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

2. ©thevisualclassroom.com Ex. 1: The coordinates of  ABC are A(4, 7), B(–2, 3), C(6, –1). Find the intersection of the medians (centroid). Find the midpoint of AC. B(–2, 3) C(6, –1) A(4, 7) D(5,3) Find the equation of the median from B to AC. M AC = (5, 3) Find slope of BD m BD = 0

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

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

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

6. ©thevisualclassroom.com The centroid is the center of gravity of the triangle.

7. ©thevisualclassroom.com 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 midpoint of BC. B(8, 3) C(6, 5) A(0, –5) Find the perpendicular bisector of BC. M BC = (7, 4) Find the slope of BC. = – 1

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

9. ©thevisualclassroom.com B(8, 3) C(6, 5) A(0, –5) Find the perpendicular bisector of AB Find the midpoint of AB M AB = (4, –1) Find the slope of AB. m AB = 1 The slope of the perpendicular bisector is –1

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

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

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