The 3 altitudes of a triangle intersect at a common point called the **orthocentre.** Its location depends on the type of triangle; unlike the median, it can be located either inside, outside, or on the vertex of a triangle.

### Acute Triangle

The orthocentre is located on the inside of the triangle.

### Right Triangle

The orthocentre is located on the vertex of the right angle.

### Obtuse Triangle

The orthocentre is located on the outside of the triangle.

## Steps to Calculate Orthocentre

- Calculate the slope of at least 2 side lengths of the triangle using the formula \(m = \cfrac{y2 - y1}{x2 - x1}\)
- Using the slopes of the side lengths, find the slopes of the altitudes by calculating the perpendicular slopes using the formula \(m = \cfrac{-1}{m}\)
- Use the slope-point formula \(y - y₁ = m(x - x₁)\) to determine the altitudes. \(m\) represents the slope of the altitude and \((x₁, y₁)\) represents the point it connects to
- Set the altitudes equal to each other to solve for the x-coordinate of the orthocentre
- Plug the x-value into one of the altitude formulas to solve for the y-coordinate

**Example**
Find the orthocentre of the following triangle:

We can find the slopes of 2 side lengths of the triangle. In this instance, we will find the slopes of sides \(\text{AB}\) and \(\text{BC}\) with \(\text{A}\) acting as Point 1, \(\text{B}\) acting as Point 2 and \(\text{C}\) acting as Point 3:

**m(ᴀʙ)** \(= \cfrac{y₂ - y₁}{x₂ - x₁}\)

**m(ᴀʙ)** \(= \cfrac{4 - 1}{3 - 1}\)

**m(ᴀʙ)** \(= \cfrac{3}{2}\)

**m(ʙᴄ)** \(= \cfrac{y₃ - y₂}{x₃ - x₂}\)

**m(ʙᴄ)** \(= \cfrac{1 - 4}{5 - 3}\)

**m(ʙᴄ)** \(= \cfrac{-3}{2}\)

Now that we have the slopes, we can determine the slopes of the altitudes by calculating the inverse slopes:

**m(altitude)AB** \(= \cfrac{-1}{3/2}\)

**m(altitude)AB** \(= \cfrac{-2}{3}\)

**m(altitude)BC** \(= \cfrac{-1}{-3/2}\)

**m(altitude)BC** \(= \cfrac{2}{3}\)
Since we have the slopes of the altitudes, we can plug these values into the slope-point formula to determine the altitudes For side length \(\text{AB}\), we will be using Point \(\text{C}\). For side length \(\text{BC}\), we will be using Point \(\text{A}\):

**altitudeAB**: \(y - 1 = \cfrac{-2}{3}(x - 5)\)

**altitudeAB**: \(y - 1 = \cfrac{-2}{3}x + \cfrac{10}{3}\)

**altitudeAB**: \(y = \cfrac{-2}{3}x - \cfrac{10}{3} + \cfrac{3}{3}\)

**altitudeAB**: \(y = \cfrac{-2}{3}x + \cfrac{13}{3}\)

**altitudeBC**: \(y - 1 = \cfrac{2}{3}(x - 1)\)

**altitudeBC**: \(y - 1 = \cfrac{2}{3}x - \cfrac{2}{3}\)

**altitudeBC**: \(y = \cfrac{2}{3}x - \cfrac{2}{3} + \cfrac{3}{3}\)

**altitudeBC**: \(y = \cfrac{2}{3}x + \cfrac{1}{3}\)

Now that we have both altitudes, we can determine the location of the orthocentre. We can set the altitudes equal to each other to solve for \(x\).

\(\cfrac{-2}{3}x + \cfrac{13}{3} = \cfrac{2}{3}x + \cfrac{1}{3}\)

\(\cfrac{-2}{3}x - \cfrac{2}{3}x = \cfrac{1}{3} - \cfrac{13}{3}\)

\((\cfrac{-3}{4})(\cfrac{-4}{3}x) = (\cfrac{-3}{4})(\cfrac{-12}{3})\)

\(x = \cfrac{36}{12} = 3\)
Now we can plug the x-coordinate into one of the altitude formulas to solve for \(y\):

\(y = (\cfrac{-2}{3})(3) + \cfrac{13}{3}\)

\(y = \cfrac{-6}{3} + \cfrac{13}{3}\)

\(y = \cfrac{7}{3}\)
Therefore, we can determine that the location of the orthocentre is \((3, \(\cfrac{7}{3}\))\).

Determine the orthocentre of △\(\text{CRT}\) with coordinates \(\text{C}(1,8)\), \(\text{R}(3, 12)\) and \(\text{T}(6,2)\).

Show Answer
First, we can draw a diagram to visualize the problem:

We can find the slopes of 2 side lengths of the triangle. In this instance, we will find the slopes of sides \(\text{CR}\) and \(\text{CT}\) with \(\text{C}\) acting as Point 1, \(\text{R}\) acting as Point 2 and \(\text{T}\) acting as Point 3:

**mCR** \(= \cfrac{y2 - y1}{x2 - x1}\)

**mCR** \(= \cfrac{12 - 8}{3 - 1}\)

**mCR** \(= \cfrac{4}{2}\)

**mCR** \(= 2\)

**mCT** \(= \cfrac{y3 - y1}{x3 - x1}\)

**mCT** \(= \cfrac{2 - 8}{6 - 1}\)

**mCT** \(= \cfrac{-6}{5}\)
Now that we have the slopes, we can determine the slopes of the altitudes by calculating the inverse slopes:

**m(altitude)CR** \(= \cfrac{-1}{2}\)

**m(altitude)CT** \(= \cfrac{-1}{-6/5} = \cfrac{5}{6}\)
Since we have the slopes of the altitudes, we can plug these values into the slope-point formula to determine the altitudes For side length CR, we will be using Point T. For side length CR, we will be using Point R:

**altitudeCR**: \(y - 2 = \cfrac{-1}{2}(x - 6)\)

**altitudeCR**: \(y - 2 = \cfrac{-1}{2}x + 3\)

**altitudeCR**: \(y = \cfrac{-1}{2}x + 5\)

**altitudeCT**: \(y - 12 = \cfrac{5}{6}(x - 3)\)

**altitudeCT**: \(y - 12 = \cfrac{5}{6}x - \cfrac{15}{6}\)

**altitudeCT**: \(y = \cfrac{5}{6}x - \cfrac{15}{6} + \cfrac{52}{6}\)

**altitudeCT**: \(y = \cfrac{5}{6}x + \cfrac{57}{6}\)

**altitudeCT**: \(y = \cfrac{5}{6}x + \cfrac{19}{2}\)

Now that we have both altitudes, we can determine the location of the orthocentre. We can set the altitudes equal to each other to solve for \(x\). In order to do so in this instance, we must set all terms to have the same denominator:

\(\cfrac{-1}{2}x + 5 = \cfrac{5}{6}x + \cfrac{19}{2}\)

\(\cfrac{-3}{6}x + \cfrac{30}{6} = \cfrac{5}{6}x + \cfrac{57}{6}\)

\(\cfrac{-3}{6}x - \cfrac{5}{6}x = \cfrac{-30}{6} + \cfrac{57}{6}\)

\((\cfrac{-8}{6}x) = (\cfrac{9}{2})\)

\(x = (\cfrac{-6}{8} \cdot \cfrac{9}{2})\)

\(x = \cfrac{-27}{8}\)
Now we can plug the x-coordinate into one of the altitude formulas to solve for y:

\(y = (\cfrac{-1}{2})(\cfrac{-27}{8}) + 5\)

\(y = \cfrac{27}{16} + \cfrac{80}{16}\)

\(y = \cfrac{107}{16}\)
Therefore, we can determine that the location of the orthocentre is \((\cfrac{-27}{8}, \cfrac{107}{16})\).