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Plane Geometry Unit 1 – Chapter 1

Plane Geometry Unit 1 – Chapter 1. Section 1.2 – Use Segments and Congruence Day 2. Section 1.2 Notes. Postulate or Axiom – a rule that is accepted without proof. Ruler Postulate The points on any line can be matched to real numbers, that real number is known as the coordinate of that point.

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Plane Geometry Unit 1 – Chapter 1

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  1. Plane GeometryUnit 1 – Chapter 1 Section 1.2 – Use Segments and Congruence Day 2

  2. Section 1.2 Notes • Postulate or Axiom – a rule that is accepted without proof. • Ruler Postulate • The points on any line can be matched to real numbers, that real number is known as the coordinate of that point. • The distance between any two points A and B, is written as AB and is the absolute value of the difference of the coordinates of A and B. Names A B x1 x2 Coordinates DISTANCE AB = |x1 – x2|

  3. Section 1.2 Notes • Segment Addition Postulate: • Between - is only when three points are collinear • If B is between A and C, then AB + BC = AC • If AB + BC = AC, then B is between A and C. AC B A C AB BC

  4. Section 1.2 Notes • Congruency: • Congruent Segments – line segments that have the same length. Equal vs. Congruence AB = CD lengths are equal AB CD Segments are congruent A B C D Mathematical Notation: Congruent

  5. Section 1.2 Notes • Using the ruler postulate... • -10 -8 -6 -4 -2 0 2 4 6 8 10 • Find: • 1) WX 2) XY 3) WY 4) ZW W Y X Z

  6. Section 1.2 Notes • Points G, H, I, J and K are collinear. GK = 24, HJ = 10, and GH = HI = IJ. Find each length • 1) HI 2) IJ 3) GH • 4) JK 5) IG 6) GK 24 K G H I J 10

  7. Section 1.2 Notes • Find RS, QS, and TV using the following information: • S is between T and V. R is between S and T. T is between R and Q. QV = 23, QT = 8, and TR = RS = SV Q V S R T 23 ? 8

  8. Section 1.2 Notes • Point J is between M and N. Find the length of JM and JN given the following information. JM = 7x + 2, MN = 64, and JN = 2x – 1. • So (7x + 2) + (2x – 1) = 64 9x + 1 = 64 • 9x = 63 • x = 7 • JM: 7(7) + 2 = 51 • JN: 2(7) – 1 = 13 N M J 64 2x – 1 7x + 2

  9. Section 1.2 Notes • Point M is between V and W. Find the length of VMand MW, given the following information. VM = 4x – 1, VW = 30, and MW = 3x + 3. • So (3x + 3) + (4x – 1) = 30 7x + 2 = 30 • 7x = 28 • x = 4 • MW: 3(4) + 3 = 15 • MV: 4(4) – 1 = 15 W V M 30 3x + 3 4x – 1

  10. CONCEPT CHECK • P. 10-11 • #3 – 6 • “Guided Practice”

  11. HOMEWORK • P. 12-13 • #6 - 30 • (Omit #12)

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