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Lesson 3-5 Menu

Write an equation in point-slope form of a line having slope as ¾ and contains the point (5, –2). Write an equation in point-slope form of a line having slope as 3 and contains the point (–2, 7).

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Lesson 3-5 Menu

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  1. Write an equation in point-slope form of a line having slope as ¾ and contains the point (5, –2). • Write an equation in point-slope form of a line having slope as 3 and contains the point (–2, 7). • Write an equation in slope-intercept form of a line having slope as –3 and contains the point (0, 2.5). • Write an equation in slope-intercept form of a line having slope as –½ and contains the point (4, –6). • Write an equation in slope-intercept form of a line passing through (1, 5) and (3, 11). Lesson 3-5 Menu

  2. Recognize angle conditions that occur with parallel lines. • Prove that two lines are parallel based on given angle relationships. Lesson 3-5 Ideas/Vocabulary

  3. Lesson 3-5 Postulates

  4. Lesson 3-5 Theorems

  5. Since RQT and SQP are vertical angles, m SQP = 77. Since mUPQ + mSQP = 103 + 77 or 180, consecutive interior angles are supplementary. So, a || b. Since mTQR + mVRQ = 77 + 100or177,consecutive interior angles are notsupplementary. So,cisnotparallel to a or b. Identify Parallel Lines Determine which lines, if any, are parallel. Answer:a || b Lesson 3-5 Example 1

  6. A • B • C • D Determine which lines, if any are parallel.I. e || fII. e || gIII. f || g I only II only III only I, II, and III Lesson 3-5 CYP 1

  7. ALGEBRA Find x and m ZYN so that || . ExploreFrom the figure, you know that m WXP = 11x – 25 and m ZYN = 7x + 35. You also know that WXP and ZYN are alternate exterior angles. Solve Problems with Parallel Lines Lesson 3-5 Example 2

  8. Plan For line PQ to be parallel to MN, the alternate exterior angles must be congruent. So, m WXP = m ZYN. Substitute the given angle measures into this equation and solve for x. Once you know the value of x, use substitution to findm ZYN. Solve m WXP = m ZYN Alternate exterior angles Solve Problems with Parallel Lines 11x – 25 = 7x + 35 Substitution 4x – 25 = 35 Subtract 7x from each side. 4x = 60 Add 25 to each side. x = 15 Divide each side by 4. Lesson 3-5 Example 2

  9. Now use the value of x to findm ZYN. m ZYN = 7x + 35 Original equation Examine Verify the angle measure by using the value of x to find m WXP. That is, 11x – 25 = 11(15) – 25 or 140. Since m WXP = m ZYN,m WXP m ZYN and || . Answer:x = 15, m ZYN = 140 Solve Problems with Parallel Lines = 7(15) + 35 x= 15 = 140 Simplify. Lesson 3-5 Example 2

  10. ALGEBRA Find x so that || . x = 60 x = 9 x = 12 x = 12 • A • B • C • D Lesson 3-5 CYP 2

  11. Given: ℓ || m Prove: r || s Prove Lines Parallel Lesson 3-5 Example 3

  12. Proof: Statements Reasons 1.1. Given 2.2. Consecutive Interior Angle Theorem 3.3. Definition of supplementary angles 4.4. Definition of congruent angles 5.5. Substitution 6.6. Definition of supplementary angles 7. 7. If consecutive interior angles are supplementary, then lines are parallel. Prove Lines Parallel Lesson 3-5 Example 3

  13. Given x || y and , do you need to use theCorresponding Angles Postulate to prove a || b? • A • B • C yes no not enough informationto determine Lesson 3-5 CYP 3

  14. Slope and Parallel Lines Determine whether p || q. slope of p: slope of q: Answer: Since the slopes are equal, p || q. Lesson 3-5 Example 4

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