## What is the Orthocentre of a Triangle?

The three altitudes of any triangle are concurrent line segments (they intersect in a single point) and this point is known as the orthocentre of the triangle. A triangle ABC is given below that has three vertices, i.e., A, B, C, and has three altitudes, AE, BF, and CD. The intersection point of these altitudes is H. The intersection point H is the orthocentre of triangle ABC.

(Image will be uploaded soon)

In this article, we will learn the orthocentre formula and how to calculate the orthocentre of any triangle.

### How to Find the Orthocentre of a Triangle?

The steps to find the coordinates of the orthocentre of a triangle are relatively simple, given that we know the coordinates of the vertices of the triangle.

Let us consider the following triangle ABC, the coordinates of whose vertices are known.

(Image will be uploaded soon)

The vertices of the triangle are A, B, and C where coordinates of these vertices are (x1,y1), (x2,y2), and (x3,y3), respectively.

Step 1: First, we will find the slopes of any two sides of the triangle (say AC and BC).

\[ m_{AC} = \frac{y_{3}-y_{1}}{x_{3}-x_{1}} \]

\[ m_{BC} = \frac{y_{3} -y_{2}}{x_{3}-x_{2}} \]

Step 2: Next, we can find the slopes of the corresponding altitudes. Remember that if two lines are perpendicular to each other, they satisfy the following equation.

Perpendicular slope of line= \[ \frac{-1}{\text{slope of the line}} = \frac{-1}{m} \]

Slope of BE= \[ \frac{-1}{m_{AC}} \]

Slope of AD= \[ \frac{-1}{m_{BC}} \]

Step 3: Next, we will use the slope-point form of the equation of a straight line to find the equations of the lines that are coincident with the altitudes BE and AD.

BE: \[\frac{y-y_{2}}{x-x_{2}} = m_{BE}\]

AD:\[ \frac{y-y_{1}}{x-x_{1}} = m_{AD} \]

Next, we can solve the equations of BE and AD simultaneously to find their solution, which gives us the coordinates of the orthocentre H.

### Properties of Orthocentre

The properties of the orthocentre are given below:

The orthocentre lies inside the triangle in an acute triangle.

The orthocentre of an obtuse triangle always lies outside of the triangle.

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

The orthocentre of the triangle divides the altitudes into different parts. The product of the parts is equivalent for all the three perpendiculars.

### Solved Examples

Question 1: Find the coordinates of the orthocenter of a triangle ABC whose vertices are A(1 ,7), B(−6, 0) and C(3, 4).

Solution:

\[m_{AC}\] = \[\frac{y_{3} - y_{1}}{x_{3} - x_{1}}\] = \[\frac{(4 - 7)}{(3-1)}\] = \[\frac{-3}{2}\] \[\Rightarrow\] \[m_{BE}\] = \[\frac{-1}{m_{AC}}\] = \[\frac{2}{3}\]

\[m_{BC}\] = \[\frac{y_{3} - y_{2}}{x_{3} - x_{2}}\] = \[\frac{(4 - 0)}{(3-(-6))}\] = \[\frac{4}{9}\] \[\Rightarrow\] \[m_{AD}\] = \[\frac{-1}{m_{BC}}\] = \[\frac{-9}{4}\]

BE: \[\frac{y - y_{2}}{x - x_{2}}\] = \[m_{BE}\] \[\Rightarrow\] \[\frac{(y - 0)}{(x-(-6))}\] = \[\frac{2}{3}\] \[\Rightarrow\] 2x - 3y + 12 = 0

AD: \[\frac{y - y_{1}}{x - x_{1}}\] = \[m_{AD}\] \[\Rightarrow\] \[\frac{(y - 7)}{(x-1)}\] = \[\frac{-9}{4}\] \[\Rightarrow\] 9x + 4y -37 = 0

Solving the equations for BE and AD , we get the coordinates of the orthocenter H as follows.

$H\left( {\frac{9}{5},\frac{{26}}{5}} \right)$

Why don’t you try to solve a problem to see if you are getting the hang of the methodology?

Question 2: Find the coordinates of the orthocentre of a triangle ABC whose vertices are A (−1, −4), B (2, −3), and C (5, 2).

Options:

(a) (10, 9)

(b) (10, −9)

(c) (9, 10)

(d) (9, −10)

Answer: (d)

Solution:

\[m_{AC}\] = \[\frac{y_{3} - y_{1}}{x_{3} - x_{1}}\] = \[\frac{(2 -(-4))}{(5-(-1))}\] = 1 \[\Rightarrow\] \[m_{BE}\] = \[\frac{-1}{m_{AC}}\] = - 1

\[m_{BC}\] = \[\frac{y_{3} - y_{2}}{x_{3} - x_{2}}\] = \[\frac{(2 -(-3))}{(5 - 2}\]] = \[\frac{5}{3}\] \[\Rightarrow\] \[m_{AD}\] = \[\frac{-1}{m_{BC}}\] = \[\frac{-3}{5}\]

BE: \[\frac{y - y_{2}}{x - x_{2}}\] = \[m_{BE}\] \[\Rightarrow\] \[\frac{(y -(- 3))}{(x - 2}\] = -1 \[\Rightarrow\] x + y + 1 = 0

AD: \[\frac{y - y_{1}}{x - x_{1}}\] = \[m_{AD}\] \[\Rightarrow\] \[\frac{(y -(- 4))}{(x -(- 1))}\] = \[\frac{-3}{5}\] \[\Rightarrow\] 3x + 5y + 23 = 0

Solving the equations for BE and AD, we get the coordinates of the orthocenter H as follows.

$H\left( {9, - 10} \right)$

### Conclusion

The above article discusses the orthocentre of a triangle and explains the steps to calculate the orthocentre. The solved examples will help students to know how to use the orthocentre formula in problems.

## FAQs on Orthocenter Formula

**1. Is the orthocentre and circumcentre of a triangle the same?**

No, the orthocentre and circumcentre of a triangle are not the same. The orthocentre of a triangle is defined as the point of concurrency of the three altitudes of the triangle. On the other side, the circumcentre of a triangle is defined as the point of intersection of the perpendicular bisector of the sides of a triangle. Orthocentre and circumcentre are significantly important in the study of the various properties of a triangle with respect to its other dimensions.

**2. Can an orthocentre be outside of a triangle?**

Yes, an orthocentre of a triangle can be outside of a triangle. For an acute angle triangle, we have seen that the orthocentre lies inside the triangle. In the case where the orthocentre lies outside of the triangle, that triangle is the obtuse angle triangle. The orthocentre lies on the vertex of the right angle for a right triangle.