MATH SOLUTIONSMATH SOLUTIONSBy: Kasi, Jessica, Bhagy,

and Janie!

QUESTION: ? QUESTION: ?

A triangle has vertices A (-4,0), B (2,6) and C (8, -4).

Determine the exact coordinates of the circumcentre

STEPS TO STEPS TO SOLUTION:SOLUTION:

• 1) FIND MIDPOINT

2) FIND SLOPE

• 3) NEGATIVE RECIPROCAL

4) EQUATION OF PERPENDICULAR BISECTOR

• 5) REPEAT.

With the With the 33 given points..... given points.....we are going to find the we are going to find the

PERPENDICULAR PERPENDICULAR BISECTORBISECTORfor each side. for each side.

Explanation:

So how do we do it?So how do we do it?

y=m(x-p)+q

Perpendicular Bisector of Perpendicular Bisector of ABAB

1) Find the midpoint of AB!

Coordinate points:A (-4,0) B (2,6) C (8, -4)

M = (-4+2), (0+6)

2 2

MIDPOINT:

Perpendicular Bisector of Perpendicular Bisector of ABAB

2) Find the slope of AB

m = 6-02-(-4)

m = 66

m =1

3) Opposite Reciprocal = -1

Equation of Perpendicular Equation of Perpendicular Bisector for ABBisector for ABy = m(x-p)+qy = -1 (x+1) + 3y = -x-1+3y = -x + 2

Perpendicular Bisector of Perpendicular Bisector of BC BC

• 1)Midpoint

• 2)Slope

• 3)Opposite Reciprocal

• 4) Equation of Perpendicular Bisector

Let’s refresh the steps:To find the perpendicular bisector for each side

Perpendicular Bisector of Perpendicular Bisector of BCBC

Coordinate points:A (-4,0) B (2,6) C (8, -4)

M = (2+8), (6+(-4)Midpoint:

2 2

= 10 , 222

= ( 5, 1)

Slope: m = -4 - 6

8 - 2= -10

6= -5

3

Negative Reciprocal: = 3

5

Equation of Equation of Perpendicular Bisector Perpendicular Bisector

for BC for BC y = 3 (x-5)+1

y = 3x -15/5 +15

5y = 3x - 3 +1

y = 3x -25

5

y=m(x-p)+q y=m(x-p)+q

Perpendicular Bisector of Perpendicular Bisector of CACA

1) Slope: M = ( 8 + ( -4) , (-4 - 0)

22M = 4 , -4

2 2M = ( 2 , -2 )

2) Midpoint: m = 0 - ( - 4)

- 4 - 8m = 4

-12 m = 1

-3

3) Negative Reciprocal m = 3

4) Equation of Perpendicular

Bisector for CA

y = 3 ( x - 2 ) - 2y = 3x - 6 -2

y = 3x - 8

Three Equations Three Equations

y = 3x - 8

AB

BC

CA

5y = 3x -2y = -x + 2

Intersection Points Intersection Points (Circumcentre) (Circumcentre) (Circumcentre) (Circumcentre) * Using Substitution *

AB & BC

y = -x + 2

5y = 3x -2

AB y = -x + 2 y = -2.5 + 2 y = -0.5

-x + 2 = 3x - 25

2 + 2 = 3x + 1x5 1

x 5 x 5

4 = 3x + 5x5 5

4 = 8x5

4 = 1.6x1.61.6 2.5 = x

Let’s check/verify! Let’s check/verify!

(2.5 , - 0.5) x y

Verify:

CA y = 3x - 8- 0.5 = 3 (2.5) - 8- 0.5 = 7.5 - 8- 0. 5 = - 0.5

Le t’s grap h it!

THE END : )