**Circumcentre of triangle worksheet : **

Here we are going to see solution of first question on the quiz Circumcentre of triangle.

**Definition :**

The point of concurrency of the perpendicular bisector of the sides of a triangle is called the circumcentre of the triangle.

(1) Find the co ordinates of the circumcentre of a triangle whose vertices are (2,-3) (8,-2) and (8,6).

(2) Find the co ordinates of the circumcentre of a triangle whose vertices are (0,4) (3,6) and (-8,-2).

**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 -----(1)**

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 ------(2)**

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)**

Let us see the solution of next problem on "Circumcentre of triangle worksheet"

**Question 2 :**

Find the co ordinates of the circumcentre of a triangle whose vertices are (0,4) (3,6) and (-8,-2).

**Solution :**

Let A (0,4), B (3,6) and C (-8,-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 (0,4) and B (3,6)

Here x₁ = 0, x₂ = 3 and y₁ = 4,y₂ = 6

= [(0 + 3)/2,(4+6)/2]

= [3/2,10/2]

= [3/2,5]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

= [(6 - 4)/(3 - 0)]

= 2/3

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

= -1/(2/3) ==> -1 x 3/2 ==>-3/2

**Equation of the perpendicular line through D:**

(y-y₁) = m (x-x₁)

Here point D is (3/2,5)

x₁ = 3/2 ,y₁ = 5

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

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

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

4 y - 20 = - 6 x + 9

6 x + 2 y - 20 - 9 = 0

6 x + 2 y - 29 = 0

Equation of the perpendicular line through D is 6 x + 2 y - 29 = 0

Now we need to find the midpoint of the side BC

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

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

Here x₁ = 3, x₂ = -8 and y₁ = 6,y₂ = -2

= [(3 + (-8))/2,(6 + (-2))/2]

= [-5/2,4/2]

= [-5/2,2]

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

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

= [(-2 - 6)/(-8 - 3)]

= (-8)/(-11)

= 8/11

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

= -1/(8/11) ==> -11/8

**Equation of the perpendicular line through E:**

(y-y₁) = m (x-x₁)

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

x₁ = -5/2 ,y₁ = 2

(y - 2) = -11/8 (x - (-5/2))

(y - 2) = -11/8 (x + 5/2)

(y - 2) = -11/16 (2 x + 5)

16 (y - 2) = -11 (2 x + 5)

16 y - 32 = - 22 x - 55

22 x + 16 y - 32 + 55 = 0

22 x + 16 y + 23 = 0

Equation of the perpendicular line through E is

**22 x + 16 y + 23 = 0 -------(1)**

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

6 x + 4 y - 29 = 0 ---------(1)

22 x + 16 y + 23 = 0 ---------(2)

(1) x 4 = > 24 x + 16 y - 116 = 0

22 x + 16 y + 23 = 0

(-) (-) (-)

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

2 x - 139 = 0 ==> 2 x = 139 ==> x = 139/2

Substitute x = 139/2 in the (1) equation we get 6 (139/2) + 4 y - 29 = 0

3 (139) + 4 y = 29

417 + 4 y = 29

4 y = 29 - 417

4 y = - 388

y = - 388/4

y = - 97

So the circumcentre of a triangle ABC is **(139/2,-97)**

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