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Medians and Perpendicular bisectors:

Centroid : Intersection of the medians of a triangle. Circumcentre : Intersection of the perpendicular bisectors of two sides. 2.10 Using Point of Intersection to Solve Problems. Medians and Perpendicular bisectors:. A (4, 7). D (5,3). B (–2, 3). C (6, –1).

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Medians and Perpendicular bisectors:

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  1. Centroid:Intersection of the medians of a triangle. Circumcentre:Intersection of the perpendicular bisectors of two sides. 2.10 Using Point of Intersection to Solve Problems Medians and Perpendicular bisectors:

  2. A(4, 7) D(5,3) B(–2, 3) C(6, –1) Ex. 1:The coordinates of DABC are A(4, 7), B(–2, 3), C(6, –1).Find the intersection of the medians (centroid). Find the equation of the median from B to AC. Find the midpoint of AC. MAC = (5, 3) Find slope of BD mBD = 0

  3. A(4, 7) D(5,3) B(–2, 3) C(6, –1) mBD = 0 y = mx + b y = 0x + b y = b y = 3 (equation of BD) Part 2: Find the equation ofthe median from A. Find the midpoint of BC.

  4. y = 3x– 5 A(4, 7) D(5,3) B(–2, 3) E(2,1) C(6, –1) Find the equation of the median from A. Find the midpoint of BC. MBC = (2, 1) Find the slope of AE. mAE = 3 y = 3x + b 1 = 3(2) + b –5 = b

  5. A(4, 7) D(5,3) B(–2, 3) E(2,1) C(6, –1) The centroid is (2.67, 3) y = 3x– 5 equation of AE equation of BD y = 3 solve by substitution 3x– 5 = 3 3x = 3 + 5 3x = 8 x = 2.67

  6. The centroid is the center of gravity of the triangle.

  7. Ex. 2:The coordinates of A(0, –5), B(8, 3), and C(6, 5). Find the circumcentre. (intersection of the perpendicular bisectors of the sides. Find the perpendicular bisector of BC. Find the midpoint of BC. C(6, 5) B(8, 3) MBC = (7, 4) Find the slope of BC. = – 1 A(0, –5)

  8. mBC= – 1 Slope of perpendicular bisector is 1 MBC = (7, 4) y = mx + b 4 = (1)7 + b 4 – 7 = b C(6, 5) – 3 = b B(8, 3) y = x – 3 (perpendicular bisector of BC) A(0, –5)

  9. Find the perpendicular bisector of AB Find the midpoint of AB MAB = (4, –1) Find the slope of AB. C(6, 5) B(8, 3) mAB = 1 The slope of the perpendicular bisector is –1 A(0, –5)

  10. The slope of the perpendicular bisector is –1 MAB = (4, –1) y = mx + b –1 = (–1)4 + b –1 + 4 = b C(6, 5) 3 = b B(8, 3) y = – x + 3 Perpendicular bisector of AB. A(0, –5)

  11. The circumcentre is (3, 0) Perpendicular bisector of AB. (i) y = – x + 3 (ii) y = x – 3 Perpendicular bisector of BC. 2y = 0 y = 0 C(6, 5) sub y = 0 into (i) B(8, 3) x = 3 A(0, –5)

  12. The circumcentre is the same distance from the three vertices of the triangle C(6, 5) B(8, 3) (3, 0) A(0, –5)

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