Circumcentre of Triangle Question9





In this page Circumcentre of triangle question 9 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 9:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (3,1) (2,2) and (2,0).

Circumcentre of triangle question9 - Solution

Let A (3,1), B (2,2) and C (2,0) 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 (2,2)

Here x₁ = 3, x₂ = 2 and y₁ = 1,y₂ = 2

                 =  [(3+2)/2,(1+2)/2]

                 =  [5/2,3/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(2-1)/(2-3)]

                = 1/(-1)

                = - 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 (5/2,3/2)

x₁ = 5/2 ,y₁ = 3/2

                 (y-3/2) = 1 (x-5/2)

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

              (2y - 3) = (2 x - 5)

                   2 x - 2 y - 5 + 3 = 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 (2,2) and C (2,0)

Here x₁ = 2, x₂ = 2 and y₁ = 2,y₂ = 0

                 =  [(2 + 2)/2,(2 + 0)/2]

                 =  [4/2,2/2]

                 = [2,1]

So the vertices of E is (2,1)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(0 - 2)/(2 -2)]

                = -2/0

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

                                                       = -1/(-2/0)

                                                       = 0/2

                                                       = 0

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (2,1)

x₁ = 2,y₁ = 1

                 (y - 1) = 0 (x - 2)

                 (y - 1) = 0

                  y - 1 = 0                     

Equation of the perpendicular line through E is  y - 1 = 0      

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

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

                 y - 1 = 0       ---------(2)

                   y = 1

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

                                                                x = 1 + 1

                                                                x = 2

So the circumcentre of a triangle ABC is (2,1) Circumcentre of triangle question9 Circumcentre of triangle question9

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Circumcentre of Triangle Question9 to Analytical Geometry
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