Circumcentre of Triangle Question1





In this page Circumcentre of triangle question1 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 1:

Find the co ordinates of the circumcentre of a triangle whose

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

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

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

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

                 =  [10/2,-3-2/2]

                 = [5,-5/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-2-(-3))/(8-2)]

                = (-2+3)/(6)

                = 1/6

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

                                                       = -1/(1/6)

                                                       = -6

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

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

x₁ = 5 ,y₁ = -5/2

                 (y-(-5/2)) = -6 (x-5)

                 (y+5/2) = -6(x-5)

              (2y + 5)/2 = (-6 x + 30)

                    2y + 5 = 2 (-6 x + 30)

                    2 y + 5 = -12 x + 60    

                         12 x + 2 y = 60 - 5

Equation of the perpendicular line through D is 6 x + 2 y = 55

Now we need to find the midpoint of the side BC

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

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

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

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

                 =  [16/2,4/2]

                 = [8,2]

So the vertices of E is (8,2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(6 - (-2))/(8 - 8)]

                = (6 + 2)/0

                = ∞

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

                                                       = 1/∞

                                                       = 0

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (2,1)

x₁ = 8 ,y₁ = 2

                 (y - 2) = 0 (x - 8)

                 (y - 2) = 0

                  y - 2 = 0                     

Equation of the perpendicular line through E is y - 2 = 0      

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

               12 x + 2 y = 55  ---------(1)

                y - 2 = 0        ---------(2)

                  y = 2

Substitute y = 2 in the first equation we get 12 x + 2 (2) = 55

                                                             12 x + 4 = 55

                                                              12 x = 55 - 4

                                                               12 x = 51

                                                                x = 51/12

                                                                x = 17/4

So the circumcentre of a triangle ABC is (17/4,2) Circumcentre of triangle question1 Circumcentre of triangle question1

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Circumcentre of Triangle Question1 to Analytical Geometry
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