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)**.

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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