Circumcentre of Triangle Question7





In this page Circumcentre of triangle question7 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 question7 - Solution

Question 7:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (-3,-9) (3,9) and (5,-8).

Let A (-3,-9), B (3,9) and C (5,-8) 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 (3,9)

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

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

                 =  [0/2,0/2]

                 = [0,0]

So the vertices of D is (0,0)

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(9-(-9))/(3-(-3)]

                = (9+9)/(3+3)

                = 18/6

                = 3

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

                                                       = -1/3

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

Here point D is (0,0)

x₁ = 0 ,y₁ = 0

                 (y-0) = -1/3 (x-0)

                 3 y  = -1 (x)         

                 x + 3 y = 0

Equation of the perpendicular line through D is  x + 3 y = 0

Now we need to find the midpoint of the side BC

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

B (3,9) and C (5,-8)

Here x₁ = 3, x₂ = 5 and y₁ = 9,y₂ = -8

                 =  [(3 + 5)/2,(9 + (-8)/2]

                 =  [8/2,1/2]

                 = [4,1/2]

So the vertices of E is (4,1/2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-8 - 9)/(5 - 3)]

                = -17/2

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

                                                       = -1/(-17/2)

                                                       = 2/17

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (4,1/2)

x₁ = 4,y₁ = 1/2

                 (y - 1/2) = 2/17 (x - 4)

                 17(2y - 1)/2 = 2 (x - 4)

                  34 y - 17 = 4 (x - 4)                     

                  34 y - 17 = 4 x - 16

                  4 x - 34 y - 16 + 17

                  4 x - 34 y + 1 = 0

                  4 x - 34 y = -1

Equation of the perpendicular line through E is  4 x - 34 y = -1   

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

                 x + 3 y = 0  ---------(1)

                4 x - 34 y = -1   ---------(2)


(1) x 4 =>  4 x + 12 y = 0

               4 x - 34 y = -1

              (-)   (+)      (+)

              --------------

                    46 y = 1

                        y = 1/46 

Substitute y = 1/46 in the first equation we get  x + 3 y = 0

                                                             x  + 3 (1/46)= 0

                                                        ( 46 x + 3) = 0 x 46

                                                               46 x + 3 = 0

                                                               46 x = -3

                                                                x  = -3/46

So the circumcentre of a triangle ABC is (-3/46,1/46) Circumcentre of triangle question7 Circumcentre of triangle question7

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Circumcentre of Triangle Question5 to Analytical Geometry
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