ORTHOCENTER OF A TRIANGLE WITH COORDINATES

It can be shown that the altitudes of a triangle are concurrent and the point of concurrence is called the orthocentre of the triangle.

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

The following steps will be useful to find circumcenter of a triangle. 

Step 1 : 

Find the equations of any two altitudes. 

Step 2 : 

Solve the two equations found in step 2 for x and y. 

The solution (x, y) is the orthocenter of the triangle given.

Example :

Find the co ordinates of the orthocenter of a triangle whose vertices are (3, 1) (0, 4) and (-3, 1).

Solution :

Let the given points be A (3, 1) B (0, 4) and C (-3, 1).

Equation of the altitude AD :

Slope of BC is 

=  [(y₂ - y₁)/(x₂ - x₁)]

Substitute (x₁, y₁)  =  (0, 4) and (x₂, y₂)  =  (-3, 1).

=  (1 - 4) / (-3 - 0)

=  (-3) / (-3)

=  1

Slope of the perpendicular line to BC is

=  -1 / slope of BC

=  -1 / 1

=  -1

Equation of the altitude BC : 

y  =  mx + b

Substitute  m  =  -1. 

y  =  -x + b -----(1)

Substitute the point A(3, 1) for (x, y) into the above equation. 

1  =  -3 + b

4  =  b

Substitute b  =  4 in (1). 

y  =  -x + 4 -----(2)

Equation of the altitude BE :

Slope of AC is 

=  [(y₂ - y₁)/(x₂ - x₁)]

Substitute (x₁, y₁)  =  (3, 1) and (x₂, y₂)  =  (-3, 1).

=  (1 - 1) / (-3 - 3)

=  0 / (-6)

=  0

Slope of the perpendicular line to AC is

=  -1 / slope of AC

=  -1 / 0

=  ∞

So, the slope of the altitude BE is undefined. 

Because slope of the altitude BE is undefined, it's a vertical line passing through B(0, 4). 

Equation of a vertical line passing through the (0, 4) is 

x  =  0

So, equation of the altitude BE  is  

x  =  0 -----(3)

Solving (2) and (3), we get

x  =  0  and  y  =  4

Therefore, the orthocenter of the triangle ABC is 

(0, 4)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.