Orthocentre with Examples and FAQs

Orthocentre

The orthocentre is the concurrency point where all the three altitudes of a triangle intersect, for any given triangle only one orthocentre exists. Altitude is the line drawn from a vertex that is perpendicular to the side opposite to that vertex. An orthocentre is a significantly important point of any triangle. The orthocentre varies for different triangle types. The orthocentre of an equilateral triangle is the centroid itself. In some triangles, the orthocentre can even lie outside the triangle. 

In ∆ABC, the altitudes are AE, BF and CD. The orthocentre of the triangle is point O.

Construction of Orthocentre

The orthocentre of a triangle can be constructed by following the steps given below:

• Drop perpendiculars from any two vertices to their respective opposite sides. These perpendiculars are altitudes of the triangle. 
• The point where these two perpendiculars intersect gives the orthocentre of that particular triangle.

Properties 

The properties of the orthocentre depend on the type of triangle. For some triangles, the orthocentre may even lie outside the triangle. The properties of orthocentre are:

1. The orthocentre of an acute triangle always lies inside the triangle.

The triangle ABC is an acute triangle, it can be seen that the orthocentre lies inside the triangle.

2. The orthocentre of an obtuse triangle, lies outside the triangle.

Triangle DEF is an obtuse triangle, hence its orthocentre lies outside the triangle.

3. The orthocentre of a right-angled triangle lies on the right-angled vertex of the triangle.

For the triangle PQR, the orthocentre is the vertex R.

4. The orthocentre of a triangle divides an altitude into different parts. The product of the lengths of all the parts is equal for all three altitudes. 

Orthocentre Formula

The orthocentre formula can help in determining the coordinates of the orthocentre of any triangle. Let’s derive the formula for the orthocentre by taking a triangle ABC.

As shown in the figure of ∆ABC

Let the coordinates of the vertices be \mathrm{A}\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right), \mathrm{C}\left(x_{3}, y_{3}\right), \mathrm{O}(x, y).

From the figure, AE, BF and CD are the altitudes of the triangle. O is the orthocentre.

Step 1: First calculate the slope of any two sides of the triangle using the formula

\text { Slope }(\mathrm{m})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

[ This is the formula for calculating the slope of the line using two points on the line\left.\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)\right]

The slope of the side BC and AC can be calculated using this formula.

\text{The slope of }\mathrm{BC}, m_{B C}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}

\text{The  slope of }\mathrm{AC},m_{A C}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}

Step 2: The perpendicular slope of a line is given by the formula,

The perpendicular slope of line =-\frac{1}{m} .

Using this formula, the slope of perpendiculars to BC and AC can be found out.

The slope of the perpendiculars to BC and AC will actually be the slope of altitudes, AE and BF respectively.

\text{The slope of } \mathrm{AE}, m_{A E}=-\frac{1}{m_{B C}}

\text{The slope of } \mathrm{BF}, m_{B F}=-\frac{1}{m_{A C}}

Step 3: Calculating the slope of the altitudes AE and BF using the point-slope formula:

\text{The slope of } \mathrm{AE}, m_{A E}=\frac{y-y_{1}}{x-x_{1}}

\text{The slope of } \mathrm{BF}, m_{B F}=\frac{y-y_{2}}{x-x_{2}}

Thus, solving the two equations for the given values will give the coordinates of the orthocentre O(x, y).

Examples

Examples 1: Name the vertices, altitudes and orthocentre of the given triangle.

Solution:

In the above figure for △ABC,

The Vertices of the triangle are A, B, and C.

Sides of the ∆ are AB, BC, AC and the Altitudes are AE, BF, CD.

The Orthocentre of the triangle is O.

Example 2: Determine the coordinates of the orthocentre of the triangle whose vertices are: A(4, 8), B(-2, 0), C(2, 4).

Solution:

Let us calculate the slope of any two sides of the triangle.

The slope of the side BC and AC can be calculated using the point-slope formula.

\text { Slope }(\mathrm{m})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

The slope of \mathrm{BC}, m_{B C}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{4-0}{2-(-2)}=\frac{4}{4}=1

The slope of \text { AC, } m_{A C}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{4-8}{2-4}=\frac{-4}{-2}=2

Now, the perpendicular slopes of the sides BC and AC.

The perpendicular slope of side BC =-\frac{1}{m_{B C}}=-\frac{1}{1}=-1 \ldots(1)

The perpendicular Slope of side AC =-\frac{1}{m_{A C}}=-\frac{1}{2} \cdots(2)

Let the coordinates of the Orthocentre be (x, y).

Now by point-slope form, the slope of the line passing through vertex A(4, 8) and orthocentre (x, y) is:

The slope of the perpendicular from vertex \mathrm{A}=\frac{y-8}{x-4}

From (2) we have the slope of the perpendicular from A=-\frac{1}{2}

\Rightarrow \frac{y-8}{x-4}=-\frac{1}{2}

\Rightarrow 2(y-8)=-(x-4)

\Rightarrow x+2 y=4+16=20 \ldots(3)

Similarly, the slope of line passing through the vertex B(-2, 0) and orthocentre (x, y) is:

Slope of perpendicular from vertex B=\frac{y-0}{x-(-2)}=\frac{y}{x+2}

From (1) we have slope of perpendicular from B = -1 

\Rightarrow \frac{y}{x+2}=-1

\Rightarrow \mathrm{y}=-(\mathrm{x}+2)

\Rightarrow \mathrm{x}+\mathrm{y}=-2 \ldots(4)

Solving equations (3) and (4)

x + 2y = 20

x +   y = -2 

–     –      +

—————–

       y = 22

∴ x = 20 – 2y = 20 – 2(22) = 20 – 44 = -24

Hence the orthocentre of the given triangle is at point (-24, 20).