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Finding the centroid and circumcentre of a triangle given the coordinates of its vertices

Finding the centroid and circumcentre of a triangle given the coordinates of its vertices. Finding the centroid. 2 Methods 1) Use the formula

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Finding the centroid and circumcentre of a triangle given the coordinates of its vertices

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  1. Finding the centroid and circumcentre of a triangle given the coordinates of its vertices

  2. Finding the centroid • 2 Methods • 1) Use the formula • 2) Find the equations of the medians and then find the point of intersection of two of them using substitution. Then check using the third equation.

  3. Example: determine the centroid of the triangle with vertices j(-4,3), K(2,-3) and L(8,3).

  4. Method 1

  5. Method 2 • 1) Equation of median • from J • MKL=(5,0) • m = (3-0)/(-4-5) = -1/3 • 0 = (-1/3)(5)+b • b = 5/3 • y = -1/3 x + 5/3

  6. Method 2 • 2) Equation of median • from L • MJK=(-1,0) • m = (3-0)/(8-(-1)) = 1/3 • 0 = (1/3)(-1)+b • b = 1/3 • y = 1/3 x + 1/3

  7. Method 2 • 3) Equation of median • from K • x = 2

  8. Method 2 • 4) • Intersection of median from L and median from J • 1/3 x + 1/ 3 = -1/3 x + 5/3 • 2/3 x = 4/3 • x = 2 • y = 1/3(2) + 1/3 = 3/3 = 1 • Therefore the centroid is (2,1) which does lie on x = 2, the third median.

  9. Finding the circumcentre • Find the equations of the perpendicular bisectors and then find the point of intersection of two of them using substitution. Then check using the third equation.

  10. Find the circumcentre of the triangle with vertices of A(4,-1) B(-6,-3) and C(2,5) • 1) Equation of right bisector • to side AB • MAB= G = (-1,-2) • mAB= (-1-(-3))/(4-(-6)) =1/5 • m = -5 • -2 = (-5)(-1)+b • b = -7 • y = -5x - 7

  11. Find the circumcentre of the triangle with vertices of A(4,-1) B(-6,-3) and C(2,5) • 2) Equation of right bisector • to side BC • MBC= E = (-2,1) • mBC= (5-(-3))/(2-(-6)) =1 • m = -1 • 1 = (-1)(-2)+b • b = -1 • y = -1x - 1

  12. Find the circumcentre of the triangle with vertices of A(4,-1) B(-6,-3) and C(2,5) • 3) Equation of right bisector • to side AC • MAC=(3,2) • mAC= (-1-(5))/(4-2)) =-3 • m = 1/3 • 2 = (1/3)(3)+b • b = 1 • y = 1/3x + 1

  13. 4) • Intersection of R.B. to AB and R.B. to BC • -5x – 7 = -x - 1 • -4x = 6 • x = -1.5 • y = -5(-1.5) - 7 = 0.5 • Therefore the circumcentreis (-1.5,0.5) which does lie on y = 1/3x + 1 , the third right bisector because • 0.5 = 1/3(-1.5) + 1

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