## Circumcentre of Triangle Question3

In this page Circumcentre of triangle question3 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 3:

Find the co ordinates of the circumcentre of a triangle whose

vertices are (2,-3) (8,-2) and (8,6).

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

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

=  [(3 + 2)/2,(4 + (-1)/2]

=  [5/2,-(4-1)/2]

= [5/2,3/2]

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

Slope of AB = [(y₂ - y₁)/(x₂ - x₁)]

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

= (-5)/(-1)

= 5

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

= -1/5

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) = -1/5 (x-5/2)

5 (2y - 3)/2 = -1(2x-5)/2

(10 y - 15) = (-2 x + 5)

10 y - 15 = -2 x + 5

2 x + 10 y = 5 + 15

2 x + 10 y = 20

Equation of the perpendicular line through D is 2 x + 10 y = 20

Now we need to find the midpoint of the side BC

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

B (2,-1) and C (4,-6)

Here x₁ = 2, x₂ = 4 and y₁ = -1,y₂ = -6

=  [(2 + 4)/2,(-1 + (-6)/2]

=  [6/2,-1-6/2]

= [3,-7/2]

So the vertices of E is (3,-7/2)

Slope of BC = [(y₂ - y₁)/(x₂ - x₁)]

= [(-6 - (-1))/(4 -2)]

= (-6 + 1)/2

= -5/2

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

= -1/(-5/2)

= 2/5

Equation of the perpendicular line through E:

(y-y₁) = m (x-x₁)

Here point E is (3,-7/2)

x₁ = 3 ,y₁ = -7/2

(y - (-7/2)) = 2/5 (x - 3)

(y + 7/2) = 2/5 (x - 3)

(2 y + 7)/2 = 2/5 (x -3)

5 (2 y + 7) = 2 x 2 (x -3)

10 y + 35  = 4 (x -3)

10 y + 35  = 4 x - 12

4 x + 10 y = 12 + 35

4 x + 10 y = 47

Equation of the perpendicular line through E is 4 x + 10 y = 47

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

2 x + 10 y = 20  ---------(1)

4 x + 10 y = 47  ---------(2)

(1) - (2)           2 x + 10 y = 20

4 x + 10 y = 47

(-)   (-)       (-)

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

-2 x   = - 27

x = 27/2

Substitute x = 27/2 in the first equation we get 2 x + 10 y = 20

2 (27/2) + 10 y = 20

27 + 10 y = 20

10 y = 20 - 27

10 y = -7

y = -7/10

So the circumcentre of a triangle ABC is (27/2,-7/10) Circumcentre of triangle question3 Circumcentre of triangle question3

HTML Comment Box is loading comments...

Circumcentre of Triangle Question3 to Analytical Geometry 