Circumcentre Of a Triangle





On this page we are going to see how to find the circumcentre of a triangle. First, let us see the definition of circumcentre

Definition:

The point of concurrency of the perpendicular bisector of the sides of a triangle is called the circumcentre of the triangle. The circumcentre is denoted by S.


Let ABC be the triangle and D,E,F are the midpoint of BC,CA and AB.We need to find the slopes of the perpendicular bisectors of BC,CA  and AB.Then we need to find the equation of the perpendicular bisectors.By solving any two equations we can get the circumcentre.

Example 1:

Find the coordinates of the circumcentre of a triangle whose vertices are  (3,1) (2,2) and (2,0)

Let A,B and C are the vertices of the triangle

Now we need to find the midpoint of the side AB

Midpoint of AB = [(x₁ + x₂)/2 , (y₁ + y₂)/2]

A (3,1) and B (2,2)

Here x₁ = 3, x₂ = 2 and y₁ = 1,y₂ = 2

                 =  [(3+2)/2,(1+2)/2]

                 =  [5/2,3/2]

                 = [5/2,3/2]

So the vertices of D is (5/2,3/2)

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(2-1)/(2-3)]

                = 1/(-1)

                = -1

Slope of the perpendicular line through D = -1/slope of AB

                                                       = -1/(-1)

                                                       = 1

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

Here point D is (5/2,3/2)

x₁ = 5/2 ,y₁ = 3/2

                 (y-3/2) = 1(x-5/2)

                 (y-3/2) = 1(x-5/2)

              (2y -3)/2 = (2x-5)/2

                    2y-3 = 2x-5

                         2x-2y= -3+5

                         2x-2y = 2

Equation of the perpendicular line through D is 2x -5 =0       

Now we need to find the midpoint of the side BC

Midpoint of BC = [(x₁ + x₂)/2 , (y₁ + y₂)/2]

B (2,2) and C (2,0)

Here x₁ = 2, x₂ = 2 and y₁ = 2,y₂ = 0

                 =  [(2+2)/2,(2+0)/2]

                 =  [4/2,2/2]

                 = [2,1]

So the vertices of E is (2,1)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(0-2)/(2-2)]

                = -2/0

Slope of the perpendicular line through  E  =  -1/slope of BC

                                                       = 1/(-2/0)

                                                       = 1/∞

                                                       = 0

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (2,1)

x₁ = 2 ,y₁ = 1

                 (y-1) = 0(x-2)

                 (y-1) = 0

                  y -1 = 0                     

Equation of the perpendicular line through E is y -1 = 0      

Now we need to solve the equations of perpendicular bisectors D and E

               2x-2y = 2       ---------(1)

                y -1 = 0       ---------(2)

                  y = 1

Substitute y = 1 in the first equation we get 2x-2(1) = 2

                                                             2x-2 = 2

                                                              2x = 2+2

                                                               2x = 4

                                                                x = 4/2

                                                                x =2

So the circumcentre of a triangle ABC is (2,1)

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Circumcentre Of A Triangle to Analytical Geometry