Geometry 201 unit 3.4
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This document discusses parallel and perpendicular lines. It begins with examples of finding distances from points to lines and using properties of perpendicular lines to prove statements. It explains that the perpendicular bisector of a segment is a line perpendicular to the segment at its midpoint. It also explains that the shortest distance from a point to a line is the perpendicular segment from the point to the line. Several examples are provided of using properties of parallel and perpendicular lines to solve problems and write two-column proofs. Key terms like perpendicular bisector are defined.
Geometry 201 unit 3.4
- 1. UNIT 3.4 PARALLEL ANDUNIT 3.4 PARALLEL AND PERPENDICULAR LINESPERPENDICULAR LINES
- 2. Warm Up Solve each inequality. 1. x – 5 < 8 2. 3x + 1 < x Solve each equation. 3. 5y = 90 4. 5x + 15 = 90 Solve the systems of equations. 5. x < 13 y = 18 x = 15 x = 10, y = 15
- 3. Prove and apply theorems about perpendicular lines. Objective
- 4. perpendicular bisector distance from a point to a line Vocabulary
- 5. The perpendicular bisector of a segment is a line perpendicular to a segment at the segment’s midpoint. The shortest segment from a point to a line is perpendicular to the line. This fact is used to define the distance from a point to a line as the length of the perpendicular segment from the point to the line.
- 6. Example 1: Distance From a Point to a Line The shortest distance from a point to a line is the length of the perpendicular segment, so AP is the shortest segment from A to BC. B. Write and solve an inequality for x. AC > AP x – 8 > 12 x > 20 Substitute x – 8 for AC and 12 for AP. Add 8 to both sides of the inequality. A. Name the shortest segment from point A to BC. AP is the shortest segment. + 8 + 8
- 7. Check It Out! Example 1 The shortest distance from a point to a line is the length of the perpendicular segment, so AB is the shortest segment from A to BC. B. Write and solve an inequality for x. AC > AB 12 > x – 5 17 > x Substitute 12 for AC and x – 5 for AB. Add 5 to both sides of the inequality. A. Name the shortest segment from point A to BC. AB is the shortest segment. + 5+ 5
- 9. Example 2: Proving Properties of Lines Write a two-column proof. Given: r || s, ∠1 ≅ ∠2 Prove: r ⊥ t
- 10. Example 2 Continued Statements Reasons 2. ∠2 ≅ ∠3 3. ∠1 ≅ ∠3 3. Trans. Prop. of ≅ 2. Corr. ∠s Post. 1. r || s, ∠1 ≅ ∠2 1. Given 4. r ⊥ t 4. 2 intersecting lines form lin. pair of ≅ ∠s lines ⊥.
- 11. Check It Out! Example 2 Write a two-column proof. Given: Prove:
- 12. Check It Out! Example 2 Continued Statements Reasons 3. Given 2. Conv. of Alt. Int. ∠s Thm. 1. ∠EHF ≅ ∠HFG 1. Given 4. ⊥ Transv. Thm. 3. 4. 2.
- 13. Example 3: Carpentry Application A carpenter’s square forms a right angle. A carpenter places the square so that one side is parallel to an edge of a board, and then draws a line along the other side of the square. Then he slides the square to the right and draws a second line. Why must the two lines be parallel? Both lines are perpendicular to the edge of the board. If two coplanar lines are perpendicular to the same line, then the two lines are parallel to each other, so the lines must be parallel to each other.
- 14. Check It Out! Example 3 A swimmer who gets caught in a rip current should swim in a direction perpendicular to the current. Why should the path of the swimmer be parallel to the shoreline?
- 15. Check It Out! Example 3 Continued The shoreline and the path of the swimmer should both be ⊥ to the current, so they should be || to each other.
- 16. Lesson Quiz: Part I 1. Write and solve an inequality for x. 2x – 3 < 25; x < 14 2. Solve to find x and y in the diagram. x = 9, y = 4.5
- 17. Lesson Quiz: Part II 3. Complete the two-column proof below. Given: ∠1 ≅ ∠2, p ⊥ q Prove: p ⊥ r Proof Statements Reasons 1. ∠1 ≅ ∠2 1. Given 2. q || r 3. p ⊥ q 4. p ⊥ r 2. Conv. Of Corr. ∠s Post. 3. Given 4. ⊥ Transv. Thm.
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