Circumcentre of Triangle Question10





In this page Circumcentre of triangle question10 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 question10 - Solution 

Question 10:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (3,1) (0,4) and (-3,1).

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

Here x₁ = 3, x₂ = 0 and y₁ = 1,y₂ = 4

                 =  [(3+0)/2,(1+4)/2]

                 =  [3/2,5/2]

                 = [3/2,5/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(4-1)/(0-3)]

                = 3/(-3)

                = -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 (3/2,5/2)

x₁ = 3/2 ,y₁ = 5/2

                 (y-5/2) = 1 (x-3/2)

                 (2y-5)/2 = 1(2x-3)/2

              (2y - 5) = (2 x - 3)

                    2 x - 2 y -3 + 5 = 0

                    2 x - 2 y + 2 = 0    

÷ by 2 => x - y = -1

Equation of the perpendicular line through D is  x -  y = -1

Now we need to find the midpoint of the side BC

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

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

Here x₁ = 0, x₂ = -3 and y₁ = 4,y₂ = 1

                 =  [(0 + (-3))/2,(4 + 1)/2]

                 =  [-3/2,5/2]

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

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(1 - 4)/(-3 - 0)]

                = -3/(-3)

                = 1

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

                                                       = -1/1

                                                       = -1

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (-3/2,5/2)

x₁ = -3/2,y₁ = 5/2

                 (y - 5/2) = -1 (x - (-3/2))

                 (2y - 5)/2 = -1 (2x + 3)/2

                  2 y - 5 = - 2 x - 3                     

                2 x + 2 y - 5 + 3 = 0

                  2x + 2 y - 2 = 0

÷ by 2 => x + y = 1

Equation of the perpendicular line through E is  x + y = 1      

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

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

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


                x - y = -1

                x + y = 1

              --------------

                2 x = 0

                  x = 0/2

                  x = 0

Substitute x = 0 in the first equation we get  0 - y = -1

                                                                - y = -1

                                                                  y = 1

So the circumcentre of a triangle ABC is (0,1) Circumcentre of triangle question10 Circumcentre of triangle question10

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Circumcentre of Triangle Question10 to Analytical Geometry
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