In this page Circumcentre of triangle question1 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 1 :**

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

**Solution :**

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)**

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