Circumcentre of Triangle Question4





In this page Circumcentre of triangle question4 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 4:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (5,6) (2,4) and (1,-3).

Let A (5,6), B (2,4) and C (1,-3) 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 (5,6) and B (2,4)

Here x₁ = 5, x₂ = 2 and y₁ = 6,y₂ = 4

                 =  [(5+2)/2,(6+4)/2]

                 =  [7/2,10/2]

                 = [7/2,5]

So the vertices of D is (7/2,5)

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

                = [(4-6)/(2-5)]

                = (-2)/(-3)

                = 2/3

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

                                                       = -1/(2/3)

                                                       = -3/2

Equation of the perpendicular line through D:

                 (y-y₁) = m (x-x₁)

Here point D is (7/2,5)

x₁ = 7/2 ,y₁ = 5

                 (y-5) = -3/2 (x-7/2)

                 (y-5) = -3/4(2x-7)

              4(y - 5) = -3(2x - 7)

                    4y - 20 = (-6 x + 21)

                    6 x + 4 y = 21 + 20

                    6 x + 4 y = 41

Equation of the perpendicular line through D is 6 x + 4 y = 41

Now we need to find the midpoint of the side BC

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

B (2,4) and C (1,-3)

Here x₁ = 2, x₂ = 1 and y₁ = 4,y₂ = -3

                 =  [(2 + 1)/2,(4 + (-3)/2]

                 =  [3/2,1/2]

                 = [3/2,1/2]

So the vertices of E is (3/2,1/2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

                = [(-3 - 4))/(1 - 2)]

                = (-7)/(-1)

                = 7

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

                                                       = -1/7


Equation of the perpendicular line through E:

                 (y-y₁) = m (x-x₁)

Here point E is (3/2,1/2)

x₁ = 3/2 ,y₁ = 1/2

                 (y - 1/2) = -1/7 (x - 3/2)

                 (2y - 1)/2 = -1/7 (2 x - 3)/2

                  7(2 y - 1) = -1 (2 x - 3)                      

                  14 y - 7 = -2 x + 3

                  2 x + 14 y - 7 - 3 = 0 

                  2 x + 14 y - 10 = 0    

                  2 x + 14 y = 10      

Equation of the perpendicular line through E is 2 x + 14 y = 10

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

              6 x + 4 y = 41      ---------(1)

              2 x + 14 y = 10       ---------(2)

(2) x 3 => 6 x + 42 y = 30

(1) - (2)           6 x + 4 y = 41  

                      6 x + 42 y = 30

                      (-)   (-)       (-)

                    ----------------    

                            - 38 y = 11

                  y = -11/38

Substitute y = -11/38 in the first equation we get 6 x + 4 (-11/38) = 41

                                                             6 x - 44/38 = 41

                                                      (228 x - 44)/38 = 41

                                                               228 x -  44 = 41 x 38

                                                                228 x - 44 = 1558

                                                                228 x = 1558 + 44

                                                                228 x = 1602

                                                                      x = 1602/228

                                                                      x = 801/114

                                                                      x = 267/38

So the circumcentre of a triangle ABC is (267/38,-11/38) Circumcentre of triangle question4 Circumcentre of triangle question4

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Circumcentre of Triangle Question4 to Analytical Geometry
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