**Orthocenter :**

It can be shown that the altitudes of a triangle are concurrent and the point of concurrence is called the orthocenter of the triangle. The orthocenter is denoted by O.

Let ABC be the triangle AD,BE and CF are three altitudes from A, B and C to BC, CA and AB respectively.

**Steps Involved in Finding Orthocenter of a Triangle :**

- Find the equations of two line segments forming sides of the triangle.
- Find the slopes of the altitudes for those two sides.
- Use the slopes and the opposite vertices to find the equations of the two altitudes.
- Solve the corresponding x and y values, giving you the coordinates of the orthocenter.

**Example 1 :**

Find the coordinates of the orthocentre of the triangle whose vertices are (3, 1), (0, 4) and (-3, 1)

**Solution :**

Now we need to find the slope of AC. From that we have to find the slope of the perpendicular line through B.

Slope of AC = (y_{2} - y_{1}) / (x_{2} - x_{1})

A (3, 1) and C (-3, 1)

here x_{1 }= 3, y_{1} = 1, x_{2 }= -3 and y_{2} = 1

= (1 - 1) / (-3 - 3)

= 0

Slope of the altitude BE = -1/ slope of AC

= 1/0

Equation of the altitude BE :

(y - y_{1}) = m (x -x_{1})

Here B (0, 4) m = 1/0

(y - 4) = (1/0) (x - 0)

(y - 4) = x / 0

x = 0

Now we need to find the slope of BC. From that we have to find the slope of the perpendicular line through D.

Slope of BC = (y_{2} - y_{1}) / (x_{2} - x_{1})

B (0, 4) and C (-3, 1)

here x_{1 }= 0, y_{1} = 4, x_{2 }= -3 and y_{2} = 1

= [(1 - 4) / (-3 - 0)]

= -3 / (-3)

= 1

Slope of the altitude AD = -1/ slope of AC

= -1/1

= -1

Equation of the altitude AD :

(y - y_{1}) = m (x -x_{1})

Here A(3, 1) m = -1

(y - 1) = -1(x - 3)

y - 1 = -x + 3

x + y = 3 + 1

x + y = 4 --------(1)

Substitute the value of x in the first equation

0 + y = 4

y = 4

So the orthocentre is (0, 4).

**Example 2 :**

Find the co ordinates of the orthocentre of a triangle whose vertices are (2, -3) (8, -2) and (8, 6).

**Solution :**

Let the given points be A (2, -3) B (8, -2) and C (8, 6)

Now we need to find the slope of AC.From that we have to find the slope of the perpendicular line through B.

Slope of AC = (y_{2} - y_{1}) / (x_{2} - x_{1})

A (2, -3) and C (8, 6)

here x_{1 }= 2, y_{1} = -3, x_{2 }= 8 and y_{2} = 6

= (6 - (-3)) / (8 - 2)

= 9/6

= 3/2

Slope of the altitude BE = -1/ slope of AC

= -1 / (3/2)

= -2/3

Equation of the altitude BE :

(y - y_{1}) = m (x - x_{1})

Here B (8, -2) m = 2/3

y - (-2) = (-2/3) (x - 8)

3(y + 2) = -2 (x - 8)

3y + 6 = -2x + 16

2x + 3y - 16 + 6 = 0

2x + 3y - 10 = 0

Now we need to find the slope of BC. From that we have to find the slope of the perpendicular line through D.

Slope of BC = (y_{2} - y_{1}) / (x_{2} - x_{1})

B (8, -2) and C (8, 6)

here x_{1 }= 8, y_{1} = -2, x_{2 }= 8 and y_{2} = 6

= (6 - (-2)) / (8 - 8)

= 8/0 = undefined

Slope of the altitude AD = -1/ slope of AC

= -1/undefined

= 0

Equation of the altitude AD:

(y - y_{1}) = m (x - x_{1})

Here A(2, -3) m = 0

y - (-3) = 0 (x - 2)

y + 3 = 0

y = -3

Substitute the value of x in the first equation

2x + 3(-3) = 10

2x - 9 = 10

2x = 10 + 9

2x = 19

x = 19/2

So, the orthocentre is (19/2,-3).

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