Circumcentre of Triangle Question3





In this page Circumcentre of triangle question3 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.

Question 3:

Find the co ordinates of the circumcentre of a triangle whose

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

Let A (3,4), B (2,-1) and C (4,-6) 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 (3,4) and B (2,-1)

Here x₁ = 3, x₂ = 2 and y₁ = 4,y₂ = -1

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

                 =  [5/2,-(4-1)/2]

                 = [5/2,3/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-1-4))/(2-3)]

                = (-5)/(-1)

                = 5

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

                                                       = -1/5

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

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

              (10 y - 15) = (-2 x + 5)

                10 y - 15 = -2 x + 5    

                 2 x + 10 y = 5 + 15

                 2 x + 10 y = 20

Equation of the perpendicular line through D is 2 x + 10 y = 20

Now we need to find the midpoint of the side BC

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

B (2,-1) and C (4,-6)

Here x₁ = 2, x₂ = 4 and y₁ = -1,y₂ = -6

                 =  [(2 + 4)/2,(-1 + (-6)/2]

                 =  [6/2,-1-6/2]

                 = [3,-7/2]

So the vertices of E is (3,-7/2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-6 - (-1))/(4 -2)]

                = (-6 + 1)/2

                = -5/2

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

                                                       = -1/(-5/2)

                                                       = 2/5

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (3,-7/2)

x₁ = 3 ,y₁ = -7/2

                 (y - (-7/2)) = 2/5 (x - 3)

                 (y + 7/2) = 2/5 (x - 3)

                 (2 y + 7)/2 = 2/5 (x -3)                     

                 5 (2 y + 7) = 2 x 2 (x -3)                     

                 10 y + 35  = 4 (x -3)                     

                 10 y + 35  = 4 x - 12

                 4 x + 10 y = 12 + 35

                 4 x + 10 y = 47

Equation of the perpendicular line through E is 4 x + 10 y = 47      

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

               2 x + 10 y = 20  ---------(1)

               4 x + 10 y = 47  ---------(2)


(1) - (2)           2 x + 10 y = 20  

                      4 x + 10 y = 47

                     (-)   (-)       (-)

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

                     -2 x   = - 27

                        x = 27/2

Substitute x = 27/2 in the first equation we get 2 x + 10 y = 20

                                                         2 (27/2) + 10 y = 20

                                                              27 + 10 y = 20

                                                               10 y = 20 - 27

                                                               10 y = -7

                                                                y = -7/10

So the circumcentre of a triangle ABC is (27/2,-7/10) Circumcentre of triangle question3 Circumcentre of triangle question3

HTML Comment Box is loading comments...




Circumcentre of Triangle Question3 to Analytical Geometry
[?]Subscribe To This Site
  • XML RSS
  • follow us in feedly
  • Add to My Yahoo!
  • Add to My MSN
  • Subscribe with Bloglines