Circumcentre of Triangle Question6





In this page Circumcentre of triangle question6 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 question6 - Solution

Question 6:

Find the co ordinates of the circumcentre of a triangle whose

vertices are  (-3,-9) (-1,6) and (3,9).

Let A (-3,-9), B (-1,6) and C (3,9) 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,-9) and B (-1,6)

Here x₁ = -3, x₂ = -1 and y₁ = -9,y₂ = 6

                 =  [(-3+(-1))/2,(-9+6)/2]

                 =  [-4/2,3/2]

                 = [-2,3/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(6-(-9))/(-1-(-3))]

                = (6+9)/(-1+3)

                = 15/2

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

                                                       = -1/(15/2)

                                                       = -2/15

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

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

x₁ = -2 ,y₁ = 3/2

                 (y-3/2) = -2/15 (x-(-2))

                 15(2y-3)/2 = -2(x+2)

              15(2y - 3) = -4 (x + 2)

              30 y - 45 = -4 x - 8

           4 x  + 30 y - 45 + 8 = 0    

           4 x + 30 y - 37 = 0

           4 x + 30 y = 37 

Equation of the perpendicular line through D is 4 x + 30 y = 37

Now we need to find the midpoint of the side BC

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

B (-1,6) and C (3,9)

Here x₁ = -1, x₂ = 3 and y₁ = 6,y₂ = 9

                 =  [(-1 + 3)/2,(6 + 9)/2]

                 =  [2/2,15/2]

                 = [1,15/2]

So the vertices of E is (1,15/2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(9 - 6)/(3 - (-1))]

                = 3/(3+1)

                = 3/4

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

                                                       = -1/(3/4)

                                                       = -4/3

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (1,15/2)

x₁ = 1,y₁ = 15/2

                 (y - 15/2) = -4/3 (x - 1)

                 3(2y - 15)/2 = -4 (x - 1)

                  6 y - 45 = -8(x - 1)                     

                 6 y - 45 = -8 x + 8

                  8 x + 6 y = 8 + 45

                  8 x + 6 y = 53

Equation of the perpendicular line through E is  8 x + 6 y = 53

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

               4 x + 30 y = 37       ---------(1)

               8 x + 6 y = 53        ---------(2)


                4 x + 30 y  = 37

(2) x 5 =>  40 x + 30 y = 265

                (-)     (-)       (-)

              -------------------

                - 36 x = -228

                  x = -228/(-36) 

                  x = 19/3

Substitute x = 19/3 in the first equation we get  4 x + 30 y = 37

                                                             4 (19/3) + 30 y= 37

                                                        (76+90y) = 37 x 3

                                                         76+90y = 111

                                                               90 y = 111 - 76

                                                               90 y =  35

                                                                y = 35/90

                                                                y = 7/18

So the circumcentre of a triangle ABC is (19/3,7/18)
Circumcentre of triangle question6 Circumcentre of triangle question6

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Circumcentre of Triangle Question6 to Analytical Geometry
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