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Properties or Parallel Lines GEOMETRY LESSON 3-1 (For help, go to page 24 or Skills Handbook, page 720.) Solve each equation. 1. x + 2x + 3x = 180 2. (w + 23) + (4w + 7) = 180 3. 90 = 2y – 30 4. 180 – 5y = 135 Write an equation and solve the problem. 5. The sum of m 1 and twice its complement is 146. Find m 1. 6. The measures of two supplementary angles are in the ratio 2 : 3. Find their measures. 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 Solutions 1. Combine like terms: 6x = 180; divide both sides by 6: x = 30 2. Combine like terms: 5w + 30 = 180; Subtract 30 from both sides: 5w = 150; divide both sides by 5: w = 30 3. Add 30 to both sides: 120 – 2y; divide both sides by 2: 60 = y 4. Subtract 180 from both sides: – 5y = –45; divide both sides by –5: y = 9 5. The complement of 1 is 90 – m 1; m 1 + 2(90 – m 1) = 146; distribute: m 1 + 180 – 2m = 146; combine like terms: 180 – m 1 = 146; subtract 180 from both sides: –m 1 = –34; multiply both sides by –1: m 1 = 34 6. Let 2x represent the measure of one of the angles and 3x represent the measure of the other angle. Then 2x + 3x = 180; combine like terms: 5x = 180; divide both sides by 5: x = 36. Then 2x = 72 and 3x = 108.0. 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 Use the diagram above. Identify which angle forms a pair of same-side interior angles with 1. Identify which angle forms a pair of corresponding angles with 1. Same-side interior angles are on the same side of transversal t between lines p and q. 4, 8, and 5 are on the same side of the transversal as but only 1 and 8 are interior. So 1 and 8 are same-side interior angles. 3-1 1, Properties or Parallel Lines GEOMETRY LESSON 3-1 (continued) Corresponding angles also lie on the same side of the transversal. One angle must be an interior angle, and the other must be an exterior angle. The angle corresponding to 1 must lie in the same position relative to line q as 1 lies relative to line p. Because 1 is an interior angle, 1 and are corresponding angles. 3-1 5 Properties or Parallel Lines GEOMETRY LESSON 3-1 Compare 2 and the vertical angle of 1. Classify the angles as alternate interior angles, same–side interior angles, or corresponding angles. The vertical angle of 1 is between the parallel runway segments. 2 is between the runway segments and on the opposite side of the transversal runway. Because alternate interior angles are not adjacent and lie between the lines on opposite sides of the transversal, 2 and the vertical angle of 1 are alternate interior angles. 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 Which theorem or postulate gives the reason that m 3 and m 3+m 3+m 2 are adjacent angles that form a straight angle. 2 = 180 because of the Angle Addition Postulate. 3-1 2 = 180? Properties or Parallel Lines GEOMETRY LESSON 3-1 In the diagram above, || m. Find m 1 and then m 2. 1 and the 42° angle are corresponding angles. Because m 1 = 42 by the Corresponding Angles Postulate. || m, Because 1 and 2 are adjacent angles that form a straight angle, m 1 + m 2 = 180 by the Angle Addition Postulate. If you substitute 42 for m 1, the equation becomes 42 + m Subtract 42 from each side to find m 2 = 138. 3-1 2 = 180. Properties or Parallel Lines GEOMETRY LESSON 3-1 In the diagram above, || m. Find the values of a, b, and c. a = 65 Alternate Interior Angles Theorem c = 40 Alternate Interior Angles Theorem a + b + c = 180 Angle Addition Postulate 65 + b + 40 = 180 b = 75 Substitution Property of Equality Subtraction Property of Equality 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 Pages 118-121 Exercises 1. PQ and SR with transversal SQ; alt. int. s 2. PS and QR with transversal SQ; alt. int. s 3. PS and QR with transversal PQ; same-side int. s 4. PS and QR with transversal SR; corr. s 5. 6. 7. 1 and 2: corr. s 3 and 4: alt. int. 5 and 6: corr. s 8. alt. int. s 1 and 2: same-side int. s 3 and 4: corr. s 5 and 6: corr. s 1 and 2: corr. s 3 and 4: same-side int. s 5 and 6: alt. int. 3-1 s s 9. a. 2 b. 1 c. corr. 10. a. Def. of b. Def. of right c. Corr. s of i lines are d. Subst. e. Def. of right f. Def. of . Properties or Parallel Lines GEOMETRY LESSON 3-1 11. m 1 = 75 because 14. 70; 70, 110 corr. s of || lines are ; m 2 = 105 because 15. 25; 65 same-side int. s of || lines are suppl. 16. 20; 100, 80 12. m 1 = 120 because 17. m 1 = m 3 = m 6 = corr. s of || lines are ; m 8 = m 9 = m 11 m 2 = 60 because = m 13 = m 15 = 52; same-side int. s of || m 2=m 4=m 5= lines are suppl. m 7 = m 10 = m 12 = m 14 = 128 13. m 1 = 100 because same-side int. s of || lines are suppl.; m 2 = 70 because alt. int. s of || lines have = measure. 3-1 18. You must find the measure of one . All s that are vert., corr., or alt. int. to that will have that measure. All other s will be the suppl. of that measure. 19. two 20. four 21. two 22. four 23. 32 Properties or Parallel Lines GEOMETRY LESSON 3-1 24. x = 76, y = 37, v = 42, w = 25 25. x = 135, y = 45 26. The s labeled are corr. s and should be . If you solve 2x – 60 = 60 – 2x, you get x = 30. This would be impossible since 2x – 60 and 60 – 2x would equal 0. 27. Trans means across or over. A transversal cuts across other lines. 28. Answers may vary. Sample: E illustrates corr. s ( 1 and 3, 2 and 4) and same-side int. s ( 1 and 2, 3 and 4); I illustrates alt. int. s ( 1 and 4, 2 and 3) and same-side int. s ( 1 and 3, 2 and 4). 29. a. alt. int. s b. He knew that alt. int. 30. a. 57 b. same-side int. 3-1 s s of || lines are . Properties or Parallel Lines GEOMETRY LESSON 3-1 31. a. If two lines are || and cut by a transversal, then same-side ext. s are suppl. b. Given: a || b Prove: 4 and 5 are suppl. 1. a || b (Given) 2. m 5 + m 6 = 180 ( Add. Post.) 3. 4 6 (Corr. s are ) 4. m 5 + m 4 = 180 (Subst.) 5. 4 and 5 are suppl. (Def. of suppl.) 32. 1. a || b (Given) 2. 1 2 (Vert. 3. 2 3 (Corr. 4. 1 33. Never; the two planes do not intersect. 34. Sometimes; if they are ||. 35. Sometimes; they may be skew. 36. Sometimes; they may be ||. 37. D 38. G are s are s .) .) 39. D 40. I 3 (Trans. Prop.) 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 41. [2] a. First show that 1 7. Then show that 7 5. Finally, show that 1 5 (OR other valid solution plan). b. 1 7 because vert. s are . 7 5 because corr. s of || lines are . Finally, by the Trans. Prop. of , 1 5. 42. 121 43. 59 44. 29.5 45. (0.5, 7) 46. (–0.5, 3.5) [1] incorrect sequence of steps OR incorrect logical argument 47. (3, 3) 48. add 4; 20, 24 49. multiply by –2; 16, –32 50. subtract 7; –5, –12 3-1 Properties or Parallel Lines GEOMETRY LESSON 3-1 In the diagram below, m || n. Use the diagram for Exercises 1–5. 1. Complete: 2. Complete: 6 and and 4 are alternate interior angles. 8 are corresponding angles. 4 3. Suppose that m 3 = 37. Find m 4. Suppose that m 1 = x + 12 and m 6. 143 5 = 3x – 36. Find x. 24 5. If a transversal intersects two parallel lines, then same-side exterior angles are supplementary. Write a Plan for Proof. Given: m || n Prove: 2 and 7 are supplementary. Show that m 2 = m 6. Then show that m 6 + m 7 = 180, and substitute m 2 for m 6. 3-1 Proving Lines Parallel GEOMETRY LESSON 3-2 (For help, go to page 24 and Lesson 2-1.) Solve each equation. 1. 2x + 5 = 27 2. 8a – 12 = 20 3. x – 30 + 4x + 80 = 180 4. 9x – 7 = 3x + 29 Write the converse of each conditional statement. Determine the truth value of the converse. 5. If a triangle is a right triangle, then it has a 90° angle. 6. If two angles are vertical angles, then they are congruent. 7. If two angles are same-side interior angles, then they are supplementary. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Solutions 1. Subtract 5 from both sides: 2x = 22; divide both sides by 2: x = 11 2. Add 12 to both sides: 8a = 32; divide both sides by 8: a = 4 3. Combine like terms: 5x + 50 = 180; subtract 50 from both sides: 5x = 130; divide both sides by 5: x = 26 4. Add –3x + 7 to both sides: 6x = 36; divide both sides by 6: x = 6 5. Reverse the hypothesis and conclusion: If a triangle has a 90° angle, then it is a right triangle. By definition of a right triangle, it is true. 6. Reverse the hypothesis and conclusion: If two angles are congruent, they are vertical angles. A counterexample is the congruent base angles of an isosceles triangle, which are not vertical angles. The converse is false. 7. Reverse the hypothesis and conclusion: If two angles are supplementary, then they are the same-side interior angles of parallel lines. A counterexample is two angles that form a straight angle. The converse is false. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. Given: 1 Prove: || m 2 Write the flow proof below of the Alternate Interior Angles Theorem as a paragraph proof. By the Vertical Angles Theorem, 3 1. 1 2, so 3 2 by the Transitive Property of Congruence. Because 3 and 2 are corresponding angles, || m by the Converse of the Corresponding Angles Postulate. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Use the diagram above. Which lines, if any, must be parallel if are supplementary? It is given that 3 and The diagram shows that 3 and 2 are supplementary. 4 and 2 are supplementary. Because supplements of the same angle are congruent (Congruent Supplements Theorem), 3 4. Because 3 and 4 are congruent corresponding angles, EC || DK by the Converse of the Corresponding Angles Postulate. 3-2 2 Proving Lines Parallel GEOMETRY LESSON 3-2 Use the diagram above. Which angle would you use with 1 to prove the theorem In a plane, if two lines are perpendicular to the same line, then they are parallel to each other (Theorem 3-6) using the Converse of the Alternate Interior Angles Theorem instead of the Converse of the Corresponding Angles Postulate? By the Vertical Angles Theorem, 2 is congruent to its vertical angle. Because 1 2, 1 is congruent to the vertical angle of the Transitive Property of Congruence. 2 by Because alternate interior angles are congruent, you can use the vertical angle of 2 and the Converse of the Alternate Interior Angles Theorem to prove that the lines are parallel. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Find the value of x for which || m. The labeled angles are alternate interior angles. If || m, the alternate interior angles are congruent, and their measures are equal. Write and solve the equation 5x – 66 = 14 + 3x. 5x – 66 = 14 + 3x 5x = 80 + 3x Add 66 to each side. 2x = 80 Subtract 3x from each side. x = 40 Divide each side by 2. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Suppose that the top and bottom pieces of a picture frame are cut to make 60° angles with the exterior sides of the frame. At what angle should the two sides be cut to ensure that opposite sides of the frame will be parallel? In order for the opposite sides of the frame to be parallel, same-side interior angles must be supplementary. Two 90° angles are supplementary, so find an adjacent angle that, together with 60°, will form a 90° angle: 90° – 60° = 30°. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Pages 125-129 Exercises 1. BE || CG; Conv. of Corr. s Post. 2. CA || HR; Conv. of Corr. s Post. 3. JO || LM; if two lines and a transversal form same-side int. s that are suppl., then the lines are ||. 4. a || b; if two lines and a 8. a || b; Conv. of Corr. s Post. transversal form sameside int. s that are 9. none suppl., then the lines are ||. 10. a || b; Conv. of Alt. Int. s Thm. 5. a || b; if two lines and a transversal form same11. || m; Conv. of Corr. side int. s that are s Post. suppl., then the lines are ||. 12. none 6. none 13. a || b; Conv. of Corr. s Post. 7. none 14. none 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 15. || m; Conv. of Alt. Int. s Thm. 24. When the frame is put together, each of the frame is a right . Two right s are suppl. By the Conv. of the Same-Side Int. s Thm., opp. sides of the frame are ||. 18. 30 19. 50 16. a. b. c. d. Def. of Given All right s are . Conv. of Corr. s 20. 59 21. 31 Post. 22. 5 17. a. b. c. 1 1 2 23. 20 d. 3 e. Conv. of Corr. 25. The corr. s are , so the lines are || by the Conv. of Corr. s Post. 26. a. Corr. s b–c. 1, 3 (any order) d. Conv. of Corr. s s 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 27. 10; m 1 = m 2 = 70 33. PL || NA by Conv. of Same-Side Int. 28. 5; m 1 = m 2 = 50 34. none 29. 2.5; m 1 = m 2 = 30 35. PN || LA by Conv. of Same-Side Int. s Thm. s Thm. 30. 1.25; m 1 = m 2 = 10 36. Answers may vary. Sample: In the diagram, AB BH and AB BD, but BH || BD. They intersect. 31. The corr. are . 37. Reflexive: a || a; false; any line intersects itself. Symmetric: If a || b, then b || a; true; b and a are coplanar and do not intersect. Transitive: In general, if a || b, and b || c, then a || c; true; however, when a || b, and b || a, it does not follow that a || a. s he draws 32. PL || NA and PN || LA by Conv. of SameSide Int. s Thm. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 38. Reflexive: a a; false; lines are two lines that intersect to form right s . Symmetric: If a b, then b a; true; b and a intersect to form right s . Transitive: If a b, and b c, then a c; false; in a plane, two lines to the same line are ||. 42. Answers may vary. Sample: 3 11; || m by Conv. of the Alt. Int. s Thm. and j || k by Conv. of Corr. s Post. 39. The corr. s are , and the oars are || by the Conv. of Corr. s Post. 43. Answers may vary. Sample: 3 and 12 are suppl.; j || k by the Conv. of Corr. s Post. 40. Answers may vary. Sample: Conv. of the Alt. Int. sThm. 3 9; j || k by 41. Answers may vary. Sample: 3 9; j || k by Conv. of the Alt. Int. sThm. and || m by Conv. of Same-Side Int. s Thm. 3-2 44. Vert. s Thm. and Conv. of Corr. s Post. Proving Lines Parallel GEOMETRY LESSON 3-2 45. It is given that || m, so 4 8 by Corr. s Post. It is also given that 12 8, so 4 12 by Trans. Prop. of . So, j || k by the Conv. of Corr. s Post. 47. 46. 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 48. 49. 50. a. Answers may vary. Sample: b. Given: a || b with transversal e, c bisects AOB, d bisects AXZ. c. Prove: c || d 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 50. (continued) d. To prove that c || d, show that 1 3. 1 3 if AOB OXZ. AOB OXZ by the Corr. s Post. e. 1. a || b (Given) 2. AOB AXZ (Corr. s Post.) s) 3. m AOB = m AXZ (Def. of 4. m AOB = m 1 + m 2; m AXZ = m 3 + m 4 ( Add. Post.) 5. c bisects AOB; d bisects AXZ. (Given) 6. m 1 = m 2; m 3 = m 4 (Def. of bisector) 7. 8. 9. 10. 11. m 1 + m 2 = m 3 + m 4 (Trans. Prop. of m 1 + m 1 = m 3 + m 3 (Subst.) 2m 1 = 2m 3 (Add. Prop.) m 1 = m 3 (Div. Prop.) c || d (Conv. of Corr. sPost.) 3-2 ) Proving Lines Parallel GEOMETRY LESSON 3-2 51. C 52. F 53. B 54. [2] a. 136 + (x + 21) = 180 so x = 23 (OR equivalent equation resulting in x = 23). 54. (continued) b. x + 21 = 2x so x = 21. Lines c and d are not || because x cannot = both 21 and 23 (OR equivalent explanation). [1] incorrect equations OR incorrect solutions 55. (continued) b. m 1 + m 3 = 180. If 2x – 38 + 6x + 18 = 180, then x = 25. The measures are 2x – 38 = 12 and 25, but 12 =/ 25. So a can’t be || to b. [3] appropriate methods, but with one computational error 55. [4] a. [2] incorrect diagram solved correctly OR correct diagram solved incorrectly 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 55. (continued) [1] correct answer (lines a and b are not ||), without work shown 58. If you are west of the Mississippi River, then you are in Nebraska. Original statement is true, converse is false. 56. m 1 = 70 since it is a 59. If a circle has a radius suppl. of the 110° . of 4 cm, then it has a m 2 = 110 since diameter of 8 cm. Both same-side int. s are are true. suppl. 60. If same-side int. s are 57. m 1 = 66 because suppl., then a line alt. int. s are . m 2 intersects a pair of || = 180 – 94 = 86 lines. Both are true. because same-side int. s are suppl. 3-2 61. If you form the past tense of a verb, then you add ed to the verb. Original statement is false, converse is false. 62. If there are clouds in the sky, then it is raining. Original statement is true, converse is false. 63. 201.1 in.2 64. 28.3 cm2 Proving Lines Parallel GEOMETRY LESSON 3-2 65. 63.6 ft2 66. 78.5 in.2 67. 6.2 m2 68. 4.5 m2 69. 17.7 ft2 70. 0.3 m2 3-2 Proving Lines Parallel GEOMETRY LESSON 3-2 Use the diagram and the given information to determine which lines, if any, are parallel. Justify your answer with a theorem or postulate. 1. 6 3 m || n by Converse of Corresponding Angles Post. 2. 1 and 4 are supplementary. a || b by Converse of Same-Side Interior Angles Theorem. 3. 2 4 No lines must be parallel. Suppose that m 1 = 3x + 10, m 2 = 3x + 14, and m 6 = x + 58 in the diagram above. 4. Find the value of x for which a || b. 24 5. Find the value of x for which m || n. 26 3-2 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 (For help, go to Lesson 1-4.) Classify each angle as acute, right, or obtuse. 1. 2. 3. Solve each equation. 4. 30 + 90 + x = 180 5. 55 + x + 105 = 180 6. x + 58 = 90 7. 32 + x = 90 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Solutions 1. The measure of the angle is 90°, so it is a right angle. 2. The measure of the angle is between 0° and 90°, so it is an acute angle. 3. The measure of the angle is between 0° and 90°, so it is an acute angle. 4. Combine like terms: 120 + x = 180; subtract 120 from both sides: x = 60 5. Combine like terms: x + 160 = 180; subtract 160 from both sides: x = 20 6. Subtract 58 from both sides: x = 32 7. Subtract 32 from both sides: x = 58 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Find m Z. 48 + 67 + m Z = 180 Triangle Angle-Sum Theorem 115 + m Z = 180 Simplify. Z = 65 Subtract 115 from each side. m 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 In triangle ABC, ACB is a right angle, and CD Find the values of a, b, and c. Find c first, using the fact that m ACB = 90 c + 70 = 90 c = 20 ACB is a right angle. Definition of right angle Angle Addition Postulate Subtract 70 from each side. 3-3 AB. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 (continued) To find a, use ADC. a+m ADC + c = 180 m ADC = 90 a + 90 + 20 = 180 a + 110 = 180 a = 70 To find b, use CDB. 70 + m CDB + b = 180 m CDB = 90 70 + 90 + b = 180 160 + b = 180 b = 20 Triangle Angle-Sum Theorem Definition of perpendicular lines Substitute 90 for m ADC and 20 for c. Simplify. Subtract 110 from each side. Triangle Angle-Sum Theorem Definition of perpendicular lines Substitute 90 for m CDB. Simplify. Subtract 160 from each side. 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Classify the triangle by its sides and its angles. The three sides of the triangle have three different lengths, so the triangle is scalene. One angle has a measure greater than 90, so the triangle is obtuse. The triangle is an obtuse scalene triangle. 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Find m m 1. 1 + 90 = 125 Exterior Angle Theorem m Subtract 90 from each side. 1 = 35 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Explain what happens to the angle formed by the back of the chair and the armrest as you lower the back of the lounge chair. The exterior angle and the angle formed by the back of the chair and the armrest are adjacent angles, which together form a straight angle. As one measure increases, the other measure decreases. The angle formed by the back of the chair and the armrest increases as you lower the back of the lounge chair. 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Pages 135-139 Exercises 1. 30 9. 70 2. 83.1 10. 30 3. 90 11. 60 4. 71 12. acute, isosceles 5. 90 13. acute, equiangular, equilateral 16. 6. x = 70; y = 110; z = 30 14. right, scalene 7. t = 60; w = 60 15. obtuse, isosceles 8. x = 80; y = 80 3-3 17. Not possible; a right will always have one longest side opp. the right . 18. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 19. 23. 27. 115.5 28. m 3 = 92; m 4 = 88 24. a. 5, 6, 8 29. x = 147, y = 33 20. b. 1 and 1 and 1 and 3 for 2 for 2 for 5 6 8 30. a = 162, b = 18 31. x = 52.5; 52.5, 52.5, 75; acute 21. c. They are vert. s . 32. x = 7; 55, 35, 90; right 25. a. 2 22. 33. x = 37; 37, 65, 78; acute b. 6 26. 123 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 34. x = 38, y = 36, z = 90; ABD: 36, 90, 54; right; BCD: 90, 52, 38; right; ABC: 74, 52, 54; acute 35. a = 67, b = 58, c = 125, d = 23, e = 90; FGH: 58, 67, 55; acute; FEH: 125, 32, 23; obtuse; EFG: 67, 23, 90; right 37. 60; 180 3 = 60 39. eight 38. Yes, an equilateral is isosc. Because if three sides of a are , then at least two sides are . No, the third side of an isosc. does not need to be to the other two. 40. 32.5 36. x = 32, y = 62, z = 32, w = 118; ILK: 118, 32, 30; obtuse 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 41. Check students’ work. Answers may vary. Sample: The two ext. s Formed at vertex A are vert. s and thus have the same measure. 42. 30 and 60 46. 103 50. 132; since the missing is 68, the largest ext. Is 180 – 48 = 132. 47. 32 48. a. 90 b. 180 c. 90 d. compl. e. compl. 43. a. 40, 60, 80 b. acute 44. 160 45. 100 51. Check students’ work. 49. a. b. s Add. -Sum 52. a. 81 b. 45, 63, 72 c. acute 53. 120 or 60 c. Trans. 54. 135 or 45 d. Subtr. 55. 90 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 56. Greater than, because 63. Answers may vary. Sample: The measure of the there are two s with ext. is = to the sum of the measures of the two measure 90 where the remote int. s . Since these s are , the s formed meridians the by the bisector of the ext. s are to each of them. equator. Therefore, the bisector is || to the included side of the remote int. s by the Conv. of the Alt Int. s Thm. 57. 1 3 64. B 58. 1 7 65. G 59. 1 66. B 1 60. 19 67. H 61. 0 62. 115 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 68. [2] a. b. The correct equation is 2x = (2x – 40) + (x – 15) with the sol. x = 55. The three int. s measure 70, 40, and 70. 69. (continued) b. 1 to 68 ; since Y is obtuse, its whole number range is from 91 to 158, allowing the measure of 1 for m M when m Y = 158. When m Y = 91, then m M = 68. [1] incorrect sketch, equation, OR solution 69. [2] a. 159; the sum of the three s of the is 180, so m Y + m M + m F = 180. Since m F = 21, m Y + m M + 21 = 180. Subtr. 21 from both sides results in m Y + m M = 159. 3-3 [1] incorrect answer to part (a) or (b) OR incorrect computation in either part Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 70. 53 71. 46 72. 7 73. 74. 3-3 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 1. A triangle with a 90° angle has sides that are 3 cm, 4 cm, and 5 cm long. Classify the triangle by its sides and angles. scalene right triangle Use the diagram for Exercises 2–6. 2. Find m 3 if m 2 = 70 and m 4 = 42. 68 3. Find m 5 if m 2 = 76 and m 3 = 90. 166 4. Find x if m 1 = 4x, m 5. Find x if m 2 = 10x, m 6. Find m 3 if m 3 = 2x + 28, and m 3 = 5x + 40, and m 1 = 125 and m 4 = 32. 30 4 = 3x – 4. 8 5 = 160. 105 3-3 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 (For help, go to Lesson 1-4 and 3-3.) Find the measure of each angle of quadrilateral ABCD. 1. 2. 3. 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Solutions 1. m DAB = 32 + 45 = 77; m 2. m DAC = m ACD = m B = 65; m D and m BCD = 70 + 61 = 131; m CAB = m B=m D = 87 BCA; by the Triangle Angle-Sum Theorem, the sum of the measures of the angles is 180, so each angle measures 180, or 60. So, m 3 m B = 60, m DAB = 60 + 60 = 120, BCD = 60 + 60 = 120, and m D = 60. 3. By the Triangle Angle-Sum Theorem m A + 55 + 55 = 180, so m m ABC = 55 + 30 = 85; by the Triangle Angle-Sum Theorem, m C + 30 + 25 = 180, so m C = 125; m ADC = 55 + 25 = 80 3-4 A = 70. The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Name the polygon. Then identify its vertices, sides, and angles. The polygon can be named clockwise or counterclockwise, starting at any vertex. Possible names are ABCDE and EDCBA. Its vertices are A, B, C, D, and E. Its sides are AB or BA, BC or CB, CD or DC, DE or ED, and EA or AE. Its angles are named by the vertices, A (or EAB or BAE), B (or ABC or CBA), C (or BCD or DCB), D (or CDE or EDC), and E (or DEA or AED). 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Classify the polygon below by its sides. Identify it as convex or concave. Starting with any side, count the number of sides clockwise around the figure. Because the polygon has 12 sides, it is a dodecagon. Think of the polygon as a star. If you draw a diagonal connecting two points of the star that are next to each other, that diagonal lies outside the polygon, so the dodecagon is concave. 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Find the sum of the measures of the angles of a decagon. A decagon has 10 sides, so n = 10. Sum = (n – 2)(180) Polygon Angle-Sum Theorem = (10 – 2)(180) Substitute 10 for n. = 8 • 180 Simplify. = 1440 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Find m X in quadrilateral XYZW. The figure has 4 sides, so n = 4. m X + m Y + m Z + m W = (4 – 2)(180) m X + m Y + 90 + 100 = 360 m X+m m m Y + 190 = 360 X+m X+m 2m m Polygon Angle-Sum Theorem Substitute. Simplify. Y = 170 X = 170 X = 170 Subtract 190 from each side. Substitute m X for m Y. Simplify. X = 85 Divide each side by 2. 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 A regular hexagon is inscribed in a rectangle. Explain how you know that all the angles labeled 1 have equal measures. Sample: The hexagon is regular, so all its angles are congruent. An exterior angle is the supplement of a polygon’s angle because they are adjacent angles that form a straight angle. Because supplements of congruent angles are congruent, all the angles marked 1 have equal measures. 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Pages 147-150 Exercises 1. yes 2. No; it has no sides. 3. No; it is not a plane figure. 4. No; two sides intersect between endpoints. 6. KCLP; sides: KC, CL, LP, PK; s : K, C, L, P 12. 1800 7. HEPTAGN; sides: HE, EP, PT, TA, AG, GN, NH; s : H, E, P, T, A, G, N 14. 3240 13. 1440 15. 180,000 16. 102 8. pentagon; convex 5. MWBFX; sides: MW, 9. decagon; concave WB, BF, FX, XM ; s: M, W, B, F, 10. pentagon; concave X 11. 1080 3-4 17. 103 18. 145 19. 37 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 20. 60, 60, 120, 120 31. 28. 21. 113, 119 22. 108; 72 23. 150; 30 32. 3 29. 33. 8 24. 160; 20 34. 13 25. 176.4; 3.6 26. 45, 45, 90 35. 18 30. 36. a. 3; 6 b. 4; 8 27. c. 5; 10 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 37. octagon; m 1 = 135; m 2 = 45 38. If you solve (n – 2) 180 = 130, n you get n = 7.2. This number is not an integer. 39. 20-80-80; 50-50-80 40. 108; 5 43. 150; 12 44. 180 – x; 360 x 45. 4 5 46. a. n • 180 b. (n – 2)180 c. 180n – 180(n – 2) = 360 d. Polygon Ext. -Sum Thm. 41. 144; 10 48. w = 72, x = 59, y = 49, z = 121; 49. x = 36, 2x = 72, 3x = 108, 4x = 144; quad. 50–53. Answers may vary. Samples are given 50. 51. 47. y = 103; z = 70; quad. 42. 162; 20 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 52. 55. Answers may vary. Sample: The figure is a convex equilateral quad. The sum of its s is 2 • 180 or 360. b. 53. 56. octagon 54. Yes; the sum of the 57. a. (20, 162), (40, 171), measures of s at the (60, 174), (80,175.5), int. point is 360. The (100, 176.4), sum of the measures (120, 177), of all the s is 180n. (140, 177.4), 180n – 360 = (160, 177.75), (n – 2)180 (180, 178), (200, 178.2) 3-4 c. It is very close to 180 d. No, two sides cannot be collinear. The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 58. a. [180(n – 2)] ÷ n = 180n – 360 = 180 – 360 . n n 62. b. As n gets larger, the size of the angles get closer to 180. The more sides it has, the closer the polygon is to a circle. 63. Not possible; opp. and adj. sides would 59. 36 overlap. 60–63. Answers may vary. Samples are given. 64. 4140 60. 65. 20 66. 225 61. Not possible; opp. sides would overlap. 67. 360 68. 157.5 3-4 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 69. 27 78. Div. Prop. 70. 10 79. Trans. Prop. 71. 120, 25 80. RT, RK 72. 50, 40 81. 73. 104, 76, 35, 69 82. Answers may vary. Sample: BR and TK. BRT, BRK 74. Distr. Prop. 75. Subst. Prop. 83. Answers may vary. Sample: BRM 76. Reflexive Prop. of 84. TRM 77. Symm. Prop. of 85. TRK 3-4 86. R The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 For Exercises 1 and 2, if the figure is a polygon, name it by its vertices and identify its sides. If the figure is not a polygon, explain why not. quadrilateral ABCD; 1. not a polygon because two sides intersect at a point other than endpoints 2. AB, BC, CD, DA 3. Find the sum of the measures of the angles in an octagon. 1080 4. A pentagon has two right angles, a 100° angle and a 120° angle. What is the measure of its fifth angle? 140 ABCDEFGHIJ is a regular decagon. 5. Find m 6. ABC. 144 XBC is an exterior angle at vertex B. Find m 3-4 XBC. 36 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 (For help, go to Page 151.) Find the slope of the line that contains each pair of points. 1. A(–2, 2), B(4, –2) 2. P(3, 0), X(0, –5) 3. R(–3, –4), S(5, –4) 4. K(–3, 3), T(–3, 1) 5. C(0, 1), D(3, 3) 6. E(–1, 4), F(3, –2) 7. G(–8, –9), H(–3, –5) 8. L(7, –10), M(1, –4) 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Solutions y –y –2 – 2 1. m = x2 – x1 = 4 – (–2) 2 1 y2 – y1 2. m = x2 – x1 = y –y –5 – 0 0–3 –4 – (–4) 3. m = x2 – x1 = 5 – (–3) 2 1 y –y 1–3 4. m = x2 – x1 = 2 1 – 3 – (–3) y2 – y1 –4 2 =– 6 3 5 –5 = –3 = 3 = 0 =0 8 –2 = , which is undefined; There is no slope. 0 = 3–1 5. m = x – x = =2 2 1 3–0 3 y2 – y1 –2 – 4 6. m = x – x = 3 – (–1) 2 1 y2 – y1 7. m = x – x 2 1 y –y –5 – (–9) = – 3 – (–8) 8. m = x2 – x1 = 2 1 –4 – (–10) 1–7 = –6 4 = 3 =– 2 4 5 = 6 =–1 –6 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Use the slope and y-intercept to graph the line y = –2x + 9. When an equation is written in the form y = mx + b, m is the slope and b is the y-intercept. In the equation y = –2x + 9, the slope is –2 and the y-intercept is 9. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Use the x-intercept and y-intercept to graph 5x – 6y = 30. To find the x-intercept, substitute 0 for y and solve for x. To find the y-intercept, substitute 0 for x and solve for y. 5x – 6y = 30 5x – 6(0) = 30 5x – 0 = 30 5x = 30 x =6 5x – 6y = 30 5(0) – 6y = 30 0 – 6y = 30 –6y = 30 y = –5 The x-intercept is 6. A point on the line is (6, 0). The y-intercept is –5. A point on the line is (0, –5). 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 (continued) Plot (6, 0) and (0, –5). Draw the line containing the two points. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Transform the equation –6x + 3y = 12 to slope-intercept form, then graph the resulting equation. Step 1: Transform the equation to slope-intercept form. –6x + 3y = 12 3y = 6x + 12 3y 6x 12 = + 3 3 3 Step 2: Use the y-intercept and the slope to plot two points and draw the line containing them. Add 6x to each side. Divide each side by 3. y = 2x + 4 The y-intercept is 4 and the slope is 2. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Write an equation in point-slope form of the line with slope –8 that contains P(3, –6). y – y1 = m(x – x1) y – (–6) = –8(x – 3) y + 6 = –8(x – 3) Use point-slope form. Substitute –8 for m and (3, –6) for (x1, y1). Simplify. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Write an equation in point-slope form of the line that contains the points G(4, –9) and H(–1, 1). Step 1: Find the slope. y –y m = x2 – x1 2 1 Use the formula for slope. 1 – (–9) m = –1 – 4 Substitute (4, –9) for (x1, y1) and (–1, 1) for (x2, y2). 10 m = –5 m = –2 Simplify. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 (continued) Step 2: Select one of the points. Write the equation in point-slope form. y – y1 = m(x – x1) y – (–9) = –2(x – 4) y + 9 = –2(x – 4) Point-slope form Substitute –2 for m and (4, –9) for (x1, y1). Simplify. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Write equations for the horizontal line and the vertical line that contain A(–7, –5). Every point on the horizontal line through A(–7, –5) has the same y-coordinate, –5, as point A. The equation of the line is y = –5. It crosses the y-axis at (0, –5). Every point on the vertical line through A(–7, –5) has the same x-coordinate, –7, as point A. The equation of the line is x = –7. It crosses the x-axis at (–7, 0). 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 1. 4. 7. 5. 8. 2. 6. 3. 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 9. 11. y = 2x + 1 13. y = –2x + 4 14. y = –2x + 4 10. 12. y = x + 1 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 15. y = – 1 x + 1 3 19. y – 5 = –1(x + 3) 25. y – 6 = 1(x – 2) 20. y + 6 = –4(x + 2) 26. y – 4 = 1(x + 4) 1 21. y – 1 = 2 (x – 6) 16. 22. y – 4 = 1(x – 0) or y–4=x 23–28. Equations may vary from the pt. chosen. Samples are given. 17. y – 3 = 2(x – 2) 23. y – 5 = 18. y + 1 = 3(x – 4) 3 (x – 0) 5 24. y – 2 = – 1 (x – 6) 2 3-5 1 (x + 1) 2 2 28. y – 10 = 3 (x – 8) 27. y – 0 = 29. a. y = 7 b. x = 4 30. a. y = –2 b. x = 3 31. a. y = –1 b. x = 0 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 32. a. y = 4 b. x = 6 36. 39. No; a line with no slope is a vertical line. 0 slope is a horizontal line. 37. 40. a. m = 0; it is a horizontal line. 33. b. y = 0 34. 41. a. Undefined; it is a vertical line. 38. a. 0.05 35. b. the cost per min c. 4.95 d. the initial charge for a call 3-5 b. x = 0 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 42–44. Answers may vary. Samples are given. 42. The eq. is in standard form; change to slope-intercept form, because it is easy to graph the eq. from that form. 43. The eq. is in slope-int. form; use slope-int. form, because the eq. is already in that form. 44. The eq. is in pointslope form; use pointslope form, because the eq. is already in that form. 46. 45. The slopes are all different and the y-intercepts are the same. 47. Check students’ work. The slopes are the same and the y-intercepts are different. 3-5 48. Lines in the Coordinate Plane GEOMETRY LESSON 3-5 49. 50. 3 1 3 1 52. 10 = 0.3, 12 = 0.083; > ; 10 12 it is possible only if the ramp zigzags. 53. The y-intercepts are the same, and the lines have the same steepness. One line rises from left to right while the other falls from left to right. 54. Answers may vary. Sample: x = 5, y – 6 = 2(x – 5), y=x+1 55. (2, 0), (0, 4); m = 51. 0 – 4 –4 = = –2 2–0 2 y – 0 = –2(x – 2), 2x + y = 4 or y = –2x + 4 56. a. y – 0 = 5 (x – 0) or y = 5 x 2 2 b. y – 5 = – 5 (x – 2) or y = – 5 x + 10 2 3-5 2 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 56. (continued) c. The abs. value of the slopes is the same, but one slope is pos. and the other is neg. One y-int. is at (0, 0) and the other is at (0, 10). 60. Yes; the slope of JK = the slope of KL. 68. C 61. y – 2 = 3(x + 2); 3x – y = –8 1 62. y – 5 = 2 (x – 5); x – 2y = –5 2 57. Yes; the slope of AB = the slope of BC. 58. No; the slope of DE =/ the slope of EF. 59. Yes; the slope of GH = the slope of HI. 67. C 63. y – 6 = 3 (x – 2); 2x – 3y = –14 64. D 65. G 66. B 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 3 69. [2] a. Line a: y = x + 12 OR equivalent equation; 2 Line b: 3y + 4x = 19 OR equivalent equation b. 72. 1620 73. 2160 74. yes 75. No; || lines never intersect, but they are not skew. point of intersection: (22, 9) [1] at least one correct eq. or graph 76. No; all obtuse two acute s . s have 77. a = 5; m MPR = 30 70. 1260 78. a = 4; m MPR = 21 71. 540 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 79. a = 2; m QPR = 8 80. a = 6; m MPQ = 21 3-5 Lines in the Coordinate Plane GEOMETRY LESSON 3-5 Three points are on a coordinate plane: A(1, 5), B(–2, –4), and C(6, –4). 1. Find the x-intercept and the y-intercept of the line 5x + 4y = –80. x-intercept: –16, y-intercept: –20 2. Write an equation in point-slope form of the line with slope –1 that contains point C. y + 4 = –1(x – 6) 3. Write an equation in point-slope form of the line that contains points A and B. y – 5 = 3(x – 1) or y + 4 = 3(x + 2) 4. Write an equation of the line that contains B and C. y = –4 5. Graph and label the equations of the lines in Exercises 2–4 above. 3-5 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 (For help, go to page 151 and Lesson 3-5.) Find the slope of the line through each pair of points. 1. F(2, 5), B(–2, 3) 2. H(0, –5), D(2, 0) 3. E(1, 1), F(2, –4) Find the slope of each line. 4. y = 2x – 5 5. x + y = 20 6. 2x – 3y = 6 7. x = y 8. y = 7 9. y = 2 x + 7 3 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Solutions 1. m = y2 – y1 x2 – x1 3–5 = –2–2 = –2 = 1 –4 2. m = y2 – y1 x2 – x1 = 0 – (–5) = 5 3. m = y2 – y1 x2 – x1 = –4–1 2–0 2 –1 2 2 –5 = 1 = –5 4. The equation is in slope-intercept form and the slope is the coefficient of x, which is 2. 5. Rewrite the equation in slope-intercept form: y = – x + 20; the slope is the coefficient of x, which is – 1. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Solutions (continued) 6. Rewrite the equation in slope-intercept form: y = 2 x – 2; the slope 3 2 is the coefficient of x, which is . 3 7. The equation in slope-intercept form is y = x + 0; the slope is the coefficient of x, which is 1. 8. The equation in slope-intercept form is y = 0x + 7; the slope is the coefficient of x, which is 0. 9. The equation is in slope-intercept form and the slope is the coefficient of x, which is 2 . 3 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Line 1 contains P(0, 3) and Q(–2, 5). Line and S(3, –10). Are lines 1 and 2 parallel? Explain. 2 contains R(0, –7) Find and compare the slopes of the lines. Slope of line Slope of line y2 – y1 5–3 y2 – y1 –10 –(–7) 3–0 1 = x –x = –2 – 0 2 1 2 = x –x = 2 1 = 2 –2 = –1 = –3 = –1 3 Each line has slope –1. The y-intercepts are 3 and –7. The lines have the same slope and different y-intercepts, so they are parallel. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Are the lines y = –5x + 4 and x = –5y + 4 parallel? Explain. The equation y = –5x + 4 is in slope-intercept form. Write the equation x = –5y + 4 in slope-intercept form. x = –5y + 4 x – 4 = –5y Subtract 4 from each side. –1x+ 4 =y 5 Divide each side by –5. 5 y=–1x+ 4 5 5 The line x = –5y + 4 has slope – 1 . 5 The line y = –5x + 4 has slope –5. The lines are not parallel because their slopes are not equal. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Write an equation in point-slope form for the line parallel to 6x – 3y = 9 that contains (–5, –8). Step 1: To find the slope of the line, rewrite the equation in slope-intercept form. 6x – 3y = 9 –3y = –6x + 9 y = 2x – 3 Subtract 6x from each side. Divide each side by –3. The line 6x – 3y = 9 has slope 2. Step 2: Use point-slope form to write an equation for the new line. y – y1 = m(x – x1) y – (–8) = 2(x – (–5)) y + 8 = 2(x + 5) Substitute 2 for m and (–5, –8) for (x1, y1). Simplify. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Line 1 contains M(0, 8) and N(4, –6). Line 2 contains P(–2, 9) and Q(5, 7). Are lines 1 and 2 perpendicular? Explain. Step 1: Find the slope of each line. m1 = slope of line m2 = slope of line y –y 1 = x2 – x1 = 2 1 –6 – 8 4–0 y –y 7–9 2 = x2 – x1 = 5 – (–2) 2 1 = –14 7 =– 4 2 = –2 = –2 7 7 Step 2: Find the product of the slopes. m1 • m2 = – 7 2 •– =1 2 7 Lines 1 and 2 are not perpendicular because the product of their slopes is not –1. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Write an equation for a line perpendicular to 5x + 2y = 1 that contains (10, 0). Step 1: To find the slope of the given line, rewrite the equation in slope-intercept form. 5x + 2y = 1 2y = –5x + 1 Subtract 5x from each side. y=–5 x+ 1 Divide each side by 2. 2 2 The line 5x + 2y = 1 has slope – 5 . 2 Step 2: Find the slope of a line perpendicular to 5x + 2y = 1. Let m be the slope of the perpendicular line. – 5 m = –1 The product of the slopes of perpendicular lines is –1. 2 m = –1 • ( – 2 ) Multiply each side by – 2 . m= 2 5 5 5 Simplify. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 (continued) Step 3: Use point-slope form, y – y1 = m(x – x1), to write an equation for the new line. y – 0 = 2 (x – 10) 5 y = 2 (x – 10) 5 Substitute 2 for m and (10, 0) for (x1, y1). 5 Simplify. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 The equation for a line containing a lead strip is y = 1 x – 9. 2 Write an equation for a line perpendicular to it that contains (1, 7). Step 1: Identify the slope of the given line. 1 y = 2x – 9 slope Step 2: Find the slope of the line perpendicular to the given line. Let m be the slope of the perpendicular line. 1 m = –1 2 m = –2 The product of the slopes of perpendicular lines is –1. Multiply each side by 2. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 (continued) Step 3: Use point-slope form to write an equation for the new line. y – y1 = m(x – x1) y – 7 = –2(x – 1) Substitute –2 for m and (1, 7) for (x1, y1). Step 4: Write the equation in slope-intercept form. y – 7 = –2(x – 1) y – 7 = –2x + 2 y = –2x + 9 Use the Distributive Property. Add 7 to each side. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 Pages 161-164 Exercises 1. 2. Yes; both slopes = – 1. 2 1 No; the slope of 1 = 3 , and the slope of 2 = 1 . 2 3. No; the slope of 1 = 3, 2 and the slope of 2 = 2. 4. Yes; both slopes = 4. 5. Yes; both slopes = 0. 3 6. Yes; the lines both have a slope of 2 but different y-intercepts. 10. No; one slope = – 4 and the other slope = –3. 7. Yes; the lines both have a slope of 3 but 11. Yes; the lines both have a slope of – 2 5 but different y-intercepts. 4 different y-intercepts. 8. Yes; the lines both have a slope of –1 but different y-intercepts. 9. No; one slope = 7 and the other slope = –7. 3-6 12. y – 3 = –2(x – 0) or y – 3 = –2x 13. y – 0 = 1 (x – 6) or 3 1 y = (x – 6) 3 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 1 2 14. y – 4 = (x + 2) 15. y + 2 = – 3 (x – 6) 2 16. Yes; the slope of 1 = – 1 , and the slope of 2 1 • 2 = –1. 2 = 2; – 19. Yes; the slope of 1 = 25. No; 1 • 2 =/ –1. 2 –1, and the slope of 2 = 1; –1 • 1 = –1. 26. Yes; 1 • (–1) = –1. 20–23. Answers may vary. Samples are given. 2 3 17. Yes; the slope of 1 = – 3 , and the slope of 2 2 ; – 3 • 2 = –1. 2 = 3 2 3 18. No; the slope of 1 = –1, and the slope of 2 4 5 = ; –1 • 4 =/ –1. 5 20. y – 6 = – 2 (x – 6) 21. y = –2(x – 4) 22. y – 4 = 1 (x – 4) 2 4 x 5 3 24. y = – x 2 23. y = 3-6 27. Yes; one is vertical and the other is horizontal. 28. No; – 1 • (–3) =/ –1. 2 2 3 29. Yes; – • = –1. 3 2 30. No; 2 • – 7 =/ –1. 7 4 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 31. slope of AB = slope of 2 CD = ; AB || CD 3 slope of BC = slope of AD = –3; BC || AD 32. slope of AB = slope of CD = – 3 ; AB || CD 4 slope of BC = slope of AD = 1; BC || AD 33. slope of AB = 1 ; slope 2 1 of CD = ; AB || CD 4 slope of BC = –1; slope of AD 5 – 1 ; 2 BC || AD 34. slope of AB = slope of CD = 0; AB || CD slope of BC = 3 and slope of AD = 3 ; 2 BC || AD 35. Answers may vary. Sample: y = 4 x + 5, 5 5 y=– x+5 4 36. No; two II lines with the same y-intercept are the same line. 3-6 37. RS and VU are horizontal with slope = 0; RS II VU; slope of RW = slope of UT = 1; RW || UT; slope of WV = slope of ST = –1; WV || ST 38. No; because no pairs of slopes have a product of –1. 39. The lines will have the same slope. Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 40. When lines are , the product of their slopes is –1. So, two lines to the same line must have the same slope. 3 41. a. y + 20 = 4 (x – 35) b. because you are given a point and can quickly find the slope 42. || 44. neither 45. 46. (7 – 9)2 + (11 – 1)2 = (13 – 3)2 + (7 – 5)2 = 47. AC: d = BD: d = AC 104 104 BD 1 1 48. slope of AC = –5; slope of BD = ; since –5 • = –1, 5 5 AC BD; midpoint AC = (8, 6); midpoint BD = (8, 6); since the midpoints are the same, the diagonals bisect each other. 43. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 49. a–b. Answers may vary. Sample: 51. B 52. I 53. C 54. [2] a. slope of line c: 1 – (–2) 3 1 = = – –4 – 2 –6 2 ; slope of line b. 0 c. The other possible locations for S are (–2, 3) and (8, 7). 50. y – 5 = 1 (x – 4) 3 [1] at least one correct slope 1 1 55. y – 3 = – 2 (x – 0) or y – 3 = – 2 x 56. y – 2 = 5 (x + 4) 3 3-6 to c: 2 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 57. y + 2 = 3 (x – 3) 4 58. Refl. Prop. of 59. Mult. Prop. of = 60. Dist. Prop. 61. Symm. Prop. Of 62. If you are in geometry class, then you are at school. 63. If you travel to Switzerland, then you have a passport. 3-6 Slopes of Parallel and Perpendicular Lines GEOMETRY LESSON 3-6 1. Are lines 1 and 2 parallel? Explain. Yes; the lines have the same slope and different y-intercepts. 2. Are the lines x + 4y = 8 and 2x + 6y = 16 parallel? Explain. No; their slopes are not equal. 3. Write an equation in point-slope form for the line parallel to –18x + 2y = 7 that contains (3, 1). y – 1 = 9(x – 3) 2 4. Are the lines y = 3 x + 5 and 3x + 2y = 10 perpendicular? Explain. Yes; the product of their slopes is –1. 5. Write an equation in point-slope form for the line perpendicular to y = – 1 x – 2 that contains (–5, –8). 6 y + 8 = 6(x + 5) 3-6 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 (For help, go to Lesson 1-5.) Use a straightedge to draw each figure. Then use a straightedge and compass to construct a figure congruent to it. 1. a segment 2. an obtuse angle 3. an acute angle Use a straightedge to draw each figure. Then use a straightedge and compass to bisect it. 4. a segment 5. an acute angle 3-7 6. an obtuse angle Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Solutions 1. 2. 3. 4. 5. 6. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Examine the diagram below. Explain how to construct 1 congruent to H. Use the method learned for constructing congruent angles. Step 1: With the compass point on point H, draw an arc that intersects the sides of H. Step 2: With the same compass setting, put the compass point on point N. Draw an arc. Step 3: Put the compass point below point N where the arc intersects HN. Open the compass to the length where the arc intersects line . Keeping the same compass setting, put the compass point above point N where the arc intersects HN. Draw an arc to locate a point. Step 4: Use a straightedge to draw line m through the point you located and point N. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Construct a quadrilateral with both pairs of sides parallel. Step 1: Draw point A and two rays with endpoints at A. Label point B on one ray and point C on the other ray. Step 2: Construct a ray parallel to AC through point B. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 (continued) Step 3: Construct a ray parallel to AC through point C. Step 4: Label point D where the ray parallel to AC intersects the ray parallel to AB. Quadrilateral ABDC has both pairs of opposite sides parallel. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 In constructing a perpendicular to line at point P, why must you open the compass wider to make the second arc? With the compass tip on A and B, the same compass setting would make arcs that intersect at point P on line . Without another point, you could not draw a unique line. With the compass tip on A and B, a smaller compass setting would make arcs that do not intersect at all. Once again, without another point, you could not draw a unique line. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Examine the construction. At what special point does RG meet line ? Point R is the same distance from point E as it is from point F because the arc was made with one compass opening. Point G is the same distance from point E as it is from point F because both arcs were made with the same compass opening. This means that RG intersects line at the midpoint of EF, and RG is the perpendicular bisector of EF. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Pages 168-170 Exercises 5–7. Constructions may vary. Samples using the following segments are shown. 3. 1. || AB || AB 5. 4. 2. 6. || AB || AB 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 7. 11. 9. AB 8. RS 10. 12. AB RS RS 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 13. 15. Construct a alt. int. then draw the || line. ; 17. 16. a–b RS 18. 14. 1 2 17–25. Constructions may vary. Samples are given. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 19. 23. 21. a. b. The sides are || and . 20. c. Check students’ work. 22. 3-7 24. a. Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 24. (continued) b. Answers may vary. Sample: 26. a–b. Check students’ work. 25. a–c. c. p || m; in a plane, two lines to a third line are ||. c. One; if the lengths for the 3 sides are given, only one is possible; many different quad. are possible because the s formed by the sides can vary. d. The sides of the smaller are half the length of the sides of the larger that they are || to. 27–28. Answers may vary. Samples are given. 27. e. Check students’ work. The quad. is a rectangle. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 28. 30. 32. 33. The quad. is a square. 29. 31. 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 34. 36. Not possible; the 39. (continued) smaller sides would b. All points on the arc meet at the midpoint of with center G are the the longer side, forming same dist. from G, so a segment. GR = GT. The is isosc. because an 37. A isosc. must have at least two sides of the 38. I same length. 35. Not possible; if 39. [2] a. a = 2b = 2c, then 2a = 2b + 2c or a = b + c. The smaller sides would meet at the midpoint of the longer side, forming a segment. [1] incorrect sketch OR incorrect classification 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 40. [2] a. I, IV, III, I b. (III): location of compass at points C and G; (I): same as c and the intersection points of with arcs drawn in (III) 41. No; the slopes are different. 42. No; the slopes are different. 43. Yes; the slopes are both – 1 . 3 44. 10 45. 8.9 [1] incorrect sequence OR incorrect location of compass point 46. EB 47. DF 3-7 Constructing Parallel and Perpendicular Lines GEOMETRY LESSON 3-7 Draw a figure similar to the one given. Then complete the construction. 1. Construct a line through D that is parallel to XY. Answers may vary. Sample given: 2. Construct a quadrilateral with one pair of parallel sides of lengths p and q. Answers may vary. Sample given: 3. Construct the line perpendicular to line m at point Z. 4. Construct the perpendicular to line n through point O. 3-7 Parallel and Perpendicular Lines GEOMETRY CHAPTER 3 Page 176 1. acute isosceles 5. m 1 = 85 because corr. s are ; m 2 = 95 because same-side int. s are suppl. 2. obtuse scalene 3. m 1 = 65 because corr. s are ; m 2 = 65 because vert. s are . 11. 5 12. 25 13. 6 6. m 1 = 70 because corr. s are ; m 2 = 110 because sameside int. s are suppl. 7. yes 4. m 1 = 85 because alt. int. s are ; m 2 8. yes = 110 because sameside int. s are suppl. 9. no 10. no 3-A 14. 75 15. Parallel and Perpendicular Lines GEOMETRY CHAPTER 3 16. 20. y + 1 = –5(x – 3) or y = –5x + 14 24. 25. || 17. Answers may vary. Sample: Z 18. Check students’ work. 21. Draw segments connecting nonconsecutive vertices. If no points of the segments are outside the polygon, then it is convex. Otherwise, the polygon is concave. 19. a. Given b. Corr. s are . c. Given 22. 109 d. Trans. Prop. e. If corr. s are , 23. x = 85; y = 100; z = 100 then the lines are ||. 3-A 26. neither 27. 28. 30