Circumcentre of Triangle Question5





In this page Circumcentre of triangle question5 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 question5 - Solution

Question 5:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (1,1) (3,4) and (5,-2).

Let A (1,1), B (3,4) and C (5,-2) 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 (1,1) and B (3,4)

Here x₁ = 1, x₂ = 3 and y₁ = 1,y₂ = 4

                 =  [(1+3)/2,(1+4)/2]

                 =  [4/2,5/2]

                 = [2,5/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(4-1)/(3-1)]

                = 3/2

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

                                                       = -1/(3/2)

                                                       = -2/3

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

Here point D is (2,5/2)

x₁ = 2 ,y₁ = 5/2

                 (y-5/2) = -2/3 (x-2)

                 3(y+5/2) = -2(x-2)

              3(2y + 5)/2 = (-2 x + 4)

                    6y + 15 = 2 (-2 x + 4)

                    6 y + 15 = -4 x + 8    

                         4 x + 6 y = 8 - 15

                         4 x + 6 y = - 7

Equation of the perpendicular line through D is 4 x + 6 y = -7

Now we need to find the midpoint of the side BC

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

B (3,4) and C (5,-2)

Here x₁ = 3, x₂ = 5 and y₁ = 4,y₂ = -2

                 =  [(3 + 5)/2,(4 + (-2))/2]

                 =  [8/2,2/2]

                 = [4,1]

So the vertices of E is (4,1)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-2 - 4)/(5 - 3)]

                = -6/2

                = -3

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

                                                       = -1/(-3)

                                                       = 1/3

Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (4,1)

x₁ = 4,y₁ = 1

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

                 3(y - 1) = 1 (x - 4)

                  3 y - 3 = x - 4                     

                  x - 3 y - 4 + 3 = 0

                  x - 3 y = 1

Equation of the perpendicular line through E is  x - 3y = 1      

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

               4 x + 6 y = - 7  ---------(1)

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


                4 x + 6 y = -7

(2) x 2 =>  2 x - 6 y = 2

              --------------

                6 x = -5

                  x = -5/6 

Substitute x = -5/6 in the first equation we get  4 (-5/6) + 6 y = -7

                                                             -10/3 + 6 y= -7

                                                        (-10+18y) = -7 x 3

                                                               18 y = -21 + 10

                                                               18 y = -11

                                                                y = -11/18

So the circumcentre of a triangle ABC is (-5/6,-11/18) Circumcentre of triangle question5 Circumcentre of triangle question5

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Circumcentre of Triangle Question5 to Analytical Geometry
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