Circumcentre: Using Point of Intersection to Solve Problems

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Presentation transcript:

Circumcentre: Using Point of Intersection to Solve Problems Intersection of the perpendicular bisectors of two sides. (Find the equation of two perpendicular bisectors then determine the point of intersection.)

Centroid: Orthocentre Intersection of the medians of a triangle. (Find the equation of two medians then determine the point of intersection.) Orthocentre Intersection of the altitudes. (Find the equation of two altitudes then determine the point of intersection.)

Ex. 1: The coordinates of DABC are A(4, 7), 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. A(4, 7) MAC = (5, 3) B(–2, 3) D(5,3) Find slope of BD C(6, –1) mBD = 0

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

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

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

The centroid is the center of gravity of the triangle.

Ex. 2: The coordinates of A(0, –5), B(8, 3), and C(6, 5) 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)

(perpendicular bisector of BC) 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)

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)

Perpendicular bisector of AB. 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 y = – x + 3 B(8, 3) Perpendicular bisector of AB. A(0, –5)

(i) y = – x + 3 (ii) y = x – 3 2y = 0 y = 0 The circumcentre is (3, 0) Perpendicular bisector of AB. (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 The circumcentre is (3, 0) A(0, –5)

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

Ex. 2 Find the orthocentre of DPQR; P(1, –7), Q(5, –1), R(–4, –1). (Intersection of two altitudes) 1) Find equation of altitude from Q Find slope of PR P(1, –7) R(–4, –1) Q(5, –1) sub (5, –1)

1) Find equation of altitude from Q sub (5, –1) R(–4, –1) Q(5, –1) P(1, –7)

The equation of the altitude at P is x = 1. 2) Find equation of altitude from P Find slope of QR The equation of the altitude at P is x = 1. P(1, –7) R(–4, –1) Q(5, –1) The slope of the altitude at P is undefined.

Equation of altitude at Q x = 1 Equation of altitude at P Solve by substitution P(1, –7) R(–4, –1) Q(5, –1) The point of intersection is (1, –4.33)