater than 90 °.

Orthocenter of a right triangle

For all right triangles, the orthocenter is located at the vertex of the right triangle. Remember that a right triangle is characterized by having an angle of 90°.

Orthocenter of an equilateral triangle

For all equilateral triangles, the orthocenter is located in the same position as the centroid of the triangle. Let us remember that equilateral triangles are characterized by having all their sides of the same length.

Formula for the orthocenter of a triangle

The orthocenter formula allows us to find the coordinates of the orthocenter of a triangle. To derive this formula we are going to use the following triangle ABC.

In this triangle, AD, BE and CF are the heights and A$latex (x_{1},~y_{1})$, B$latex (x_{2},~y_{2})$, C$latex (x_{3},~y_ {3})$ are the vertices. Point O is the orthocenter.

Let’s start by calculating the slope of the sides of the triangle using the slope formula:

$latex m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

To calculate the slope of the heights of the triangle, we can consider the altitudes to be the lines perpendicular to the sides. In addition, we remember that we can calculate the line perpendicular to another line in the following way:

slope of the perpendicular line $latex =-\frac{1}{m}$

where m is the slope of the original line.

Now, let’s represent the slope of line AC as $latex m_{AC}$ and we have:

$latex m_{AC}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$

And we also have:

$latex m_{BC}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$

Therefore, the slopes of the heights are:

Slope of BE: $latex m_{BE}=-\frac{1}{m_{AC}}$

Slope of AD: $latex m_{AD}=-\frac{1}{m_{BC}}$

If we use the point-slope form of a straight line, we can find the equations of the lines that pass through BE and AD. Then, we have:

$latex m_{BE}=\frac{y-y_{2}}{x-x_{2}}$

$latex m_{AD}=\frac{y-y_{1}}{x-x_{1}}$

If we know the values of $latex (x_{1},~y_{1})$ and $latex (x_{2},~y_{2})$, we can solve the equations for and y, which are the coordinates of the orthocenter.

Finding the orthocenter of a triangle graphically

If we don’t know the coordinates of the vertices of the triangle, we can find the orthocenter graphically. The orthocenter of a triangle can be graphed geometrically by plotting two heights of the triangle and finding their point of intersection.

We need to determine the perpendicular segments (heights) from two vertices to the opposite sides. Therefore, we follow the following steps using a compass:

Step 1: We use vertex B as the center and use a radius equal to BC. With that radius, we draw an arc on side AC to get the point E.

Step 2: We use vertex C with the same radius equal to BC and draw an arc on side AB to form point D.

Step 3: We draw intersecting arcs from B and D using a radius equal to BD to form point F and we draw the segment CF.

Step 4: We draw intersecting arcs from C and E using a radius equal to CE to form point G and draw the segment BG.

Step 5: Find the point of intersection of segments CF and BG.

CF and BG are perpendicular to sides AB and AC respectively. This means that they are two heights of the triangle. Therefore, the point of intersection is the orthocenter of the triangle.

Orthocenter of a triangle – Examples with answers

The following examples show how to find the orthocenter of a triangle using the orthocenter formula.

EXAMPLE 1

Find the slopes of the sides of the triangle below that has vertices A(5, 7), B(2, 3), and C(6, 4).

Solution: We use the slope formula to find the slopes of each side of the triangle with the following coordinates:

• $latex (x_{1},~y_{1})=(5, ~7)$
• $latex (x_{2},~y_{2})=(2,~3)$
• $latex (x_{3},~y_{3})=(6,~4)$

Slope of AC:

$latex m_{AC}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$

$latex m_{AC}=\frac{4-7}{6-5}$

$latex m_{AC}=-3$

Slope of BA:

$latex m_{BA}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

$latex m_{BA}=\frac{7-3}{5-2}$

$latex m_{BA}=\frac{4}{3}$

Slope of BC:

$latex m_{BC}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$

$latex m_{BC}=\frac{4-3}{6-2}$

$latex m_{BC}=\frac{1}{4}$

EXAMPLE 2

Using the slopes found in Example 1, determine the coordinates of the orthocenter of the triangle.

Solution: We have to start by determining the slopes of the perpendicular lines. We have the following slopes:

• $latex m_{AE}=$ perpendicular to BC
• $latex m_{BF}=$ perpendicular to AC
• $latex m_{CD}=$ perpendicular to AB

Remembering that the slope of a perpendicular line is equal to $latex -\frac{1}{m}$, where, m is the slope of the original line, we have:

$latex m_{AE}=-4$

$latex m_{BF}=\frac{1}{3}$

$latex m_{CD}=-\frac{3}{4}$

Now, we can get the equations of the perpendicular lines using the point-slope form of a line: $latex y-y_{1}=m(x-x_{1})$.

Using the point C=(6, 4) and the slope of CD, we have:

$latex y-4=-\frac{3}{4}(x-6)$

$latex 4(y-4)=-3(x-6)$

$latex 4y-16=-3x+18$

$latex 3x+4y=34$

Using the point B=(2, 3) and the slope of BF, we have:

$latex y-3=\frac{1}{3}(x-2)$

$latex 3(y-3)=x-2$

$latex 3y-9=x-2$

$latex -x+3y=7$

We only need two equations to find the point of intersection. Using any method we can solve the system of two equations and find the solution $latex x=\frac{74}{18},~ y=\frac{55}{13}$.

This solution represents the point of intersection of the lines. Therefore, the coordinates of the orthocenter are $latex (\frac{74}{18},~\frac{55}{13})$.