# Circumcentre of Triangle Question8

In this page Circumcentre of triangle question 8 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 question8 - Solution

Question 8:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (4,5) (4,2) and (-2,2).

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

Here x₁ = 4, x₂ = 4 and y₁ = 5,y₂ = 2

=  [(4+4)/2,(5+2)/2]

=  [8/2,7/2]

= [4,7/2]

So the vertices of D is (4,7/2)

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

= [(2-5)/(4-4)]

= -3/0

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

= -1/(-3/0)

=  0/3

=  0

Equation of the perpendicular line through D:

(y-y₁) = m (x-x₁)

Here point D is (4,7/2)

x₁ = 4 ,y₁ = 7/2

(y-7/2) = 0 (x-4)

(2y-7)/2 = 0

(2y - 7) = 0

2 y - 7 = 0

Equation of the perpendicular line through D is 2 y - 7 = 0

Now we need to find the midpoint of the side BC

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

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

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

=  [(4 + 2)/2,(-2 + 2)/2]

=  [6/2,0/2]

= [3,0]

So the vertices of E is (3,0)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

= [(2 - 2)/(-2 - 4)]

= 0/(-6)

= 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 (4,1)

x₁ = 3,y₁ = 0

(y - 0) = 1/0 (x - 3)

0 = 1 (x - 3)

x - 3 = 0

Equation of the perpendicular line through E is  x - 3 = 0

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

2 y - 7 = 0  ---------(1)

x - 3 = 0     ---------(2)

x = 3

2 y = 7

y = 7/2

So the circumcentre of a triangle ABC is (3,7/2) Circumcentre of triangle question8 Circumcentre of triangle question8

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Circumcentre of Triangle Question8 to Analytical Geometry 