Circumcentre of Triangle Question2





In this page Circumcentre of triangle question2 we are going to see solution of first question.

Definition:

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

Circumcentre of triangle question2 - solution

Question 2:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (0,4) (3,6) and (-8,-2).

Let A (0,4), B (3,6) and C (-8,-2) be the vertices of triangle.

Now we need to find the midpoint of the side AB

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

A (0,4) and B (3,6)

Here x₁ = 0, x₂ = 3 and y₁ = 4,y₂ = 6

                 =  [(0 + 3)/2,(4+6)/2]

                 =  [3/2,10/2]

                 = [3/2,5]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(6 - 4)/(3 - 0)]

                = 2/3

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

                                                       = -1/(2/3)

                                                       = -1 x 3/2

                                                       = -3/2

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

Here point D is (3/2,5)

x₁ = 3/2 ,y₁ = 5

                 (y - 5) = -3/2 (x - 3/2)

                 y - 5 = -3/4 (2x - 3)

             4 (y - 5) = -3 (2x - 3)

              4 y - 20  = - 6 x + 9 

              6 x + 2 y - 20 - 9 = 0

              6 x + 2 y - 29 = 0

Equation of the perpendicular line through D is 6 x + 2 y - 29 = 0

Now we need to find the midpoint of the side BC

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

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

Here x₁ = 3, x₂ = -8 and y₁ = 6,y₂ = -2

                 =  [(3 + (-8))/2,(6 + (-2))/2]

                 =  [-5/2,4/2]

                 =  [-5/2,2]

So the vertices of E is (-5/2,2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-2 - 6)/(-8 - 3)]

                = (-8)/(-11)

                = 8/11

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

                                                       = -1/(8/11)

                                                       = -11/8

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (-5/2,2)

x₁ = -5/2 ,y₁ = 2

                 (y - 2) = -11/8 (x - (-5/2))

                 (y - 2) = -11/8 (x + 5/2)

              (y - 2) = -11/16 (2 x + 5)                     

              16 (y - 2) = -11 (2 x + 5)                     

               16 y - 32 = - 22 x - 55 

               22 x + 16 y - 32 + 55 = 0

               22 x + 16 y + 23 = 0

Equation of the perpendicular line through E is 22 x + 16 y + 23 = 0      

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

                6 x + 4 y - 29 = 0   ---------(1)

            22 x + 16 y + 23 = 0   ---------(2)


(1) x 4 = >   24 x + 16 y - 116 = 0

                  22 x + 16 y + 23 = 0

                  (-)      (-)     (-)  

                 ---------------------

                  2 x - 139 = 0

                          2 x = 139

                             x = 139/2

Substitute x = 139/2 in the first equation we get  6 (139/2) + 4 y - 29 = 0

                                                             3 (139) +  4 y = 29

                                                              417 + 4 y = 29

                                                               4 y = 29 - 417

                                                             4 y  = - 388

                                                                y = - 388/4

                                                                y = - 97

So the circumcentre of a triangle ABC is (139/2,-97) Circumcentre of triangle question2 Circumcentre of triangle question2

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Circumcentre of Triangle Question2 to Analytical Geometry
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